ਹਿਸਾਬ ਨੂੰ ਭਾਵੇਂ ਮੁਸ਼ਕਿਲ ਵਿਸ਼ੇ ਦਾ ਦਰਜ਼ਾ ਦਿੱਤਾ ਜਾਂਦਾ ਰਿਹਾ ਹੈ ਪ੍ਰੰਤੂ ਸਮਝਣ ਤੋਂ ਬਾਦ ਹਿਸਾਬ ਦਾ ਕੋਈ ਵੀ ਸਵਾਲ ਹੱਲ ਕਰਨਾ ਉਨ੍ਹਾਂ ਹੀ ਆਸਾਨ ਹੁੰਦਾ ਹੈ ਜਿੰਨ੍ਹਾਂ ਕਿ ਕਿਸੇ ਵਿਅਕਤੀ ਲਈ ਆਪਣੀ ਮਾਤਭਾਸ਼ਾ ਬੋਲਣਾ। ਫਿਰ ਵੀ ਇਸ ਲਈ ਮੁੱਢਲੀ ਸ਼ਰਤ ਇਹ ਹੈ ਕਿ ਹਿਸਾਬ ਪ੍ਰਤੀ ਡਰ ਦੀ ਭਾਵਨਾਂ ਮਨ ਵਿੱਚ ਬਿਲਕੁਲ ਨਾਂ ਹੋਵੇ। ਮੈਂ ਪੂਰਣ ਵਿਸ਼ਵਾਸ਼ ਨਾਲ ਕਹਿ ਸਕਦਾ ਹਾਂ ਕਿ ਅਜਿਹੇ ਡਰ ਦੀ ਭਾਵਨਾਂ ਕੇਵਲ ਉਨ੍ਹਾ ਵਿਅਕਤੀਆਂ ਦੇ ਮਨ ਵਿੱਚ ਬੈਠਦੀ ਹੈ ਜਿਹੜੇ ਜੋੜ, ਘਟਾਓ, ਗੁਣਾ ਅਤੇ ਭਾਗ ਵਰਗੀਆਂ ਹਿਸਾਬ ਦੀਆਂ ਮੁੱਢਲੀਆਂ ਕ੍ਰਿਆਵਾਂ ਵਿੱਚ ਕਮਜੋਰ ਹੁੰਦੇ ਹਨ। ਇਸ ਲਈ ਜੂਨੀਅਰ ਅਤੇ ਸੀਨੀਅਰ ਵਿਦਿਆਰਥੀਆਂ ਨੂੰ ਸਭ ਤੋਂ ਪਹਿਲਾਂ ਅਜਿਹੀਆਂ ਕਮਜ਼ੋਰੀਆਂ ਨੂੰ ਦੂਰ ਕਰਨ ਦੀ ਕੋਸਿ਼ਸ਼ ਕਰਨੀ ਚਾਹੀਦੀ ਹੈ।
ਭਾਗ ਕਰਨ ਦੇ ਵੀ ਸੌਖੇ ਤਰੀਕੇ ਸਮਝੇ ਜਾ ਸਕਦੇ ਹਨ। ਖਾਸ ਕਰਕੇ 9 ਨਾਲ ਭਾਗ ਦੇਣ ਸਮੇਂ ਦਸ਼ਮਲਵ (Decimal) ਵਿੱਚ ਉੱਤਰ ਪਤਾ ਕਰਨਾ ਜਿੰਨਾ ਕੁ ਸਧਾਰਣ ਤਰੀਕੇ ਨਾਲ ਮੁਸ਼ਕਿਲ ਹੈ, ਸਾਡੇ ਸੁਝਾਏ ਤਰੀਕੇ ਨਾਲ ਉਨਾਂ ਹੀ ਆਸਾਨ। 7 / 9 ਦਾ ਉੱਤਰ ‘10’ ਦਸ਼ਮਲਵ ਤੱਕ ਪਤਾ ਕਰਨ ਲਈ ਕਿੰਨਾਂ ਸਮਾਂ ਲੱਗਣ ਨੂੰ ਤੁਸੀਂ ਜਾਇਜ਼ ਕਹਿ ਸਕਦੇ ਹੋ। ਫਿਰ ਘੱਟ ਤੋਂ ਘੱਟ ਸਮੇਂ ਬਾਰੇ ਵੀ ਸੋਚੋ। ਕੀ ‘2’ ਸੈਕੰਡ ਵਿੱਚ ਇਹ ਸੰਭਵ ਹੈ? ਮੈਨੂੰ ਇਹ ਸਮਾਂ ਵੀ ਜਿ਼ਆਦਾ ਲੱਗਦਾ ਹੈ। ਇਸ ਦਾ ਉੱਤਰ ਹੈ 0.777777777777777777 ਜਾਂ ਫਿਰ 0.7 ਦੇ ਉੱਪਰ ਬਾਰ।
ਆਓ ਇਸ ਤੋਂ ਅੱਗੇ ਵੀ ਕੁਝ ਹੋਰ ਸਮਝਣ ਦੀ ਕੋਸਿ਼ਸ਼ ਕਰੀਏ। ਹੇਠ ਦਿੱਤੀਆਂ ਉਦਾਹਰਣਾ ਇਸ ਮੰਤਵ ਵਿੱਚ ਸਹਾਇਤਾ ਕਰ ਸਕਦੀਆਂ ਹਨ।
4 / 9
= 0.4444444444444444444
14 / 9
= 1.555555555555555555555555
123 / 9
= 13.6666666666666666666666666
ਇੱਥੇ 1, ਫਿਰ 1+2, ਫਿਰ 1+2+3, ਦਸ਼ਮਲਵ ਤੋਂ ਬਾਦ ਅੰਕ ਵਾਰ ਵਾਰ ਦੁਹਰਾਇਆ ਜਾਂਦਾ ਹੈ।
57 / 99
= 0.5757575757575757
2 / 99
= 0.020202020202020202020202
124 / 99
= 1.252525252525252525252525
ਇੱਥੇ 1, ਫਿਰ 1+24
ਤੁਸੀਂ ਆਪ ਇੱਕ ਵਾਰ 61 / 99, 8 / 9 ਅਤੇ 433 / 9999 ਕਰ ਕੇ ਵੇਖੋ।
ਮਜ਼ਾ ਆਇਆ?
ਸਾਡੇ ਸਕੂਲਾਂ ਦੀ ਸ਼੍ਰੇਣੀ 4 ਵਿੱਚ ਹੀ ਕਿਸੇ ਰਕਮ ਦੀ ਵੰਡਣਯੋਗਤਾ (Divisibility)ਦੇ ਨਿਯਮਾ ਦੀ ਚਰਚਾ ਹੈ। ਵੰਡਣਯੋਗਤਾ (Divisibility) ਦਾ ਅਰਥ ਹੈ ਕਿ ਉਸ ਅੰਕ ਨੂੰ ਵਿਸ਼ੇਸ਼ ਨੰਬਰ ਨਾਲ ਭਾਗ ਦੇਣ ਤੇ ‘ਬਾਕੀ’ (Remainder) ‘0’ ਆਵੇ। ਆਮ ਕਰਕੇ ਵਿਦਿਆਰਥੀ ਇਸ ਨੂੰ ਹਿਸਾਬ ਦੀ ਕਿਤਾਬ ਦੇ ਅਭਿਆਸ ਤੱਕ ਹੀ ਮਹੱਤਵ ਦਿੰਦੇ ਹਨ ਜਦੋਂ ਕਿ ਇਸ ਦੀ ਵਰਤੋਂ ਅਗਲੀਆਂ ਸ਼੍ਰੇਣੀਆਂ ਵਿੱਚ ਵਧੇਰੇ ਲਾਭ ਲਈ ਕੀਤੀ ਜਾਂ ਸਕਦੀ ਹੈ। ਭਿੰਨਾਂ ਨੂੰ ਆਪਣੇ ਸਭ ਤੋਂ ਛੋਟੇ ਰੂਪ ਵਿੱਚ ਪ੍ਰਗਟ ਕਰਨਾਂ (Minimise ਕਰਨਾਂ) ਉਕਤ ਨਿਯਮਾਂ ਤੋਂ ਬਿਨਾਂ ਲਗਭਗ ਅਸੰਭਵ ਹੈ। ਸਿੱਧੇ ਜਾਂ ਅਸਿੱਧੇ ਰੂਪ ਇਸ ਦੀ ਵਰਤੋਂ ਅਨੇਕਾਂ ਵਾਰ ਸਾਡੇ ਹਿਸਾਬ ਦੇ ਸਵਾਲਾਂ ਦਾ ਹਿੱਸਾ ਬਣਦੀ ਹੈ।
2 ਨਾਲ ਵੰਡਣਯੋਗਤਾ (Divisibility) ਪਤਾ ਕਰਨ ਲਈ ਇਹ ਵੇਖਣਾ ਜ਼ਰੂਰੀ ਹੈ ਕਿ ਅੰਤਿਮ ਅੱਖਰ ਜਿਸਤ (Even) ਹੋਵੇ ਭਾਵ 0,2,4,6 ਜਾਂ 8 ਹੋਵੇ। ਹੇਠ ਲਿਖੀਆਂ ਉਦਾਹਰਣਾ ਨਾਲ ਜਿ਼ਆਦਾ ਚੰਗੀ ਤਰਾਂ ਸਮਝਿਆ ਜਾ ਸਕਦਾ ਹੈ:
6 8 5 2 4 0 7 8 2 2 ਜਿਸਤ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
6 8 5 2 4 0 7 8 8 ਜਿਸਤ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
6 8 5 2 4 0 7 7 ਟਾਂਕ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
6 8 5 2 4 0 0 ਜਿਸਤ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
6 8 5 2 4 4 ਜਿਸਤ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
6 8 5 2 2 ਜਿਸਤ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
6 8 5 5 ਟਾਂਕ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
6 8 8 ਜਿਸਤ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
6 6 ਜਿਸਤ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
3 ਨਾਲ ਵੰਡਣਯੋਗਤਾ (Divisibility) ਪਤਾ ਕਰਨ ਲਈ ਇਹ ਵੇਖਣਾ ਜ਼ਰੂਰੀ ਹੈ ਕਿ ਦਿੱਤੇ ਹੋਏ ਨੰਬਰ ਦੇ ਅੰਕਾ ਦਾ ਜੋੜ ‘3’ ਤੇ ਵੰਡਿਆ ਜਾਂਦਾ ਹੋਵੇ। ਅਸੀਂ ਇਸ ਨੂੰ ਹੋਰ ਵੀ ਸਰਲ ਕੀਤਾ ਹੈ। ਅੰਕਾਂ ਨੂੰ ਜੋੜਣ ਸਮੇਂ 3, 6 ਅਤੇ 9 ਨੂੰ ਨਹੀਂ ਜੋੜਿਆ ਜਾਂਦਾ। ਹੇਠ ਲਿਖੀਆਂ ਉਦਾਹਰਣਾ ਨਾਲ ਜਿ਼ਆਦਾ ਚੰਗੀ ਤਰਾਂ ਸਮਝਿਆ ਜਾ ਸਕਦਾ ਹੈ:
6 8 5 2 8 + 5 + 2 = 15 ਅਤੇ 1 + 5 = 6 ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘3’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
4 0 7 8 4 + 0 + 7 + 8 = 19 ਅਤੇ 1 + 9 = 10 ਅਤੇ 1 + 0 = 1 ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘3’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
5 2 4 0 5 + 2 + 4 + 0 = 11 ਅਤੇ 1 + 1 = 2 ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘3’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
3 2 9 1 2 + 1 = 3 ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘3’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
6 8 3 9 8 ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘3’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
