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#29860
Satya
Participant

Challenge Question 6: Multiplication by 111

I think it works this way –

a b c * 111 =

a (a+b) (a+b+c) (b+c) c.

So 234 * 111 = 2 (2+3) (2+3+4) (3+4) 4 = 25974

Similarly 4 digit number * 111 –>

a b c d * 111 =  a (a+b) (a+b+c) (b+c+d) (c+d) d

So 1234 * 111 = 1 (1+2) (1+2+3) (2+3+4) (3+4) 4 = 136974

2345 * 111 = 260295 (there are carry on’s).

2 digit number * 111 –>

a b * 111 = a (a+b) (a+b) b

applying this to 51 * 111 = 5 6 6 1

72 * 111 = 7 9 9 2

99 * 111 = 9 (18) (18) 9 = 10 9 8 9

Please correct if my observation has any limitations. Thank you!

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