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I am trying to solve the following problem:

Digital sum is defined as the sum of the decimal digits of an integer. E.g. Digital sum of 385 = 3+8+5 = 16. Among 35 two-digit integers ($ \ge 10 $), show that there are 3 integers that share the same digital sum

I tried to find the smallest possible digit sum that any two-digit integer ($\ge 10$). This would clearly be $ 1 + 0 = 10 $. The greatest would be $ 9 + 9 = 18 $. Thus, there are 18 possible digital sums. By pigeonhole, $ \lceil\frac{35}{18}\rceil = 2 $. This means that there are at least 2 similar digital sums. I don't know what is incorrect with my method. Could anyone please advise me?

Donald
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2 Answers2

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There are $90$ two-digit numbers (ranging between $10$ and $99$).

Only $1$ of them has a digit-sum of $1$ ($10$).

Only $1$ of them has a digit-sum of $18$ ($99$).

For the other $88$ numbers, there are $16$ possible digit-sums (ranging between $2$ and $17$).

So the largest amount of numbers such that each digit-sum appears exactly $2$ times is $32$.

We can add $10$ and $99$, but that will only bring us to $34$ numbers.

Adding the $35$th number will result with a digit-sum which appears $3$ times.

barak manos
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Hint: For some of the digital sums, it is impossible to have more than one number with that sum. Taking into account that you can have at most one number with each of those sums, see how many numbers you must have split among the remaining sums.

Eric Wofsey
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  • Only $ 1 $ and $ 18 $ can be produced by one two-digit number. But how does that change the number of sums? I still have 18 possible sums? – Donald Nov 10 '16 at 07:48
  • How many of your numbers must have a sum different from $1$ and $18$? And how many possible sums are there for those numbers? – Eric Wofsey Nov 10 '16 at 07:54
  • Oh! I see what you mean now. I get 33 numbers and 16 different sums. That means there are $ \lceil\frac{33}{16}\rceil = 3 $ similar digital sums. – Donald Nov 10 '16 at 07:57