9 1 3 3 1 ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘3’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
4 ਨਾਲ ਵੰਡਣਯੋਗਤਾ (Divisibility) ਪਤਾ ਕਰਨ ਲਈ ਆਮ ਕਰਕੇ ਦਿੱਤੇ ਹੋਏ ਨੰਬਰ ਦੇ ਅਖਰਿਲੇ ਦੋ ਅੰਕਾਂ ਨੂੰ ‘4” ਤੇ ਭਾਗ ਦੇ ਕੇ ਪਤਾ ਕੀਤਾ ਜਾਂਦਾ ਹੈ ਪ੍ਰੰਤੂ ਅਸੀਂ ਅਜਿਹਾ ਨਹੀਂ ਕਰਦੇ। ਅਸੀਂ ‘ਦੋ ਵਾਰ ਜਿਸਤ’ (EE Even Even) ਦੇ ਆਪਣੇ ਬਣਾਏ ਨਿਯਮ ਨਾਲ ਇਸ ਦਾ ਪਤਾ ਲਾਉਂਦੇ ਹਾਂ। ਹੇਠ ਲਿਖੀਆਂ ਉਦਾਹਰਣਾ ਨਾਲ ਜਿ਼ਆਦਾ ਚੰਗੀ ਤਰਾਂ ਸਮਝਿਆ ਜਾ ਸਕਦਾ ਹੈ:
6 8 5 2 4 0 7 8 2 2 ਜਿਸਤ ਹੈ ਇਸ ਦਾ ਅੱਧ 1, 8+1 =9 ਟਾਂਕ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
6 8 5 2 4 0 7 8 8 ਜਿਸਤ ਹੈ ਇਸ ਦਾ ਅੱਧ 4, 4+7 =11 ਟਾਂਕ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
6 8 5 2 4 0 7 7 ਟਾਂਕ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
6 8 5 2 4 0 0 ਜਿਸਤ ਹੈ ਇਸ ਦਾ ਅੱਧ 0, 0+4 =4 ਜਿਸਤ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
6 8 5 2 4 4 ਜਿਸਤ ਹੈ ਇਸ ਦਾ ਅੱਧ 2, 2+2 =4 ਜਿਸਤ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
6 8 5 2 2 ਜਿਸਤ ਹੈ ਇਸ ਦਾ ਅੱਧ 1, 1+5 =6 ਜਿਸਤ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
6 8 8 ਜਿਸਤ ਹੈ ਇਸ ਦਾ ਅੱਧ 4, 4+6 =10 ਜਿਸਤ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘2’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
5 ਨਾਲ ਵੰਡਣਯੋਗਤਾ (Divisibility) ਪਤਾ ਕਰਨ ਲਈ ਆਮ ਕਰਕੇ ਦਿੱਤੇ ਹੋਏ ਨੰਬਰ ਦੇ ਅਖਰਿਲਾ ਅੰਕ ‘5’ ਜਾਂ ‘0’ ਹੋਵੇ ਤਾਂ ਉਹ ਨੰਬਰ ‘5’ ਤੇ ਵੰਡਣਯੋਗ (Divisible) ਹੁੰਦਾ ਹੈ। ਹੇਠ ਲਿਖੀਆਂ ਉਦਾਹਰਣਾ ਨਾਲ ਜਿ਼ਆਦਾ ਚੰਗੀ ਤਰਾਂ ਸਮਝਿਆ ਜਾ ਸਕਦਾ ਹੈ:
6 8 5 2 4 0 7 5 2 ਅੰਤਿਮ ਅੱਖਰ 2 ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘5’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
6 8 5 2 4 0 7 5 ਅੰਤਿਮ ਅੱਖਰ 5 ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘5’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
6 8 5 2 4 0 7 ਅੰਤਿਮ ਅੱਖਰ 7 ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘5’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
6 8 5 2 4 0 ਅੰਤਿਮ ਅੱਖਰ 0 ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘5’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
6 8 5 2 4 ਅੰਤਿਮ ਅੱਖਰ 4 ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘5’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
6 8 5 2 ਅੰਤਿਮ ਅੱਖਰ 2 ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘5’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
6 8 5 ਅੰਤਿਮ ਅੱਖਰ 5 ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘5’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
6 ਨਾਲ ਵੰਡਣਯੋਗਤਾ (Divisibility) ਪਤਾ ਕਰਨ ਲਈ ਇਹ ਵੇਖਣਾ ਜ਼ਰੂਰੀ ਹੈ ਕਿ ਦਿੱਤੇ ਹੋਏ ਨੰਬਰ ਦੇ ਅੰਕਾ ਦਾ ਜੋੜ ‘3’ ਤੇ ਵੰਡਿਆ ਜਾਂਦਾ ਹੋਵੇ ਅਤੇ ਉਸ ਦਾ ਅੰਤਿਮ ਅੱਖਰ ਜਿਸਤ (Even) ਹੋਵੇ। ਇਹ ਦੋਵੇਂ ਸ਼ਰਤਾਂ ਪੁਰੀਆ ਹੋਣਾ ਜ਼ਰੂਰੀ ਹੈ। ਅਸੀਂ ਇਸ ਨੂੰ ਵੀ ਹੋਰ ਸਰਲ ਕੀਤਾ ਹੈ। ਅੰਕਾਂ ਨੂੰ ਜੋੜਣ ਸਮੇਂ 3, 6 ਅਤੇ 9 ਨੂੰ ਨਹੀਂ ਜੋੜਿਆ ਜਾਂਦਾ। ਹੇਠ ਲਿਖੀਆਂ ਉਦਾਹਰਣਾ ਨਾਲ ਜਿ਼ਆਦਾ ਚੰਗੀ ਤਰਾਂ ਸਮਝਿਆ ਜਾ ਸਕਦਾ ਹੈ:
6 8 5 2 8 + 5 + 2 = 15 ਅਤੇ 1 + 5 = 6 ਅਤੇ ਅੰਤਿਮ ਅੱਖਰ ਜਿਸਤ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘6’ ਤੇ ਵੰਡਣਯੋਗ ਹੈ।
4 0 7 8 4 + 0 + 7 + 8 = 19 ਅਤੇ 1 + 9 = 10 ਅਤੇ 1 + 0 = 1 ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘6’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
5 2 4 0 5 + 2 + 4 + 0 = 11 ਅਤੇ 1 + 1 = 2 ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘6’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
3 2 9 1 2 + 1 = 3 ਅਤੇ ਅੰਤਿਮ ਅੱਖਰ ਟਾਂਕ ਹੈ ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘6’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
6 8 3 9 8 ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘6’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
9 1 3 3 1 ਇਸ ਲਈ ਇਹ ਨੰਬਰ ‘6’ ਤੇ ਵੰਡਣਯੋਗ ਨਹੀਂ ਹੈ।
ਜਿਥੋਂ ਤੱਕ Prime Numbers ਜਿਵੇਂ ਕਿ 7,13,17,19,23,29,31 ਆਦਿ ਦਾ ਸਵਾਲ ਹੈ, ਕਿਊਮੈਥਸ ਵਿੱਚ ਇਨ੍ਹਾਂ ਅੱਖਰਾਂ ਨਾਲ ਵੰਡਣਯੋਗਤਾ ਵੀ ਆਸਾਨੀ ਨਾਲ ਇੱਕੋ ਜਿਹੇ ਢੰਗ ਨਾਲ ਪਤਾ ਲਗਾਈ ਜਾ ਸਕਦੀ ਹੈ।
ਕਿੳੈਮੈਥਸ ਰਾਹੀਂ 5 ‹ 39, 11 ‹ 19, 24 ‹ 49 ਜਾਂ 8 ‹ 29 ਦਾ ਉੱਤਰ ਵੀ ਬਿਨਾਂ ਕਿਸੇ ਪ੍ਰੇਸ਼ਾਨੀ ਦੇ ਸੰਭਵ ਹੈ ਜੋ ਅਗਲੇ ਲੇਖਾਂ ਵਿੱਚ ਪਾਠਕਾਂ ਦੇ ਸਾਹਮਣੇ ਪੇਸ਼ ਕੀਤਾ ਜਾਵੇਗਾ।
ਬਾਕੀ ਅਗਲੇ ਲੇਖ ਵਿੱਚ ............................
