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The following question was asked in a competitive exam for which I am preparing and I was unable to solve it (in fact I am completely clueless about it).

So, I am asking for help here.

Given a number with 1998 digits which is divisible by 9. Let x be sum of its digits and y be sum of digits of x and z be sum of digits of y. Find value of z.

a 9

b 1998

c 27

d none of these

I am completely clueless on how to approach this and would be really thankful for any help received.

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  • $\begingroup$ You can place bounds on the number of digits in each number. $\endgroup$
    – Bass
    Nov 13 '20 at 18:41
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To calculate the answer, all we need are some bounds on the order of magnitude of $x$ and $y$. (In particular, we won't be looking at the answer options, nor will we need to resort to any kind of reasoning "from the assumed uniqueness of the answer", which is questionable at best.)

Here's all it takes:

The starting number has 1998 digits. Therefore $x$ is at most

$9\times1998$, which is some five-digit number.

Therefore, $y$ is at most

$9\times5 = 45$, which is a 2-digit number.

Therefore, $z$ is at most

$9\times2$, which is 18.

Finally, $y$ cannot actually be 99 (which is larger than 45), so $z$ cannot be 18. The only possible value for $z$ is then

9, because the starting number was divisible by 9, and the digit sum of any number divisible by 9 is always divisible by 9

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  • $\begingroup$ Well... it's obvious that it doesn't work and it isn't one of the options, but (rot13) mreb vf nyfb qvivfvoyr ol avar ;) $\endgroup$
    – boboquack
    Nov 15 '20 at 1:42
  • $\begingroup$ @boboquack yeah, thought about adding that, but then I'd have needed to say something useless about negative numbers too, so I decided to leave both out :-) $\endgroup$
    – Bass
    Nov 15 '20 at 6:22
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My answer is

9 because

If there are 1998 digits their sum must be between 1 and (1998 x 9) = 17982
So the sum of the digits of that number is between 1 and (4 x 9) = 36
So the sum of the digits in that number is between 1 and (2 + 9) = 11
This rules out the answers 1998 and 27, which can never be true.

Now, if the number consists of 1998 ones, that is divisible by 9.
The digit sum is 1998.
The sum of these digits is 27.
Finally the sum of those digits is 9.

So 'none of these' cannot be correct as there is at least one example.
Leaving 9 as the answer.

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    $\begingroup$ Or it could be none of these, which is an option for the answer... $\endgroup$
    – Amorydai
    Nov 13 '20 at 18:49
  • $\begingroup$ @Amorydai ah I didn't notice that. Edited. $\endgroup$ Nov 13 '20 at 18:50
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    $\begingroup$ $x$ could be $9999$, then $y$ would be $36$, though this does change the fact that $z=\#$. Otherwise, the logic is correct. $\endgroup$ Nov 13 '20 at 20:44
  • $\begingroup$ @DanielMathias good spot, thanks. I presume you mean doesn't change z. $\endgroup$ Nov 13 '20 at 20:45
  • $\begingroup$ The sum of digits operation maintains the remainder on division by $9$. This is the justification for the technique of casting out $9$s. As we are given the original number is divisible by $9$, the final sum will be, too. That guarantees the result is $9$. $\endgroup$ Nov 15 '20 at 1:13
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Well, given that

a number is divisible by 9 if and only if the sum of it digits is 9, we can work backwards, i.e. assume $z = 9$, then assume $y = 9$, then assume $x = 9$,

and we only need to find

a number with 1998 digits whose sum of digits is 9, e.g. a number starting with nine 1s and the rest 0s, i.e. $111111111 \cdot 10^{1989}$.

The format of question seems to imply the other answers must be wrong.

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  • $\begingroup$ I'm not following your answer. In the first step, you assume z = 9. Can I assume z = 27 (the other choice) and then continue with your logic? (assume z = 27, then assume y = 9, then assume x = 9). Am I misunderstanding something? $\endgroup$
    – fejyesynb
    Nov 13 '20 at 19:26
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    $\begingroup$ No, I'm just proving answer a) is right, not that the others are wrong. If $z = 27$, $y$ can't be 9 (but it could be 999). $\endgroup$
    – Glorfindel
    Nov 13 '20 at 19:27
  • $\begingroup$ Sorry, my mistake, we're working backwards so it can't be larger than the one before $\endgroup$
    – fejyesynb
    Nov 13 '20 at 19:28
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Consider the number: 9 followed by 1997 zeros ($9*10^{1997}$). This is trivially divisible by 9 (result is 1 followed by 1997 zeros or ($1*10^{1997}$).

The sum of digits is 9 ($9 + 0 + 0 + ... + 0$). Therefore, a is a possible answer.

Does this exclude the possiblity of other choices being correct?


Because the setting is a competitive exam, we can assume there is only one correct choice (or if they made a mistake, this answer to this question wouldn't matter). Either way, speed and correctness is important in this setting, but not for coming up with a proof the other choices are wrong.

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    $\begingroup$ Why am I being downvoted? Because the setting is a competitive exam, we can assume there is only one correct choice (or if they made a mistake, this answer to this question wouldn't matter). Either way, speed and correctness is important in this setting, but not for coming up with a proof the other choices are wrong. $\endgroup$
    – fejyesynb
    Nov 13 '20 at 18:50
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    $\begingroup$ Reasoning from the assumption of a unique solution is a mortal sin. $\endgroup$
    – Bass
    Nov 13 '20 at 19:21
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    $\begingroup$ I agree with you, fejyesynb, and suggest that: 1) copy the explanation about expediency from your comment into the post -- it is entirely pertinent for the reason you say, that this is the quickest route to answering the test question; 2) don't mention downvoting in your edit but do leave your comment as is; 3) if your post is edited downvoters will be allowed to change their votes if they, hopefully, happen to notice. @Bass's point is valid in terms of solving the test question as a puzzle here but i think your answer is valid in terms of the question in context. $\endgroup$
    – humn
    Nov 13 '20 at 19:26

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