Find all three-digit numbers whose squares end in 1001.

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The first method. Let $\overline{a b c}=100 a+10 b+c$ be the desired number. Then $\overline{a b c^{2}}=100000 a^{2}+2000 a b+100 b^{2}+$ $200 a c^{2}+20 b c^{2}+c^{2}$.

The last digit of this number depends only on $c^{2}$. Since $c^{2}$ ends in 1, then $c=1$ or 9. The second-to-last digit depends on the expression

$20 b c+c^{2}=20 b+1$ or $180 b+81$. By trying all possibilities, we find that the ending 01 can only be satisfied by three pairs of values $(b, c):(5,1),(0,1)$ or $(4,9)$. Thus, the square of the desired number has the form: $10000 a^{2}+10200 a+$ $2601,10000 a^{2}+200 a+1$ or $10000 a^{2}+9800 a+1201$.

By trying all possibilities, we find that in the first case no value of $a$ satisfies the condition, in the second case $a$ $=5$, and in the third case $-a=7$.

The second method. Let $n$ be the desired number. Since $n^{2}$ ends in 1001, then $n^{2}-1=(n-1)(n+1)$ ends in 1000, meaning it is divisible by 1000. Since exactly one of the numbers $n-1$ or $n+1$ can be divisible by 5, this number must also be divisible by 125. Both of these numbers are even, so the one that is divisible by 125 is also divisible by 250.

Since the desired number is a three-digit number, we need to check what the products end in in seven possible cases: $248 \cdot 250 ; 250 \cdot 252,498 \cdot 500,500 \cdot 502,748 \cdot 750,750 \cdot 752,998 \cdot 1000$. Only two of them end in 1000: 1) $500 \cdot 502=251000$, then $n=501$; 2) $748 \cdot 750=561000$, then $n=749$.

## Answer

$501,749$.

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