ਸਾਹਿਤਕ ਇੰਟਰਨੈੱਟ ਮੈਗਜ਼ੀਨ "ਸ਼ਬਦ ਸਾਂਝ" 'ਤੇ ਆਪਜੀ ਦਾ ਨਿੱਘਾ ਸਵਾਗਤ ਹੈ । ਪੰਜਾਬੀ ਸਾਹਿਤ ਦੇ ਵਿਕਾਸ ਤੇ ਪ੍ਰਸਾਰ ਲਈ ਇਹ ਇੱਕ ਨਿਮਾਣਾ ਜਿਹਾ ਉੱਦਮ ਹੈ । "ਸ਼ਬਦ ਸਾਂਝ" ਦਾ ਕਿਸੇ ਵੀ ਰਾਜਨੀਤਿਕ, ਵਪਾਰਿਕ ਜਾਂ ਧਾਰਮਿਕ ਸੰਸਥਾ ਨਾਲ ਕੋਈ ਸਿੱਧਾ ਜਾਂ ਅਸਿੱਧਾ ਸੰਬੰਧ ਨਹੀਂ ਹੈ । ਆਪ ਜੀ ਦੇ ਹੁੰਗਾਰੇ ਦੀ ਭਰਪੂਰ ਆਸ ਰਹੇਗੀ ।
ARITHMATIC – LET’S MAKE IT A FUN (5)
‘Supplementary’ is used in many of the Vedic Math Sutras. The Supplementary of a number is the number which when added to the original number gives us a ‘round’ number, that means 10, 100, 1000, 10000, 100000 or any number made up of ‘1’ and zero(es). For example 47 is added to 53 to find 100, therefore 47 is the Supplementary of 53. Similarly when 61846 is added to 38154 to make it 100,000, 61846 is the Supplementary for 38154. The use of Supplementary is very useful in addition, subtraction and multiplication. To find the supplementary of a number, the following SUTRA of Vedic Math can be very helpful:
ALL FROM 9 LAST FROM 10
The sutra says that such a digit is added individually to all the digits of a number so that all but the last digit comes to ‘9’ and the last as ‘10’, the new number is the supplementary of the given number or we can say that by adding the two numbers, we will get an answer consisting of ‘1’ and ‘0’ or ‘0’es. To understand the concept, let’s discuss the following examples:
2 3 4 1 6 7 3 8 2 9 8 4
7 6 6 8 3 2 6 1 7 1 1 6
1 0 0 0 1 0 0 0 0 0 0 0 1 0 0
In the first example 2+7=9, 3+6=9 and 4+6=10, in the second example 1+8-9, 6+3=9, 7+2=9, 3+6=9, 8+1=9, 2+7=9 and 9+1=10 while in the third 1+8=9 and 4+6=10. When the original number is added to the supplementary so found the answer is always made up of ‘1’ and zeroes.
Let’s try to understand the use of supplementary in Vedic formulae for multiplication. This is understandable that smaller the ‘supplementary’ easier will be the solution. This is not so that if the ‘supplementary’ is larger, it can’t be applied in these formulae but there will be no use of such short cut that is difficult than the ‘long cut’. That is why these formulae are most effective where there is smaller ‘supplementary’, that means for numbers those are near to 10, 100, 1000, 10000, 100000 i.e. 9992, 999989, 91, 985 etc.
First of all, we will discuss squaring with the help of supplementary:
Example: 997 x 997 The answer will be in 3+3=6 (***,***) digits.
997 + 3 = 1000 So supplementary of 997 is 3
First step: 997 – 3 = 994 First three digits 994,***
Second step: 3 x 3 = 9 Last three digits 994,009
So 997 x 997 = 994,004
Example: 99991 x 99991 The answer will be in 5+5=10 (*****,*****) digits.
99991 + 9 = 100000 So supplementary of 99991 is 9
First step: 99991 – 9 = 99982 First five digits 99982,*****
Second step: 9 x 9 = 81 Last three digits 99982,00081
So 99991 x 99991 = 99982,00081
Try to multiply 9995 x 9995 and 9999989 X 9999989.
Now we will discuss multiplying a number by equal number ‘9’:
Example: 36254 X 99999 There will be an answer in 5+5=10 (*****,*****) digits.
First Step: 36254 – 1 = 36254 First five digits 36253,***** (Always deduct ‘1’ )
Second Step: Find the supplement for 36254 (ALL FROM 9 LAST FROM 10)
36254 + 63746 = 100000 Last five digits 36253,63746
So 36254 X 99999 = 36253,63746
Example: 47 X 99 There will be an answer in 2+2=4 (**,**) digits.
First Step: 47 – 1 = 46 First two digits 46,** (Always deduct ‘1’ )
Second Step: Find the supplement for 47 (ALL FROM 9 LAST FROM 10)
47 + 53 = 100 Last five digits 46,53
So 47 X 99 = 46,53
One more to multiply two such numbers are 10, 100, 1000, 10000, 100000 etc.
Example: 993 X 996 There will be answer in 3+3=6 (***,***).
9 9 3 9 9 6
0 0 7 0 0 4
First step: 993 – 4 = 989 First three digits 989,*** (Deduct crosswise)
Second Step: 7 X 4 = 28 Last three digits 989,028
So 993 X 996 = 989,028
Example: 999991 X 999998 There will be an answer in 6+6=12 (******,******) digits.
9 9 9 9 9 1 9 9 9 9 9 8
0 0 0 0 0 9 0 0 0 0 0 2
First step: 999991 – 2 = 999989 First six digits 999989,****** (Deduct Crosswise)
Second Step: 9 X 2 = 18 Last six digits 999989,000018
So 999991 X 999998 = 999989,000018
Example: 9975 X 9988 There will be answer in 4+4=8 digits
9 9 7 5 9 9 8 8
0 0 2 5 0 0 1 2
First step: 9975 – 12 = 9963 First 4 digits 9963,**** (Deduct Crosswise)
Second step: 25 X 12 = 300 Last 4 digits 9963,0300
So 9975 X 9988 = 9963,0300
If there is number having repeated numbers near 100,1000,10000 or any such number, the given number is multiplied by with help of a supplementary.
Example: 95959595 X 17 There will be answer in 8+2=10 (**,******,**) digits.
95 + 5 = 100 So supplementary for 95 is 5.
First step: 17 -1 = 16 First two digits 16,******,**
Second step: 17 X 5 = 85 and 100 – 85 = 15 Last two digits 16,******,15
Third step: 16 + 15 = 31 6 digits in the middle 16,313131,15
So 95959595 X 17 = 16,313131,15
Example: 9898 X 36 There will be answer in 4+2=6 (**,**,**) digits.
98 + 2 = 100 So supplementary for 98 is 2.
First step: 36 -1 = 35 First two digits 35,**,**
Second step: 36 X 2 = 72 and 100 – 72 = 28 Last two digits 35,**,28
Third step: 35 + 28 = 63 6 digits in the middle 35,63,28
So 9898 X 36 = 35,63,28
Such kind of small and large multiplications are helpful in generating confidence for mathematics among the students.
I can understand that everything can’t be explained just by writing down some words. Therefore I would like to request the readers to feel free to write me any question at info@qmaths.com or just open my site qmaths.com.
Wait for much more in the coming articles ………………….
ALL FROM 9 LAST FROM 10
The sutra says that such a digit is added individually to all the digits of a number so that all but the last digit comes to ‘9’ and the last as ‘10’, the new number is the supplementary of the given number or we can say that by adding the two numbers, we will get an answer consisting of ‘1’ and ‘0’ or ‘0’es. To understand the concept, let’s discuss the following examples:
2 3 4 1 6 7 3 8 2 9 8 4
7 6 6 8 3 2 6 1 7 1 1 6
1 0 0 0 1 0 0 0 0 0 0 0 1 0 0
In the first example 2+7=9, 3+6=9 and 4+6=10, in the second example 1+8-9, 6+3=9, 7+2=9, 3+6=9, 8+1=9, 2+7=9 and 9+1=10 while in the third 1+8=9 and 4+6=10. When the original number is added to the supplementary so found the answer is always made up of ‘1’ and zeroes.
Let’s try to understand the use of supplementary in Vedic formulae for multiplication. This is understandable that smaller the ‘supplementary’ easier will be the solution. This is not so that if the ‘supplementary’ is larger, it can’t be applied in these formulae but there will be no use of such short cut that is difficult than the ‘long cut’. That is why these formulae are most effective where there is smaller ‘supplementary’, that means for numbers those are near to 10, 100, 1000, 10000, 100000 i.e. 9992, 999989, 91, 985 etc.
First of all, we will discuss squaring with the help of supplementary:
Example: 997 x 997 The answer will be in 3+3=6 (***,***) digits.
997 + 3 = 1000 So supplementary of 997 is 3
First step: 997 – 3 = 994 First three digits 994,***
Second step: 3 x 3 = 9 Last three digits 994,009
So 997 x 997 = 994,004
Example: 99991 x 99991 The answer will be in 5+5=10 (*****,*****) digits.
99991 + 9 = 100000 So supplementary of 99991 is 9
First step: 99991 – 9 = 99982 First five digits 99982,*****
Second step: 9 x 9 = 81 Last three digits 99982,00081
So 99991 x 99991 = 99982,00081
Try to multiply 9995 x 9995 and 9999989 X 9999989.
