When positive integer n is divided by 3 the remainder is 1. Is n even?

(1) n is divisible by 5
(2) n + 5 is divisible by 3

A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
D. Either statement BY ITSELF is sufficient to answer the question.
E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.

Statement (1) tells us that n is divisible by 5.
On one hand 25, an odd number, is divisible by 5 and when 25 is divided by 3 the remainder is 1.
On the other hand 10, an even number, is divisible by 5 and when 10 is divided by 3 the remainder is 1.
Therefore we can not know for sure if n is even or not. Statement (1) by itself is insufficient.

Statement (2) tells us that n + 5 is divisible by 3.
On one hand when 25, an odd number, is divided by 3 the remainder is 1 and 25 + 5 equals to 30. 30 is divisible by 3.
On the other hand when 10, an even number, is divided by 3 the remainder is 1 and 10 + 5 equals to 15. 15 is divisible by 3.
Therefore we can not know for sure if n is even or not. Statement (2) by itself is insufficient.

Statement (1) and (2) taken together are also insufficient, because both 25 (odd) and 10 (even) fit all the criteria.

The answer is (E).

If you couldn’t pick numbers showing that both statements are insufficient another way to solve this question is to represent each statement in terms of some other non-negative integer.

“When n is divided by 3 the remainder is 1.” means that n = 3k + 1, where k is some non-negative number. n + 5 = 3k + 1 + 5 = 3k + 6 = 3(k + 1) is divisible by 3. Therefore statement (2) doesn’t give us any new information. If k is odd then n is even. If k is even then n is odd. Since k can be both Statement (2) alone is insufficient.

Statement (1), “n is divisible by 5”, means that n = 5m, where m is a positive integer.
Combining both statements we get 5m = 3k + 1.
Clearly m is not divisible by 3. When m is divided by 3 the remainder can be 1 or 2
If the remainder is 1 then m = 3f + 1, where f is some non-negative integer. So 5(3f + 1) = 3k + 1. 15f + 5 = 3k +1. k = 5f + 4/3 is not an integer. Therefore remainder cannot be 1.

So the remainder is 2 then m = 3f + 2, where f is some non-negative integer. n = 5m = 5(3f + 2) = 15f + 10.
We see that n = 15f + 10 satisfies all the statements. If f is odd then n is odd, if f is even then n is even. Therefore the answer is (E).
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hi,here you said "Clearly m is not divisible by 3".how is that concluded? pl. explain.

This is a very hard problem indeed.

First, let me answer your question.

"5m = 3k + 1. Clearly m is not divisible by 3." Why is it so?

Variant 1: 3k + 1 is not divisible by 3, so 5m is not and therefore m is not.

Variant 2: Suppose m is divisible by 3. Then on the left side we would have a number 5m which is divisible by 3. On the right side – number (3k + 1) which is not. So they could not be equal. Therefore m can not be divisible by 3.
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It can be very useful in many number theory questions to interpret statements like "a number is divisible by 3" or "the remainder when dividing k by 5 is 2. " in the way of

n = kb + m.

In this way you can clearly see the constant integers.

For example, "n is divisible by 4" tells us that n = 4k, where k is some integer. Using such denotation we can easily see that:

n² = (4k)² = 16k² is divisible by 16.
n/2 = 4k/2 = 2k is divisible by 2.
5n = 5 × 4k = 20k is divisible by 20

Another example: "when n is divided by 6 the remainder is 3" tells us that n = 6k + 3, where k is some integer. Using such denotation we can easily see that:

n = 6k + 3 = 3(2k + 1) is divisible by 3.

n² – 9 = (6k + 3)² – 9 = 9(2k + 1)² – 9 = 9(4k² + 4k + 1 – 1) = 9(4k² + 4k) = 9 × 4 × k × (k + 1) is divisible by 72 = 9 × 4 × 2, because k × (k + 1) is divisible by 2 as the product of two consecutive numbers (one of two must be even).

There are many more examples of how using this simple technique we can achieve significant results!