Difference between revisions of "1990 AJHSME Problems/Problem 4"

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==Problem==
 
==Problem==
  
Which of the following could '''not''' be the units digit <nowiki>[one's digit]</nowiki> of the square of a whole number?
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Which of the following could '''not''' be the units digit <nowiki>[ones digit]</nowiki> of the square of a whole number?
  
 
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8</math>
 
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8</math>

Latest revision as of 15:57, 29 September 2017

Problem

Which of the following could not be the units digit [ones 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 $\boxed{\text{E}}$.

See Also

1990 AJHSME (ProblemsAnswer KeyResources)
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

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