Now we will discuss multiplying a number by equal number ‘9’:
Example: 36254 X 99999 There will be an answer in 5+5=10 (*****,*****) digits.
First Step: 36254 – 1 = 36254 First five digits 36253,***** (Always deduct ‘1’ )
Second Step: Find the supplement for 36254 (ALL FROM 9 LAST FROM 10)
36254 + 63746 = 100000 Last five digits 36253,63746
So 36254 X 99999 = 36253,63746
Example: 47 X 99 There will be an answer in 2+2=4 (**,**) digits.
First Step: 47 – 1 = 46 First two digits 46,** (Always deduct ‘1’ )
Second Step: Find the supplement for 47 (ALL FROM 9 LAST FROM 10)
47 + 53 = 100 Last five digits 46,53
So 47 X 99 = 46,53
One more to multiply two such numbers are 10, 100, 1000, 10000, 100000 etc.
Example: 993 X 996 There will be answer in 3+3=6 (***,***).
9 9 3 9 9 6
0 0 7 0 0 4
First step: 993 – 4 = 989 First three digits 989,*** (Deduct crosswise)
Second Step: 7 X 4 = 28 Last three digits 989,028
So 993 X 996 = 989,028
Example: 999991 X 999998 There will be an answer in 6+6=12 (******,******) digits.
9 9 9 9 9 1 9 9 9 9 9 8
0 0 0 0 0 9 0 0 0 0 0 2
First step: 999991 – 2 = 999989 First six digits 999989,****** (Deduct Crosswise)
Second Step: 9 X 2 = 18 Last six digits 999989,000018
So 999991 X 999998 = 999989,000018
Example: 9975 X 9988 There will be answer in 4+4=8 digits
9 9 7 5 9 9 8 8
0 0 2 5 0 0 1 2
First step: 9975 – 12 = 9963 First 4 digits 9963,**** (Deduct Crosswise)
Second step: 25 X 12 = 300 Last 4 digits 9963,0300
So 9975 X 9988 = 9963,0300
If there is number having repeated numbers near 100,1000,10000 or any such number, the given number is multiplied by with help of a supplementary.
Example: 95959595 X 17 There will be answer in 8+2=10 (**,******,**) digits.
95 + 5 = 100 So supplementary for 95 is 5.
First step: 17 -1 = 16 First two digits 16,******,**
Second step: 17 X 5 = 85 and 100 – 85 = 15 Last two digits 16,******,15
Third step: 16 + 15 = 31 6 digits in the middle 16,313131,15
So 95959595 X 17 = 16,313131,15
Example: 9898 X 36 There will be answer in 4+2=6 (**,**,**) digits.
98 + 2 = 100 So supplementary for 98 is 2.
First step: 36 -1 = 35 First two digits 35,**,**
Second step: 36 X 2 = 72 and 100 – 72 = 28 Last two digits 35,**,28
Third step: 35 + 28 = 63 6 digits in the middle 35,63,28
So 9898 X 36 = 35,63,28
Such kind of small and large multiplications are helpful in generating confidence for mathematics among the students.
I can understand that everything can’t be explained just by writing down some words. Therefore I would like to request the readers to feel free to write me any question at info@qmaths.com or just open my site qmaths.com.
Wait for much more in the coming articles ………………….
ARITHMATIC – LET’S MAKE IT A FUN (4)
Like addition, multiply is one of the important operations in arithmetic. In Vedic Math, there are many short cuts for multiply. To understand the multiply, if we say that it is a simple way of addition, there will be nothing wrong in it. Most of the times, I ask my students ‘Why? Why? 9 multiplied by 7 always give 63? Why not anything else? The answer is quite simple :
9
+ 9
+ 9
+ 9
+ 9
+ 9
+ 9
63
We can say that if 9 is added for 7 times, it gives 63 or we can say that multiply is a short cut for addition. Though it is exactly correct even then we can’t ignore the importance of multiplication. Just suppose that if we write 67 for 84 times and then add instead of multiplying 67 with 84, will there be no wastage of time and efforts? Here the multiply will be suitable. Forget this, if we are to multiply 2345671289 and 876956743, it would be near impossible to solve it through addition.
Now we will discuss different methods of multiplying. In shools, the children are taught to multiply the numbers with the same tradition method. That is :
51
X26
--------
306
102X
--------
1326
--------
This is an example of multiply without ‘carry’ and the multiplication with ‘carry’ is somewhat more tough. QMaths is such a method of teaching Mathematics that “Why and How” of every sum is made clear taking into account the basic of every sum of mathematics. We shall discuss many methods of multiplying being used in QMaths so that the basic concept of multiplying could be understood in a better way.
Let us try to understand the system adopted by us to multiply any two digit number with any other two digit number:
56
X 73
--------
3518 ( 7X5 = 35 and 6 X 3 = 18)
15X ( 5X3 = 15 CROSS MULTIPLY )
42X ( 6X7 = 42 CROSS MULTIPLY )
--------
4088
--------
Second Example:
82
X 67
--------
4814 ( 8X6 = 48 and 2X7 = 14)
56X ( 8X7 = 56 CROSS MULTIPLY )
12X ( 6X2 = 12 CROSS MULTIPLY )
--------
5494
--------
Now try to multiply 46 X 78, 37 X 82 and 58 X 27 and see if this method is easier than the common one.
Let us have a look at another interesting method, the ‘Box Method’ of multiplying. If 36 is to be multiplied by 72, the following method can also be applied:
One more example: to multiply 63 by 28
One step ahead: to multiply 346 by 27
Now try to multiply 26 X 78, 317 X 82 and 88 X 27 and see if this method is interesting and easier than the common one.
To solve the multiplication sum, another formula that we call ‘STAR FORMULA’ can also be used. Let’s try to understand that one also:
2 4 6
3 1 5
2 4 6
X 3 1 5
------------
6 4 2 6 0
1 3 2 3
------------
7 7 4 9 0
------------
In this case, we write the ‘carry’ below the previous number as given underline in the given examples. To understand in a more accurately, go for another solution with this formula:
1 7 3
X 2 4 6
------------
2 8 0 4 8
1 4 5 1
------------
4 2 5 5 8
------------
Another more method based on the basic concept of multiplication can be understood by the rules of ‘cutting lines’. Let’s discuss:
Other than the above, there are many methods of multiplying. We have discussed here five different methods of multiplying. If we teach the students multiply with the help of all the above given methods, it is sure the students would be more interested in mathematics.
We can discuss here dozens of shortcuts related only to multiply but shortage of time and space does not allow us to do so. Even then we would discuss Vedic Math SUTRAs to multiply 999994 x 999992 and 674532 x 999999. In addition to these, we would discuss formulas to solve in single line 9998 x 9998 x 9998, 4343434343 x 98 and 999992 x 9996 which are not covered in Vedic Math.
I can understand that it is not possible to convey every thing just by writing about that, so I would request the readers to visit my website www.qmaths.com and pose any question as and when they like.
More in the coming articles…………………….
9
+ 9
+ 9
+ 9
+ 9
+ 9
+ 9
63
We can say that if 9 is added for 7 times, it gives 63 or we can say that multiply is a short cut for addition. Though it is exactly correct even then we can’t ignore the importance of multiplication. Just suppose that if we write 67 for 84 times and then add instead of multiplying 67 with 84, will there be no wastage of time and efforts? Here the multiply will be suitable. Forget this, if we are to multiply 2345671289 and 876956743, it would be near impossible to solve it through addition.
Now we will discuss different methods of multiplying. In shools, the children are taught to multiply the numbers with the same tradition method. That is :
51
X26
--------
306
102X
--------
1326
--------
This is an example of multiply without ‘carry’ and the multiplication with ‘carry’ is somewhat more tough. QMaths is such a method of teaching Mathematics that “Why and How” of every sum is made clear taking into account the basic of every sum of mathematics. We shall discuss many methods of multiplying being used in QMaths so that the basic concept of multiplying could be understood in a better way.
Let us try to understand the system adopted by us to multiply any two digit number with any other two digit number:
56
X 73
--------
3518 ( 7X5 = 35 and 6 X 3 = 18)
15X ( 5X3 = 15 CROSS MULTIPLY )
42X ( 6X7 = 42 CROSS MULTIPLY )
--------
4088
--------
Second Example:
82
X 67
--------
4814 ( 8X6 = 48 and 2X7 = 14)
56X ( 8X7 = 56 CROSS MULTIPLY )
12X ( 6X2 = 12 CROSS MULTIPLY )
--------
5494
--------
Now try to multiply 46 X 78, 37 X 82 and 58 X 27 and see if this method is easier than the common one.
Let us have a look at another interesting method, the ‘Box Method’ of multiplying. If 36 is to be multiplied by 72, the following method can also be applied:
One more example: to multiply 63 by 28
One step ahead: to multiply 346 by 27
Now try to multiply 26 X 78, 317 X 82 and 88 X 27 and see if this method is interesting and easier than the common one.
To solve the multiplication sum, another formula that we call ‘STAR FORMULA’ can also be used. Let’s try to understand that one also:
2 4 6
3 1 5
2 4 6
X 3 1 5
------------
6 4 2 6 0
1 3 2 3
------------
7 7 4 9 0
------------
In this case, we write the ‘carry’ below the previous number as given underline in the given examples. To understand in a more accurately, go for another solution with this formula:
1 7 3
X 2 4 6
------------
2 8 0 4 8
1 4 5 1
------------
4 2 5 5 8
------------
Another more method based on the basic concept of multiplication can be understood by the rules of ‘cutting lines’. Let’s discuss:
Other than the above, there are many methods of multiplying. We have discussed here five different methods of multiplying. If we teach the students multiply with the help of all the above given methods, it is sure the students would be more interested in mathematics.
