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Number Theory Level 3

The number (8).(888888........88) is the product of two factors as seen, where the second factor with a larger number of 8's has "K" digits all being 8. Find K if the sum of the digits of the product of the above two numbers is 1000.

Note:- This question is not an original.


The answer is 991.

Krishna Ar by Krishna Ar , 21, India

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1 solution

Antony Diaz
May 28, 2014

First, we can see that: 888 x 8 = 7104 8888 x 8 = 71104 88888 x 8 = 711104 888x8=7104\\ 8888x8=71104\\ 88888x8=711104\\

So, to induction I suppose that the product 8..888 8..888 with K chiffres is equal to 711 .. 104, with ( k − 2 ) − 1 ′ s (k-2) -1's . Then I have that 8 x 88...8 = 8 x ( 8 x 10 k + 88..8 ) = 6 , 4 x 10 k + 1 + 711..104 = 711...104 8x88...8=8x(8x{ 10 }^{ k }+88..8)\\ =6,4x{ 10 }^{ k+1 }+711..104\\ =711...104

with ( k − 2 ) − 1 ′ s (k-2) -1's .

So if the sum of digits is 1000, thats mean that 1000 = 7 + 4 + 1 ( k − 2 ) 989 = k − 2 991 = k 1000=7+4+1(k-2)\\ 989=k-2\\ 991=k

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Very good solution :D, I did it with two solutions, the first one very similar and the second with powers of ten; but I can share my second solution because I thought the sum of the digits was 100 , so I got it wrong :(

Isaac Jiménez - 6 years, 11 months ago

Format your L a T e X LaTeX better, its 888 \times 8 not 888x8 so it becomes 888 × 8 888\times8 .

Joshua Ong - 6 years, 7 months ago

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