7
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Find three positive integers with the following two properties:

  1. The sum of any two of them has digit sum less than 15
  2. The sum of all three integers has digit sum more than 200
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  • $\begingroup$ By sum do you mean multiplication? $\endgroup$ – Deepthinker101 Aug 19 at 14:40
  • $\begingroup$ Exclusively that is $\endgroup$ – Deepthinker101 Aug 19 at 14:41
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Here is a possible answer which I think suggests the general strategy

4554554554554554554554554554554554554555
5455455455455455455455455455455455455455
5545545545545545545545545545545545545545

The pairwise sums are

10010010010010010010010010010010010010010
10100100100100100100100100100100100100100
11001001001001001001001001001001001001000 (all digit sums are 14)

While the overall sum is

15555555555555555555555555555555555555555 (digit sum 201)

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  • $\begingroup$ Ah, you beat me to it. $\endgroup$ – Jaap Scherphuis Aug 19 at 15:03
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Take these three numbers:

$$a = 4444444444444\ 5555555555555\ 5555555555555\ 5\\b = 5555555555555\ 4444444444444\ 5555555555555\ 5\\c = 5555555555555\ 5555555555555\ 4444444444444\ 5$$ They all consist of three blocks of 13 digits, of which one is all fours and the rest all fives, followed by an extra five.

The pairwise sums are:

$$a+b = 1\ 0000000000000\ 0000000000000\ 1111111111111\ 0\\b+c = 1\ 1111111111111\ 0000000000000\ 0000000000000\ 0\\c+a = 1\ 0000000000000\ 1111111111111\ 0000000000000\ 0$$ The sum of any two of them gives a number with a leading $1$ and block of thirteen $1$s, for a digit sum of $14$.

The sum of all three is:

$$a+b+c = 1\ 5555555555555\ 5555555555555\ 5555555555555\ 5$$ which is a one followed by $40$ fives, for a digit sum of $201$.

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