We can discuss here dozens of shortcuts related only to multiply but shortage of time and space does not allow us to do so. Even then we would discuss Vedic Math SUTRAs to multiply 999994 x 999992 and 674532 x 999999. In addition to these, we would discuss formulas to solve in single line 9998 x 9998 x 9998, 4343434343 x 98 and 999992 x 9996 which are not covered in Vedic Math.
I can understand that it is not possible to convey every thing just by writing about that, so I would request the readers to visit my website www.qmaths.com and pose any question as and when they like.
More in the coming articles…………………….
ARITHMATIC – LET’S MAKE IT A FUN (3)
Though mathematics is treated as a difficult subject but after understanding the basic concept of a math sum, it is as simple to solve the sum as speaking the mother tongue by the student. Even then the first condition for this is that there should not be any kind of phobia for math. I can say with a strong confidence that this kind of phobia can be there only in such people those are weak in the basic math operations like addition, subtraction, multiplication and division. So first of all, the junior as well as senior students should try to get rid of these weaknesses and be perfect in these operations.
Many easy methods of division are also there. Especially if a number is to be divided by ‘9’ or 9’s to find the answer in decimals, our method is as easy as the traditional method is difficult. To solve 7 ‹ 9, just guess how much time is required. Also guess about the minimum time to solve this question. Can it be as less as 2 seconds? I think it is a longer time. Its answer is 0.777777777777777777777 or 0.7 and a bar above 7 to show its non-termination.
Let’s try to understand this division method with the help of examples:
4 ‹ 9
= 0.4444444444444444444
14 ‹ 9
= 1.555555555555555555555555
123 ‹ 9
= 13.6666666666666666666666666
Put ‘1, then 1+2, then 1+2+3 and repeat the decimal number time and again.
57 ‹ 99
= 0.5757575757575757
2 ‹ 99
= 0.020202020202020202020202
124 ‹ 99
= 1.252525252525252525252525
Here ‘1’, then 1+24. (Because ‘99’ has two digits)
Try yourself to solve 61 ‹ 99 and 433 ‹ 9999
Enjoyed it?
Divisibility rules for certain numbers start appearing in the school syllabus right from class four. To define divisibility, we can say that if dividing the given figure by a certain numbers gives NO (‘0’) REMAINDER, then the figure is divisible by the number. Most of the times, the students pay attention only for the text book exercise but its it is more important in the higher classes. To show the fraction in its minimum form is possible only with these rules. Use of these rules are there in arithmetic calculation, may be directly or indirectly.
To find whether a given number is divisible by ‘2’ or not, the last digit is taken into account, if the last number is ‘even’, the number is divisible by ‘2’ and if ‘odd’, the number is ‘not’ divisible by ‘2’. The rule can be understood in a better way through the following examples:
6 8 5 2 4 0 7 8 2 2 is EVEN so this number is divisible by ‘2’
6 8 5 2 4 0 7 8 8 is EVEN so this number is divisible by ‘2’
6 8 5 2 4 0 7 7 is ODD so this number is NOT divisible by ‘2’
6 8 5 2 4 0 0 is EVEN so this number is divisible by ‘2’
6 8 5 2 4 4 is EVEN so this number is divisible by ‘2’
6 8 5 2 2 is EVEN so this number is divisible by ‘2’
6 8 5 7 is ODD so this number is NOT divisible by ‘2’
6 8 8 is EVEN so this number is divisible by ‘2’
6 6 is EVEN so this number is divisible by ‘2’
To find whether a given number is divisible by ‘3’ or not, the sum of the digits in the number is taken into account, if the sum is a multiple of ‘3’, the number is divisible by ‘3’ and if not a multiple of ‘3’, the number is ‘not’ divisible by ‘3’. In QMaths, it is easier. We do not add 3,6,9 while finding the sum of digits. The rule can be understood in a better way through the following examples:
6 8 5 2 8 + 5 + 2 = 15 and 1 + 5 = 6, so 6852 is divisible by ‘3’.
4 0 7 8 4 + 0 + 7 + 8 = 19 and 1 + 9 = 10 and 1 + 0 = 1 so 4078 is NOT divisible by ‘3’
5 2 4 0 5 + 2 + 4 + 0 = 11 and 1 + 1 = 2 so 5240 is not divisible by ‘3’.
3 2 9 1 2 + 1 = 3 so 3291 is divisible by ‘3’.
6 8 3 9 8 so 6839 is NOT divisible by ‘3’
9 1 3 3 1 so 9133 is NOT divisible by ‘3’
To find whether a given number is divisible by ‘4’ or not, the last two digits are taken into account, if the last two digits are divisible by ‘4’, the number is divisible by ‘4’ and if not, the number is ‘not’ divisible by ‘4’. In QMaths, it is easier. We use our formula of DOUBLE EVEN (EE).
THE RULE SAYS: If last digit is EVEN, find half of the digit and add it to second last number, if the answer is again EVEN, the number is divisible by ‘4’, if there is any ODD in these two tests, the number is not divisible by ‘4’
The rule can be understood in a better way through the following examples:
6 8 5 2 4 0 7 8 2 2 is EVEN, half of ‘2’ is ‘1’, 8+1=9 that is ODD, so this number is not divisible by ‘4’
6 8 5 2 4 0 7 8 8 is EVEN, half of ‘8’ is ‘4’, 7+4=11 that is ODD, so this number is not divisible by ‘4’
6 8 5 2 4 0 7 7 is ODD, so this number is not divisible by ‘4’
6 8 5 2 4 0 0 is EVEN, half of ‘0’ is 0, 4+0=4 that is EVEN, so this number is divisible by ‘4’
6 8 5 2 4 4 is EVEN, half of ‘4’ is 2, 2+4=6 that is EVEN, so this number is divisible by ‘4’
6 8 5 2 2 is EVEN, half of ‘2’ is 1, 5+1=6 that is EVEN, so this number is divisible by ‘4’.
6 8 5 5 is ODD, so this number is not divisible by ‘4’
6 8 8 is EVEN, half of ‘8’ is ‘4’, 6+4=10 that is EVEN, so this number is divisible by ‘4’
To find whether a given number is divisible by ‘5’ or not, the last digit is taken into account, if the last digit is either ‘5’ or ‘0’, the number is divisible by ‘5’ and if not, the number is ‘not’ divisible by ‘5’.
6 8 5 2 4 0 7 5 2 the last digit is not ‘5’ or ‘0’, so the number is NOT divisible by ‘5’
6 8 5 2 4 0 7 5 the last digit is ‘5’, so the number is divisible by ‘5’
6 8 5 2 4 0 7 the last digit is not ‘5’ or ‘0’, so the number is NOT divisible by ‘5’
6 8 5 2 4 0 the last digit is ‘0’, so the number is divisible by ‘5’
6 8 5 2 4 the last digit is not ‘5’ or ‘0’, so the number is NOT divisible by ‘5’
6 8 5 2 the last digit is not ‘5’ or ‘0’, so the number is NOT divisible by ‘5’
6 8 5 the last digit is ‘5’, so the number is divisible by ‘5’
To find whether a given number is divisible by ‘6’ or not, both the rules for divisibility by ‘2’ and ‘3’ is taken into account, if the sum of digits is a multiple of ‘3’ and the last digit of the number is EVEN, the number is divisible by ‘6’ and if not a multiple of ‘3’ or the last digit of the number is ODD, the number is ‘not’ divisible by ‘6’. In QMaths, it is easier. We do not add 3,6,9 while finding the sum of digits. The rule can be understood in a better way through the following examples:
6 8 5 2 8 + 5 + 2 = 15 and 1 + 5 = 6 and ‘2’ is EVEN, so 6852 is divisible by ‘6’
4 0 7 8 4 + 0 + 7 + 8 = 19 and 1 + 9 = 10 and 1 + 0 = 1, so 4078 is NOT divisible by ‘6’
5 2 4 0 5 + 2 + 4 + 0 = 11 and 1 + 1 = 2 so 3291 is NOT divisible by ‘6’
3 2 9 1 2 + 1 = 3 but last digit ‘1’ is odd, so 3291 is not divisible by ‘6’
6 8 3 9 8, so 6839 is NOT divisible by ‘6’
9 1 3 3 1, so 9133 is NOT divisible by ‘6’
As far as divisibility by prime numbers such as 7,13,17,19,23,29,31 etc. it is possible through a very easy method in QMaths.
In QMaths, to solve 5 ‹ 39, 11 ‹ 19 or 8 ‹ 29 is again very easy and it would be explained in the coming articles ………………
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Many easy methods of division are also there. Especially if a number is to be divided by ‘9’ or 9’s to find the answer in decimals, our method is as easy as the traditional method is difficult. To solve 7 ‹ 9, just guess how much time is required. Also guess about the minimum time to solve this question. Can it be as less as 2 seconds? I think it is a longer time. Its answer is 0.777777777777777777777 or 0.7 and a bar above 7 to show its non-termination.
Let’s try to understand this division method with the help of examples:
4 ‹ 9
= 0.4444444444444444444
14 ‹ 9
= 1.555555555555555555555555
123 ‹ 9
= 13.6666666666666666666666666
Put ‘1, then 1+2, then 1+2+3 and repeat the decimal number time and again.
57 ‹ 99
= 0.5757575757575757
2 ‹ 99
= 0.020202020202020202020202
124 ‹ 99
= 1.252525252525252525252525
Here ‘1’, then 1+24. (Because ‘99’ has two digits)
Try yourself to solve 61 ‹ 99 and 433 ‹ 9999
Enjoyed it?
Divisibility rules for certain numbers start appearing in the school syllabus right from class four. To define divisibility, we can say that if dividing the given figure by a certain numbers gives NO (‘0’) REMAINDER, then the figure is divisible by the number. Most of the times, the students pay attention only for the text book exercise but its it is more important in the higher classes. To show the fraction in its minimum form is possible only with these rules. Use of these rules are there in arithmetic calculation, may be directly or indirectly.
