ABCDEF - a six-digit number. All digits are different and are arranged from left to right in ascending order. The number is a perfect square.

Determine what this number is.

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The number ABCDEF is the square of a three-digit number.

## Solution

Let $\mathrm{ABCDEF}=\mathrm{XYZ}^{2}$. Clearly, $\mathrm{F}$ is the units digit of $\mathrm{Z}^{2}$. Since the digits A, B, C, D, E, F are in ascending order, $\mathrm{A} \leq 4$,

$\mathrm{E} \geq 5$, and $\mathrm{F} \geq 6$.

If $A=4$, then $\mathrm{ABCDEF}=456789$, which is not a square.

A square cannot end in 7 or 8. If $\mathrm{F}=6$, the number ABCDEF can only be 123456, but this number is not a square. Therefore, $\mathrm{F}=9$, and $\mathrm{Z}$ is either 3 or 7.

The squares of 93 and 97 end in 49 and 09, respectively, so Y cannot be 9.

Since $\mathrm{A} \leq 3$, $\mathrm{X}^{2}2$, because $2 \mathrm{YZ}^{2}$ is a five-digit number.

1) $\mathrm{X}=6$. Then 360000 < ABCDEF < 400000, and the digits cannot be in ascending order.
2) $\mathrm{X}=5$. $5 \mathrm{YZ}^{2} \geq 250000$, so in this case, $\mathrm{ABCDEF}$ is either 256789 (but this is not a square) or not less than 345678, which means $\mathrm{Y} \geq 8$. But the numbers $583^{2}$ and $587^{2}$ do not fit.
3) $\mathrm{X}=4,160000 \leq 4 \mathrm{YZ}^{2}<250000$. Therefore, A=2 (otherwise, the digits cannot be arranged in ascending order), $B \geq 3$, so $\mathrm{Y} \geq 8$. But the numbers $483^{2}$ and $487^{2}$ do not fit.
4) $\mathrm{X}=3$. Then $\mathrm{A}=1, \mathrm{Y} \geq 5$. By trial, we find the only solution $367^{2}=134689$.

## Answer

134689.