Solving for $X$: Find $X$ When $X=Y^2$

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The problem involves finding a 6-digit number $X$ represented as $\overline{abcdef}$, where $Y$, a 3-digit number represented as $\overline{def}$, satisfies the equation $X = Y^2$. The solution requires identifying values of $Y$ such that $Y^2$ results in a 6-digit number, specifically ensuring that the last three digits of $X$ match the digits of $Y$. The valid range for $Y$ is from 317 to 999, as these values yield 6-digit squares.

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Albert1
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$X=\overline{abcdef}$ is a 6-digit number,

$Y=\overline{def}$ is a 3-digit number,

if $X=Y^2$ find $X$
 
Last edited:
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Albert said:
$X=\overline{abcdef}$ is a 6-digit number,

$Y=\overline{def}$ is a 3-digit number,

if $X=Y^2$ find $X$

let $X = 1000m + Y$
we have $X = Y^2$ or $Y^2-Y= 1000m$ or $Y(Y-1) = 1000m = (125a)(8b)$ where $m = ab$
now $Y,Y-1$ are co-primes so one of them has to be 125a and another 8b and a and b are coprimes or one of them is 1
a has to be odd ( else 125 cannot be coprime to 8b) and < 8 as $Y < 1000$
so we get
$Y=125a, Y-1 = 8b$ trying a = 1,3,5,7 we get $Y= 625, Y-1 = 624 = 8 * 78$
or
$Y=8b, Y-1 = 125a$ we get $Y = 376$
so there are 2 values of X $ 376^2 = 141376$ and $625^2=390625$
 

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