BTGmoderatorDC wrote:If n is a positive integer, is n the square of an integer?
(1) |n - 5m| = 2 for some integer m.
(2) |n - 7p| = 2 for some integer p.
OA A
Source: Manhattan Prep
Given: n is a positive integer.
We have to determine whether n the square of an integer or a perfect square number.
Let's take each statement one by one.
(1) |n - 5m| = 2 for some integer m.
=> n - 5m = 2 or n - 5m = -2
Taking n - 5m = 2, we get n = 2 + 5m, where m is any integer such that n is a positive integer.
We see that for m = 0, we have n = 2, a non-perfect square number
Determining that at some value of m, n becomes a perfect square number is difficult.
Let's analyze n = 2 + 5m.
Note that for any positive integer value of m, the units digit of m is either 0 or 5. This way, the units digit of n = 2 + 5m is either 2 or 7.
None of the perfect squares has either 2 or 7 as its units digit, thus, n is a non-perfect square number.
Let's now analyze n = -2 + 5m if it is a perfect square.
Again, note that for any positive integer value of m, the units digit of m is either 0 or 5. This way, the units digit of n = -2 + 5m is either 8 or 3.
None of the perfect squares has either 3 or 8 as its units digit, thus, n is a non-perfect square number. The answer is No. Unique answer. Sufficient.
(2) |n - 7p| = 2 for some integer p.
=> n - 7p = 2 or n - 7p = -2
Taking n - 7p = 2, we get n = 2 + 7p, where p is any integer such that n is a positive integer.
We see that for p = 0, we have n = 2, a non-perfect square number
Unlike in Statement 1, determining that at some value of p, n becomes a perfect square number is not difficult.
At p = 2, we have n = 2 + 7p => n = 9, a perfect square. No unique answer. Insufficient.
The correct answer:
A
Hope this helps!
-Jay
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