For which of the following values of \(n\) is \(\frac{120+n}{n}\) NOT an integer?

(A) \(3\) (B) \(4\) (C) \(5\) (D) \(6\) (E) [math]7[/math]

As usual, there are several approaches, but all of them incorporate process of elimination – always a good idea when solving a NOT question. Also a good idea with “must be,” “could be,” and Roman numeral questions.

**GOOD ENOUGH**: Checking division by hand:

\(123\div3 = 41, 124\div4 = 31, 125\div5 = 25, 126\div6 = 21\)

at this point you can stop because only one answer remains – (E).

Obviously, calculating these may be a bit time consuming, but doing this sort of division mentally is a process you should cultivate. It’s not easy, but it’s not some form of magic either. Anyone can learn to do it. Here is my mental process for \(123\div3\).

\(123 = 100 + 20 + 3 = 120 + 3\) Now divide both terms by \(3\) and you get \(40 + 1 = 41\)

**BETTER**: Divisibility rules. The answer is not (A) because \(123\) is divisible by \(3\) (the sum of the digits is divisible by \(3\)). The answer is not (B) because \(124\) is divisible by \(4\) (the number formed by the last two digits is divisible by \(4\)). The answer is not (C) because \(125\) is divisible by \(5\) (it ends in \(5\)). The answer is not (D) because \(126\) is divisible by \(6\) (it’s divisible by \(2\) and \(3\)). We’ve eliminated (A) though (D), therefore the answer must be (E) .

Both of these are efficient ways to solve this problem. The second one is “better” than the first because at some point the numbers could be large enough that dividing mentally would be impractical, and doing long division would be time consuming. You need to know your divisibility rules, and know them well.

Divisibility Rules is a pdf that summarizes the divisibility rules for 2, 3, 4, 5, 6, 8, and 10. It ends with an exercise so you can check your understanding. There are rules for 7 and 11 and your welcome to learn them as well as many other rules here.

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