To find whether a given number is divisible by ‘2’ or not, the last digit is taken into account, if the last number is ‘even’, the number is divisible by ‘2’ and if ‘odd’, the number is ‘not’ divisible by ‘2’. The rule can be understood in a better way through the following examples:
6 8 5 2 4 0 7 8 2 2 is EVEN so this number is divisible by ‘2’
6 8 5 2 4 0 7 8 8 is EVEN so this number is divisible by ‘2’
6 8 5 2 4 0 7 7 is ODD so this number is NOT divisible by ‘2’
6 8 5 2 4 0 0 is EVEN so this number is divisible by ‘2’
6 8 5 2 4 4 is EVEN so this number is divisible by ‘2’
6 8 5 2 2 is EVEN so this number is divisible by ‘2’
6 8 5 7 is ODD so this number is NOT divisible by ‘2’
6 8 8 is EVEN so this number is divisible by ‘2’
6 6 is EVEN so this number is divisible by ‘2’
To find whether a given number is divisible by ‘3’ or not, the sum of the digits in the number is taken into account, if the sum is a multiple of ‘3’, the number is divisible by ‘3’ and if not a multiple of ‘3’, the number is ‘not’ divisible by ‘3’. In QMaths, it is easier. We do not add 3,6,9 while finding the sum of digits. The rule can be understood in a better way through the following examples:
6 8 5 2 8 + 5 + 2 = 15 and 1 + 5 = 6, so 6852 is divisible by ‘3’.
4 0 7 8 4 + 0 + 7 + 8 = 19 and 1 + 9 = 10 and 1 + 0 = 1 so 4078 is NOT divisible by ‘3’
5 2 4 0 5 + 2 + 4 + 0 = 11 and 1 + 1 = 2 so 5240 is not divisible by ‘3’.
3 2 9 1 2 + 1 = 3 so 3291 is divisible by ‘3’.
6 8 3 9 8 so 6839 is NOT divisible by ‘3’
9 1 3 3 1 so 9133 is NOT divisible by ‘3’
To find whether a given number is divisible by ‘4’ or not, the last two digits are taken into account, if the last two digits are divisible by ‘4’, the number is divisible by ‘4’ and if not, the number is ‘not’ divisible by ‘4’. In QMaths, it is easier. We use our formula of DOUBLE EVEN (EE).
THE RULE SAYS: If last digit is EVEN, find half of the digit and add it to second last number, if the answer is again EVEN, the number is divisible by ‘4’, if there is any ODD in these two tests, the number is not divisible by ‘4’
The rule can be understood in a better way through the following examples:
6 8 5 2 4 0 7 8 2 2 is EVEN, half of ‘2’ is ‘1’, 8+1=9 that is ODD, so this number is not divisible by ‘4’
6 8 5 2 4 0 7 8 8 is EVEN, half of ‘8’ is ‘4’, 7+4=11 that is ODD, so this number is not divisible by ‘4’
6 8 5 2 4 0 7 7 is ODD, so this number is not divisible by ‘4’
6 8 5 2 4 0 0 is EVEN, half of ‘0’ is 0, 4+0=4 that is EVEN, so this number is divisible by ‘4’
6 8 5 2 4 4 is EVEN, half of ‘4’ is 2, 2+4=6 that is EVEN, so this number is divisible by ‘4’
6 8 5 2 2 is EVEN, half of ‘2’ is 1, 5+1=6 that is EVEN, so this number is divisible by ‘4’.
6 8 5 5 is ODD, so this number is not divisible by ‘4’
6 8 8 is EVEN, half of ‘8’ is ‘4’, 6+4=10 that is EVEN, so this number is divisible by ‘4’
To find whether a given number is divisible by ‘5’ or not, the last digit is taken into account, if the last digit is either ‘5’ or ‘0’, the number is divisible by ‘5’ and if not, the number is ‘not’ divisible by ‘5’.
6 8 5 2 4 0 7 5 2 the last digit is not ‘5’ or ‘0’, so the number is NOT divisible by ‘5’
6 8 5 2 4 0 7 5 the last digit is ‘5’, so the number is divisible by ‘5’
6 8 5 2 4 0 7 the last digit is not ‘5’ or ‘0’, so the number is NOT divisible by ‘5’
6 8 5 2 4 0 the last digit is ‘0’, so the number is divisible by ‘5’
6 8 5 2 4 the last digit is not ‘5’ or ‘0’, so the number is NOT divisible by ‘5’
6 8 5 2 the last digit is not ‘5’ or ‘0’, so the number is NOT divisible by ‘5’
6 8 5 the last digit is ‘5’, so the number is divisible by ‘5’
To find whether a given number is divisible by ‘6’ or not, both the rules for divisibility by ‘2’ and ‘3’ is taken into account, if the sum of digits is a multiple of ‘3’ and the last digit of the number is EVEN, the number is divisible by ‘6’ and if not a multiple of ‘3’ or the last digit of the number is ODD, the number is ‘not’ divisible by ‘6’. In QMaths, it is easier. We do not add 3,6,9 while finding the sum of digits. The rule can be understood in a better way through the following examples:
6 8 5 2 8 + 5 + 2 = 15 and 1 + 5 = 6 and ‘2’ is EVEN, so 6852 is divisible by ‘6’
4 0 7 8 4 + 0 + 7 + 8 = 19 and 1 + 9 = 10 and 1 + 0 = 1, so 4078 is NOT divisible by ‘6’
5 2 4 0 5 + 2 + 4 + 0 = 11 and 1 + 1 = 2 so 3291 is NOT divisible by ‘6’
3 2 9 1 2 + 1 = 3 but last digit ‘1’ is odd, so 3291 is not divisible by ‘6’
6 8 3 9 8, so 6839 is NOT divisible by ‘6’
9 1 3 3 1, so 9133 is NOT divisible by ‘6’
As far as divisibility by prime numbers such as 7,13,17,19,23,29,31 etc. it is possible through a very easy method in QMaths.
In QMaths, to solve 5 ‹ 39, 11 ‹ 19 or 8 ‹ 29 is again very easy and it would be explained in the coming articles ………………
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ARITHMATIC – LET’S MAKE IT A FUN (2)
In the last article of this series we read how some of the math operations can be made very interesting. Nobody is dull in mathematics. The problem is that when a student start lacking in a particular topic of math, he loses confidence. If such things are repeated in his school life, a strong conception develops in him/her that math is not his/her cup of tea or it is impossible for him/her. Math then becomes a stranger for the student. With the passage of time, the student dishearten about this subject that he could not get rid of the phobia for math. QMaths is such an effort so that the basic operations of arithmetic can be made easy to understand as well as interesting.
Let’s try to understand easy short cuts for multiplying or dividing a number by ‘5’.
To multiply a number by ‘5’, we should make the number as half or we can say that divide the number by ‘2’. Then we should see whether the last number is ‘Even’ or ‘Odd’. If even put ‘0’ at the last and if odd then put ‘5’ at the last. The following examples can help to understand the shortcut:
286402 X 5
= 1432010
(Half of 2 is 1, 8 is 4, 6 is 3, 4 is 2, 0 is 0, 2 is 1 and last digit ‘2’ is ‘Even’, so ‘0’ is put at the last of the answer)
4163287 X 5
= 20816435
To understand in a better way, treat it as 4,16,32,8,7.
(Half of 4 is 2, 16 is 08, 32 is 16, 8 is 4 and 7 is 3 with 1 as remainder. The last digit ‘7’ is odd, so ‘5’ has been put at the last of the answer.
5460123 X 5
= 27300615
To understand in a better way, treat it as 54,6,0,12,3.
(Half of 54 is 27, 6 is 3, 0 is 0, 12 is 06, 3 is 1 with ‘1’ as remainder. The last digit ‘3’ is ‘odd’, so ‘5’ has been put at the last of the answer.
Let’s try:
642,082,246,288,640,866,244,826,884,260,024,846,282,864 X 5
Is this multiply sum still difficult for you?
Now we would try to understand the division of a number by ‘5’:
To divide a given number by ‘5’, the easiest way is to double the given number or we can say that multiply the number by ‘5’. Then put a decimal (.) leaving the last digit of the answer. The following examples can help to understand the shortcut:
142032 ‹ 5 =
28406.4
(Double of 1 is 2, 4 is 8, 2 is 4, 0 is 0, 3 is 6 and 2 is 4. A decimal has been placed before the last digit 4.
174232625 ‹ 5 =
34846525.0 or 34846525
To understand in a better way, treat it as 17,4,2,3,26,25.
(Double of 17 is 34, 4 is 8, 2 is 4, 3 is 6, 26 is 52, 25 is 50. A decimal has been placed before the last digit ‘0’ or we can write the number without ‘0’.
1262741835 ‹ 5 =
252548367.0 or 252548367
To understand in a better way, treat it as 1, 26, 27, 4, 18, 35
(Double of 1 is 2, 26 is 52, 27 is 54, 4 is 8, 18 is 36, 35 is 70. A decimal has been placed before the last digit ‘0’ or we can write the answer without decimal.
4230,1134,2314,0021,2113,2442,1423,4012,2331,2132,1230,4112,4001 ‹ 5
Is this division sum still difficult for you?
Let’s try to understand another short cut of multiply. If a figure is to multiplied by ‘11’, then the following shortcut can be used:
35 X 11 = 3 (3+5) 5 = 385
52 X 11 = 5 (5+2) 2 = 572
Go ahead a step more:
234536 X 11
= 2(2+3)(3+4)(4+5)(5+3)(3+6)6
= 2579896
417235180345127 X 11
= 4(4+1)(1+7)(7+2)(2+3)(3+5)(5+1)(1+8)(8+0)(0+3)(3+4)(4+5)(5+1)(1+2)(2+7)7
= 45895869837963971
Try this yourself:
235416317 X 11
Similarly it is very easy to multiply a figure by 25 or 125 if we use the basic concept of multiplication and use the easy short cuts available with Qmaths.
