2005 Indonesia MO Problems/Problem 2

Problem

For an arbitrary positive integer $n$ , define $p(n)$ as the product of the digits of $n$ (in decimal). Find all positive integers $n$ such that $11 p(n)=n^2-2005$ .

Solution

First we will prove that $n$ must have 2 digits. Afterward, we'll let $n = 10a + b$ and will solve for $n$ .

Lemma: $n$ must have 2 digits
First, if $n < 10$ , then $n^2 - 2005$ is a negative number, so $n$ must have more than 1 digit.

Next, we will show that $n$ can not have 3 digits. Since 2005 is not a perfect square, $n$ can not have any zeroes. That means the minimum value of $n$ that has $k$ digits is $\frac{10^k - 1}{9}$ . The maximum value of $p(n)$ with three digits is $9^k$ . For the case where $k = 3$ , the minimum number is $111$ and the maximum value of $p(n)$ is $729$ . Since $111^2 - 2005 > 11 \cdot 729$ , all three-digit values of $n$ more than 111 have $11 p(n) < n^2 - 2005$ , so $n$ can not be a 3-digit number.

Finally, we will use induction to show that if $n$ can not have four or more digits. If $n$ has $k$ digits, where $k \ge 4$ , then the maximum value of $p(n)$ is $9^k$ while the minimum value of $n$ is $10^{k-1}$ . For the base case, since $11 \cdot 9^4 < (10^3)^2 - 2005$ , we would have $11 p(n) < n^2 - 2005$ for all four digit numbers, so no four digit number would work.

Assume that $11 \cdot 9^k < (10^{k-1})^2 - 2005$ , where $k \ge 4$ . Multiplying both sides by 100 and adding 2005(99) results in $\begin{align*} 110 \cdot 9^k + 2005(99) &< (10^k)^2 - 2005 \\ 11(10 \cdot 9^k + 2005(9)) &< (10^k)^2 - 2005 \\ 11(9 \cdot 9^k + 9^k + 2005(9)) &< (10^k)^2 - 2005 \\ 11 \cdot 9^{k+1} + 11(9^k + 2005(9)) &< (10^k)^2 - 2005 \\ 11 \cdot 9^{k+1} &< (10^k)^2 - 2005. \end{align*}$ With the induction step complete, we've shown that $11 \cdot 9^k < (10^{k-1})^2 - 2005$ if $k \ge 4$ . Since $11 p(n) \le 11 \cdot 9^k$ and $(10^{k-1})^2 - 2005 \le n^2 - 2005$ , we have $11 p(n) < n^2 - 2005$ , so $n$ can not be number with four or more digits. Thus, $n$ must be a two-digit number. $\blacktriangleright$

From the Lemma, let $n = 10a + b$ . Substitution results in $\begin{align*} 11ab &= (10a+b)^2 - 2005 \\ 11ab &= 100a^2 + 20ab + b^2 - 2005 \\ 2005 &= 100a^2 + 9ab + b^2. \end{align*}$ Taking the equation modulo 4, we have $ab + b^2 \equiv 1 \pmod{4}$ . That means $a \equiv 0 \pmod{4}$ . Let $a = 4a_1$ , resulting in $2005 = 1600a_1^2 + 36a_1b + b^2.$ If $a_1 > 1$ , then $1600a_1^2 > 2005$ , so $a_1 = 1$ and $a = 4$ . Substitution results in $\begin{align*} 2005 &= 1600 + 36b + b^2 \\ 0 &= b^2 + 36n - 405 \\ &= (b-9)(b+45) \end{align*}$ The only feasible value of $b$ is $9$ , so the only positive integer such that $11 p(n) = n^2 - 2005$ is $\boxed{49}$ . When plugging the value in, the equality holds.

2005 Indonesia MO (Problems)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8	Followed by Problem 3
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2005 Indonesia MO Problems/Problem 2

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