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gmatt1476
If n = p^2 and p is a prime number greater than 5, what is the units digit of n^2 ?

A. 1
B. 3
C. 4
D. 7
E. 9

PS37402.01

If p = 7, then n = 49, and the units digit of 49^2 is 1.

If p = 11, then n = 121, and the units digit of 121^2 is 1.

Alternate solution:

Since n = p^2, then n^2 = p^4. Since p is a prime number greater than 5, the units digit of p could be 1, 3, 7 or 9, but the units digit of p^4 will always be 1 (since 1^4 = 1, 3^4 = 81, 7^4 = 2401 and 9^4 = 6561).

Answer: A
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gmatt1476
If n = p² and p is a prime number greater than 5, what is the units digit of n² ?

A. 1
B. 3
C. 4
D. 7
E. 9

Let's choose a value of p that satisfies the condition that p is a prime number greater than 5.
So, it COULD be the case that p = 7

So, n = p² = 7² = 49

If n = 49, then n² = 49² = ----1

Answer: A

Cheers,
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gmatt1476
If n = p^2 and p is a prime number greater than 5, what is the units digit of n^2 ?

A. 1
B. 3
C. 4
D. 7
E. 9

PS37402.01

I will use the method of testing numbers to arrive at the correct answer:

1. Let's assume \(p = 7\) (We are given that \(p\) is a prime greater than \(5\))
\(p^2 = 7^2 = 49 = n\)
\(n^2 = 49 * 49\) has units digit \(1\) (\(9 * 9 = 8\)1)

2. Let's assume \(p = 11\)
\(p^2 = 11^2 = 121 = n\)
\(n^2 = 121 * 121\) has units digit \(1\) (\(1 * 1 = \)1)

3. Let's assume \(p = 13\)
\(p^2 = 13^2 = 169 = n\)
\(n^2 = 169 * 169\) has units digit \(1\) (\(9 * 9 = 8\)1)

We can safely assume that \(n^2\) will have units digit of 1

Ans. A
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Another solution (if you keep looking for a method as I do, instead of relying on plugging numbers)

P^2 has 3 factors.

[Rule] So, P^2 or n is a perfect square.

[Rule] Since all perfect squares end with the unit digit as 1, 4, 5, 6, 9, 0

Hence, n^2 will result in units digit as 1, 6, 5, 6, 1, 0

Answer = A


(Please correct me if there is any issue with my process.)
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