Are you interested to multiply 99999995 to 99999992 within 10 seconds? If so please wait for our coming articles. It may be possible for you to solve it in less than 10 seonds.
We shall also discuss methods to find the divisibility of a number not only by 2,3,4,5,6,8,9,10,11,12,15 but also the prime number like 7,13,17,23,29,31,37,41 etc. with very easy and non-conventional methods.
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Surinder Bharti Tiwari
QMaths Center of Learning
www.qmaths.com
email : info@qmaths.com
ARITHMATIC – LET’S MAKE IT A FUN (1)
Mathematics phobia has become a world phenomenon. In my coaching classes throughout my three decade experience, I have found that there are basically two types of mathematics students. First, those who are very good and really brilliant at Mathematics, they may be God gifted or they can be hard workers but they have a very good understanding of the subject. Second, students who are weak at mathematics and they have a strong belief that MATH is not their cup of tea. They had developed a Mathematics phobia within them. Some of them even had made serious efforts in their starting years of school education to get rid of this kind of phobia; they were unable to understand the subject. For these students reasons can be many but never a curse of GOD. (As some people suggest) They can be treated for their weakness at Mathematics by infusing confidence and making arithmetic a fun.
To be brilliant at Mathematics, the first requirement is confidence to learn Mathematics. In most of the cases, loss of self-confidence to be able to learn mathematics is the main reason. In higher classes, especially after 8th standard, it is too hard to change a dull to brilliant in Mathematics. Still, I believe, IT IS NEVER TOO LATE and NOTHING IS IMPOSSIBLE, WE SHOULD BELIEVE IT POSSIBLE.
I am quoting here some of the Mathematics shortcuts that can make Mathematics a fun game but only for those who practice it, just reading can do no better. These also develop confidence for learning Mathematics.
One of the area of MATHEMATICS that generally troubles the students is fraction. I would like to mention two very simple things about FRACTIONS. For addition or subtraction of mixed fractions, first of all, mixed fractions has to be converted into improper fraction and then by finding the LCM of the denominators of the given fractions, the fractions are added or subtracted as the case may be. If there are two fractions to be added or subtracted, we NEED NOT TO change these into improper fraction. Just follow the below given system:
Here we add 2 and 11 to find 13, then add the fraction. By cross multiplying we find 3X1= 3 and 2X7=14, adding we get 3 + 14 = 17. Multiply the denominators i.e. 7X3=21, so the answer is
Had there been 345 instead of 2 or 678 instead of 11, our method would remain the same and we would just add the two whole numbers (345, 678) to find 1023, rest the answer of the fraction remains the same. In this way we can save up to 95% of our time in solving such fractional sums. In case of subtraction, the method remains the same. There is also easy method to add, subtract, add and subtract more than two fractional numbers. There is even multiply method of such fractional numbers without changing mixed fractions into improper fractions. These kinds of methods have proved very helpful in boosting the confidence in MATHEMATICS and understanding the basic concept of FRACTION.
Squaring of halves and quarters (Fractions) is as easy as anything. In case of halves, simply multiply the whole number with the next number (Number + 1) and put ¼ after the product and the answer is with you. Example:
8½ X 8½ = (8 X 9) ¼ = 72¼
Or 4½ 4½ = 20¼
In case of quarters, square the whole number and add half of the number, then put 1/16 after the product. Example:
8¼ X 8¼ = (8 X 8 +8/2) 1/16 = (64 + 4) 1/16 = 68⅟16
4¼ X 4¼ = 18⅟16
I would like to mention one incidence. I was teaching QMaths at a summer camp at Major Ajaib Singh Senior Secondary School, Jiwan Wala, Faridkot and I was asked by a class four student if I remember table for 61. I told him that I do and after few moments everybody sitting there would be able to write it without much problem. Here is the method:
Write table for six but from RIGHT TO LEFT (As Urdu is written)
61 6 1
122 12 2
183 18 3
244 24 4
305 30 5
366 36 6
427 42 7
488 48 8
549 54 9
610 61 0
We can write table for any number ending with ONE (1) in this way, it just helps the child to understand the basic concept of ARITHMETIC and also infuse confidence. Another Mohali student who find game in this method, boast of his table knowledge by telling his uncle that he can write table for 201. Notice the confidence he gained!
ਅੰਕ ਗਣਿਤ ਦੀ ਖੇਡ - 2
ਲੇਖਾਂ ਦੀ ਇਸ ਲੜੀ ਦੇ ਪਿਛਲੇ ਅੰਕ ਵਿੱਚ ਅਸੀਂ ਪੜ੍ਹਿਆ ਕਿ ਕਿਸ ਤਰਾਂ ਹਿਸਾਬ ਦੀਆਂ ਕੁਝ ਕ੍ਰਿਆਵਾਂ ਨੂੰ ਬਹੁਤ ਦਿਲਚਸਪ ਬਣਾਇਆ ਜਾ ਸਕਦਾ ਹੇ। ਕੋਈ ਵੀ ਹਿਸਾਬ ਵਿੱਚ ਕਮਜੋਰ ਨਹੀਂ ਹੁੰਦਾ ਸਮੱਸਿਆ ਕੇਵਲ ਇੰਨੀ ਹੁੰਦੀ ਹੈ ਕਿ ਇੱਕ ਵਾਰ ਕਿਸੇ ਸਵਾਲ ਵਿੱਚ ਪਿੱਛੇ `ਰਹਿਣ ਤੇ ਜਾਂ ਫਿਰ ਸਵਾਲ ਸਮਝ ਨਾਂ ਆਉਣ ਤੇ ਇੱਕ ਵਾਰ ਮਨ ਵਿੱਚ ਇਹ ਭਾਵਨਾਂ ਬੈਠ ਜਾਵੇ ਕਿ ਮੈਨੂੰ ਤਾਂ ਹਿਸਾਬ ਆ ਹੀ ਨਹੀਂ ਸਕਦਾ ਤਾਂ ਇਹ ਵਿਸ਼ਾ ਉਸ ਵਿਅਕਤੀ ਲਈ ਓਪਰਾ ਬਣ ਜਾਂਦਾ ਹੈ ਅਤੇ ਸਮੇਂ ਦੇ ਨਾਲ ਨਾਲ ਇਹ ਭਾਵਨਾਂ ਇੰਨੀ ਘਰ ਕਰ ਜਾਂਦੀ ਹੈ ਕਿ ਫਿਰ ਇਸ ਵਿਚੋਂ ਨਿੱਕਲ ਪਾਉਣਾ ਮੁਸ਼ਕਲ ਲੱਗਣ ਲੱਗ ਜਾਂਦਾ ਹੈ। ਕਿਊਮੈਥਸ (www.qmaths.com) ਇੱਕ ਅਜਿਹੀ ਕੋਸਿ਼ਸ਼ ਹੈ ਜਿਸ ਰਾਹੀਂ ਹਿਸਾਬ ਦੀਆਂ ਮੁੱਢਲੀਆਂ ਤੋਂ ਗੁੰਝਲਦਾਰ ਕ੍ਰਿਆਵਾਂ ਨੂੰ ਸਰਲ ਤੇ ਸਮਝਣਯੋਗ ਬਣਾਇਆ ਗਿਆ ਹੈ।
ਹੁਣ ਅਸੀਂ 5 ਨਾਲ ਗੁਣਾ ਜਾ 5 ਨਾਲ ਭਾਗ ਦੇਣ ਦੇ ਆਸਾਨ ਤਰੀਕੇ ਨੂੰ ਸਮਝਣ ਦੀ ਕੋਸਿ਼ਸ਼ ਕਰੀਏ।
5 ਨਾਲ ਕਿਸੇ ਅੰਕ ਨੂੰ ਗੁਣਾ ਕਰਨ ਲਈ ਸਭ ਤੋਂ ਵਧੀਆ ਤੇ ਸੌਖਾ ਤਰੀਕਾ ਹੈ ਕਿ ਅਸੀਂ ਉਸ ਨੰਬਰ ਨੂੰ ਅੱਧਾ ਕਰ ਦੇਈਏ ਜਾਂ ਇਸ ਤਰਾਂ ਸਮਝੋ ਕਿ 2 ਤੇ ਭਾਗ ਕਰੀਏ। ਫਿਰ ਉਸ ਨੰਬਰ ਦਾ ਅੰਤਿਮ ਅੰਕ ਵੇਖੀਏ, ਜੇਕਰ ਜਿਸਤ (Even) ਹੋਵੇ ਤਾਂ ਅੰਤ ਵਿੱਚ ‘0’ ਲਾਓ ਅਤੇ ਜੇਕਰ ਟਾਂਕ (Odd) ਹੈ ਤਾਂ ‘5’ ਲਾਓ। ਹੇਠਾਂ ਦਿੱਤੀਆਂ ਉਦਾਹਰਣਾਂ ਰਾਹੀਂ ਇਹ ਵਧੇਰੇ ਸਮਝਿਆ ਜਾ ਸਕਦਾ ਹੈ।
286402 X 5
= 1432010
( 2 ਦਾ ਅੱਧ 1, 8 ਦਾ ਅੱਧ 4, 6 ਦਾ ਅੱਧ 3, 4 ਦਾ ਅੱਧ 2, 0 ਦਾ ਅੱਧ 0, 2 ਦਾ ਅੱਧ 1 ਹੈ ਅਤੇ ਅੰਤਿਮ ਅੰਕ ‘2’ ਜਿਸਤ (Even) ਹੈ ਇਸ ਲਈ ਅਖੀਰ ਵਿੱਚ ‘0’ ਲਾਈ ਗਈ ਹੈ।)
4163287 X 5
= 20816435
ਆਸਾਨ ਬਣਾਉਣ ਲਈ ਇਸ ਨੂੰ 4,16,32,8,7 ਸਮਝੋ।
( 4 ਦਾ ਅੱਧ 2, 16 ਦਾ ਅੱਧ 08, 32 ਦਾ ਅੱਧ 16, 8 ਦਾ ਅੱਧ 4, 7 ਦਾ ਅੱਧ 3 ਇੱਕ ਬਾਕੀ ਅਤੇ ਅੰਤਿਮ ਅੰਕ ‘7’ ਟਾਂਕ (Odd) ਹੈ ਇਸ ਲਈ ਅਖੀਰ ਵਿੱਚ ‘5’ ਲਾਇਆ ਗਿਆ ਹੈ।)
5460123 X 5
= 27300615
ਆਸਾਨ ਬਣਾਉਣ ਲਈ ਇਸ ਨੂੰ 54,6,0,12,3 ਸਮਝੋ।
( 54 ਦਾ ਅੱਧ 27, 6 ਦਾ ਅੱਧ 3, 0 ਦਾ ਅੱਧ 0, 12 ਦਾ ਅੱਧ 06, 3 ਦਾ ਅੱਧ 1 ਇੱਕ ਬਾਕੀ ਹੈ ਅਤੇ ਅੰਤਿਮ ਅੰਕ ‘3’ ਟਾਂਕ (Odd) ਹੈ ਇਸ ਲਈ ਅਖੀਰ ਵਿੱਚ ‘5’ ਲਾਇਆ ਗਿਆ ਹੈ।)
642,082,246,288,640,866,244,826,884,260,024,846,282,864 X 5
ਕੀ ਤੁਹਾਨੂੰ ਇਹ ਗੁਣਾ ਅਜੇ ਵੀ ਮੁਸ਼ਕਿਲ ਲੱਗਦੀ ਹੈ?
