1990 AJHSME Problems/Problem 4

Problem

Which of the following could not be the unit's digit [one's digit] of the square of a whole number?

$\text{(A)}\ 1 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution

We see that $1^2=1$ , $2^2=4$ , $5^2=25$ , and $4^2=16$ , so already we know that either $\text{E}$ is the answer or the problem has some issues.

For integers, only the units digit affects the units digit of the final result, so we only need to test the squares of the integers from $0$ through $9$ inclusive. Testing shows that $8$ is unachievable, so the answer is $\boxed{\text{E}}$ .

1990 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1990 AJHSME Problems/Problem 4

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