7
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A math teacher thought of a positive integer of two digits. She wants her two intelligent students Hanna and Charlie determine the exact number thought. For this, Hanna is privately told how many positive divisors the number has; Charlie, the sum of the digits of the number.

A brief conversation between Hanna and Charlie is transcribed below:

  • Charlie: I can not determine the number;
  • Hanna: Neither can I, but I can tell if it is even or odd;
  • Charlie: Now I know what the number is;
  • Hanna: You know? Then I also know.

Suppose that students are honest and there is perfect logic in all the conclusions drawn. Determine the number thought by the teacher, justifying your answer.

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  • $\begingroup$ I have rephrased your answer for clarity, but I want to make sure that I've done so correctly. Was Hanna told how many positive divisors the number has (as I put), or was she told what positive divisors the number has? Also, check to make sure that I didn't change anything else in my rephrasing. $\endgroup$ – Cameron Buie Dec 11 '12 at 21:21
  • $\begingroup$ @CameronBuie You are correctly, thank's. $\endgroup$ – Henfe Dec 11 '12 at 21:24
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The number is $80$. Charlie was told the sum of digits is $8$. Hanna was told there are $10$ divisors: the possible numbers with $10$ divisors are $48$ and $80$, and both are even, hence Hanna's first statement.

Charlie's candidates for the number are $8, 17, 26, 35, 44, 53, 62, 71, 80$. $8$ has $4$ divisors, but so does $15$; $17$ has $2$ divisors, but so does $2$; and so on for all the other candidates except $80$, which has $10$ divisors, and the only other integer in $\{1,\ldots,99\}$ with $10$ divisors is $48$ which is also even. Hence Charlie's second statement.

Hanna now knows the number can't be $48$, because if Charlie had been told the sum of digits was $12$ he couldn't decide whether the number was $48$ or $66$: $66$ has $8$ divisors, and all possible numbers with $8$ divisors ($24, 30, 40, 42, 54, 56, 66, 70, 78, 88$) are even. Hence Hanna's second statement.

EDIT: This was for a version of the problem where the possible numbers were $1$ to $99$. It is not correct for the version where the possible numbers are $10$ to $99$. I'll post a solution for that version if time permits.

EDIT: Where the possible numbers are $10$ to $99$, the number is not $80$, because after Hanna's first statement Charlie would not know that the number is not $17$ (all the possible numbers with $2$ divisors are odd).

Instead, the number in this case is $30$. Charlie was told the sum of digits is $3$. Hanna was told there are $8$ divisors; all possible numbers with $8$ divisors are even, hence Hanna's first statement.

Now Charlie's candidates for the number are $12, 21, 30$. $12$ has $6$ divisors, as does $45$; $21$ has $4$ divisors, as does $15$; so Charlie knows it must be $30$. Hence Charlie's second statement. The only sums of digits that would have allowed Charlie to make this second statement (without knowing the answer at the beginning statement) are $2$ (with answer $11$), $3$, $14$ (with answer $59$), and $17$ (with answer $89$).

Now Hanna knows the answer is $30$, which is the only number in the intersection of $\{24, 30, 40, 42, 54, 56, 66, 70, 78, 88\}$ and $\{11,30,59,89\}$. Hence Hanna's second statement.

How did I come up with the answer? A process of elimination. After Charlie's first statement the possible numbers are $11,12,\ldots,98$. After Hanna's first the possibilities are restricted to those with $2$, $3$, $8$, $10$, or $12$ divisors ($40$ possibilities). After Charlie's second statement the possibilities are restricted to $11, 30, 59, 89$. And of those all but $30$ have $2$ divisors.

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  • $\begingroup$ How do you know that Charlie was told the sum of digits is $8$, and that Hanna was told that there are $10$ divisors? $\endgroup$ – chubbycantorset Dec 12 '12 at 2:51
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Take a look at this: http://www.qbyte.org/puzzles/p003s.html , this is a pretty popular puzzle/question in logic.

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The number is $2^6$.

Hanna knows that it has 2 a divider-- a number less than 100 with 6 dividers (7 if you count the number itself*) has to have 2 one of them. And Charlie, figuring how she concluded that, sees 64 is the only one with 6 (96 also satisfies this part) divisors and digits adding up to the number given to him.

  • Edited to add the part in first parantheses.
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  • $\begingroup$ $2^6$ has seven divisors (including $64$) and is the only number with seven divisors, so it cannot be right. $\endgroup$ – Ross Millikan Dec 11 '12 at 21:31
  • $\begingroup$ Well, work it for 7 then. this is a solution. Thanks for the warning though. $\endgroup$ – ashley Dec 11 '12 at 21:54
  • $\begingroup$ But for 7, there is only 64, so Hanna would know it at her first remark. Certainly the first remark implies that 2 is a factor. $\endgroup$ – Ross Millikan Dec 11 '12 at 22:42
  • $\begingroup$ $96=2^5\times3$. Hanna can't chose between $96$ and $64$ by herself. $\endgroup$ – ashley Dec 11 '12 at 22:48
  • $\begingroup$ Still $6$ divisors BTW, not $7$. Those multiplied giving the number itself. Fell quick for your first comment. $\endgroup$ – ashley Dec 11 '12 at 22:52
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In the 10..99 version, the number is 30. Charlie doesn't know the number in the first round since 21 (for example) has the same digit sum. Hanna knows that it's even, though, since the first odd number with 8 divisors is 3*5*7 > 99. Charlie now knows that the number is 30, since only {11, 30, 59, 89} are possible for Hanna to know even/odd status on by just their number of divisors (with or without the restriction to 11..98 based on his first response). Then of these Hanna can pick out the one with 8 rather than 2 divisors.

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