ਆਓ ‘5’ ਨਾਲ ਭਾਗ ਕਰਨ ਦਾ ਆਸਾਨ ਤਰੀਕੇ ਨੂੰ ਸਮਝਣ ਦੀ ਕੋਸਿ਼ਸ਼ ਕਰੀਏ।
5 ਨਾਲ ਕਿਸੇ ਅੰਕ ਨੂੰ ਭਾਗ ਦੇਣ ਲਈ ਸਭ ਤੋਂ ਵਧੀਆ ਤੇ ਸੌਖਾ ਤਰੀਕਾ ਹੈ ਕਿ ਅਸੀਂ ਉਸ ਨੰਬਰ ਨੂੰ ਦੁੱਗਣਾ ਕਰ ਦੇਈਏ ਜਾਂ ਇਸ ਤਰਾਂ ਸਮਝੋ ਕਿ 2 ਤੇ ਗੁਣਾ ਕਰੀਏ। ਫਿਰ ਉਸ ਉੱਤਰ ਦੇ ਅੰਤਿਮ ਅੰਕ ਨੂੰ ਛੱਡ ਕੇ ਦਸ਼ਮਲਵ (Decimal Point) ਲਾ ਦਿਓ। ਹੇਠਾਂ ਦਿੱਤੀਆਂ ਉਦਾਹਰਣਾਂ ਰਾਹੀਂ ਇਹ ਵਧੇਰੇ ਸਮਝਿਆ ਜਾ ਸਕਦਾ ਹੈ।
142032 ‹ 5 =
28406.4
( 1 ਦਾ ਦੁੱਗਣਾ 2, 4 ਦਾ ਦੁੱਗਣਾ 8, 2 ਦਾ ਦੁੱਗਣਾ 4, 0 ਦਾ ਦੁੱਗਣਾ 0, 3 ਦਾ ਦੁੱਗਣਾ 6 ਅਤੇ 2 ਦਾ ਦੁੱਗਣਾ 4, ਅਖੀਰਲਾ ਅੰਕ ‘4’ ਛੱਡ ਕੇ ਡੈਸੀਮਲ, Decimal Point ਲਾਓ)
174232625 ‹ 5 =
34846525.0 ਜਾਂ 34846525
ਆਸਾਨ ਬਣਾਉਣ ਲਈ ਇਸ ਨੂੰ 17,4,2,3,26,25 ਸਮਝੋ।
( 17 ਦਾ ਦੁੱਗਣਾ 34, 4 ਦਾ ਦੁੱਗਣਾ 8, 2 ਦਾ ਦੁੱਗਣਾ 4, 3 ਦਾ ਦੁੱਗਣਾ 6, 26 ਦਾ ਦੁੱਗਣਾ 52 ਅਤੇ 25 ਦਾ ਦੁੱਗਣਾ 50, ਅਖੀਰਲਾ ਅੰਕ ‘0’ ਛੱਡ ਕੇ ਡੈਸੀਮਲ, Decimal Point ਲਾਓ)
1262741835 ‹ 5 =
252548367.0 ਜਾਂ 252548367
ਆਸਾਨ ਬਣਾਉਣ ਲਈ ਇਸ ਨੂੰ 1,26,27,4,18,35 ਸਮਝੋ।
( 1 ਦਾ ਦੁੱਗਣਾ 2, 26 ਦਾ ਦੁੱਗਣਾ 52, 27 ਦਾ ਦੁੱਗਣਾ 54, 4 ਦਾ ਦੁੱਗਣਾ 8, 18 ਦਾ ਦੁੱਗਣਾ 36 ਅਤੇ 35 ਦਾ ਦੁੱਗਣਾ 70, ਅਖੀਰਲਾ ਅੰਕ ‘0’ ਛੱਡ ਕੇ ਡੈਸੀਮਲ, Decimal Point ਲਾਓ)
4230,1134,2314,0021,2113,2442,1423,4012,2331,2132,1230,4112,4001 ‹ 5
ਕੀ ਤੁਹਾਨੂੰ ਇਹ ਭਾਗ ਅਜੇ ਵੀ ਮੁਸ਼ਕਿਲ ਲੱਗਦੀ ਹੈ?
ਇਸੇ ਤਰਾਂ ਗੁਣਾ ਦਾ ਇੱਕ ਹੋਰ ਤਰੀਕਾ ਸਮਝਣ ਦੀ ਕੋਸਿ਼ਸ਼ ਕਰਦੇ ਹਾਂ। ਕਿਸੇ ਸੰਖਿਆ ਨੂੰ 11 ਨਾਲ ਗੁਣਾ ਕਰਨਾ ਹੋਵੇ ਤਾਂ ਹੇਠ ਲਿਖਿਆ ਤਰੀਕਾ ਵਰਤਿਆ ਜਾ ਸਕਦਾ ਹੈ।
35 X 11 = 3 (3+5) 5 = 385
52 X 11 = 5 (5+2) 2 = 572
ਇਸ ਤੋਂ ਅੱਗੇ ਹੋਰ:
234536 X 11
= 2(2+3)(3+4)(4+5)(5+3)(3+6)6
= 2579896
417235180345127 X 11
= 4(4+1)(1+7)(7+2)(2+3)(3+5)(5+1)(1+8)(8+0)(0+3)(3+4)(4+5)(5+1)(1+2)(2+7)7
= 45895869837963971
ਲਓ ਫਿਰ ਇੱਕ ਕੋਸਿ਼ਸ਼ ਤੁਸੀਂ ਵੀ ਕਰੋ:
235416317 X 11
ਇਸੇ ਤਰਾਂ ਕਿਸੇ ਰਕਮ ਨੂੰ 25 ਜਾ 125 ਨਕਲ ਗੁਣਾ ਕਰਨਾਂ ਵੀ ਕਾਫੀ ਆਸਾਨ ਹੈ ਜੇਕਰ ਅਸੀਂ ਅੰਕਗਣਿਤ ਦੇ ਮੁੱਢਲੇ ਨਿਯਮਾਂ ਦੀ ਵਰਤੋਂ ਕਰਦੇ ਹੋਏ ਸਰਲ ਤਰੀਕਿਆਂ ਦੀ ਵਰਤੋਂ ਕਰੀਏ।
ਕੀ ਤੁਹਾਨੂੰ 99999995 ਅਤੇ 99999992 ਦੀ ਗੁਣਾ 10 ਸੈਕੰਡ ਵਿੱਚ ਕਰਨ ਵਿੱਚ ਦਿਲਚਸਪੀ ਹੈ ਤਾਂ ਅਗਲੇ ਲੇਖਾ ਦਾ ਇੰਤਜ਼ਾਰ ਕਰੋ। ਹੋ ਸਕਦਾ ਹੈ ਤੁਸੀਂ 10 ਸੈਕੰਡ ਤੋਂ ਵੀ ਪਹਿਲਾਂ ਹੱਲ ਕਰਨ ਵਿੱਚ ਕਾਮਯਾਬ ਹੋ ਜਾਵੋ।
ਅਗਲੇ ਲੇਖਾਂ ਵਿੱਚ ਅਸੀਂ 2,3,4,5,6,8,9,10,11,12,15 ਆਦਿ ਤੋਂ ਇਲਾਵਾ 7,13,14,17,23,29,31,37,41 ਆਦਿ ਵਰਗੇ ਪ੍ਰਾਈਮ ਨੰਬਰਾਂ ਦੀ ਵੰਡਣਯੋਗਤਾ ਲਈ ਸਰਲ ਤੇ ਆਸਾਨ ਤਰੀਕਿਆਂ ਦੀ ਚਰਚਾ ਕਰਾਂਗੇ।
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