3. A natural number $n$ is such that $n+2$ is divisible by 3, and $n+3$ is divisible by 4. Which of the following statements are guaranteed to be true, i.e., hold for all $n$ that meet the condition?

a) $n+4$ is divisible by 5;

b) $n+5$ is divisible by 6;

c) $n+6$ is divisible by 7;

d) $n+7$ is divisible by 8.
Answer. Only b).

Solution. First, note that b) works: $n+5=(n+2)+3-$ is divisible by $3, n+5=(n+3)+2-$ is divisible by 2. Since the number is divisible by 2 and by 3, it is also divisible by 6.

Now let's prove that none of the other options work. For this, in each case, we need to provide an example of $n$ that meets the condition but for which the statement does not hold. We will use the same example, $n=13$, for which the condition is met: $n+2=15$ is divisible by $3, n+3=16$ is divisible by 4, but

a) $n+4=17$ is not divisible by 5;
c) $n+6=19$ is not divisible by 7;
d) $n+7=20$ is not divisible by 8.

Remark. The second part could have been solved differently. When $n=1$, all statements are true. If we increase $n$ by 12 (which was done in the example), then each of the numbers $n+2, n+3, \ldots, n+7$ will increase by 12. When increased by 12, divisibility by 3 and 4 is preserved, but not by 5, 7, and 8.