Digits and Perfect Squares

Question

Let $a, b, c$ be distinct digits. Consider a two-digit number '$ab$' and a three-digit number '$ccb$', both defined under the usual decimal number system. If $(ab)^2 = ccb > 300$, then the value of $b$ is

Accepted Answer

A. We need $(ab)^2 = ccb > 300$. The last digit of $(ab)^2$ is $b$, so $b^2$ must end in $b$. Possible values for $b$ are $0, 1, 5, 6$. Since $a, b, c$ are distinct digits, let's test candidates for $ab$ between $18$ and $31$ (since $18^2 = 324$ and $31^2 = 961$). Check $21^2 = 441$. Here $a=2, b=1, c=4$, which are all distinct, and $ccb = 441 > 300$. Thus $b = 1$.

Answer

1

Answer

0

Answer

5

Answer

6

Digits and Perfect Squares

Explanation

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