"If a Positive Integer n Is Divisible by Both 5 and 7..." — GMAT® Worked Solution

Source: Official Guide for GMAT® Review, 11th Edition

If a positive integer $n$ is divisible by both 5 and 7, then $n$ must also be divisible by which of the following?

I. 12

II. 35

III. 70

(A) None

(B) I only

(C) II only

(D) I and II

(E) II and III

Try it before reading on.

Step 1: Write Divisibility as a Fraction

Divisible means the division comes out to an integer, with nothing left over. Written as a fraction, that gives the setup this question turns on:

$\frac{n}{5 \times 7} = \text{integer}$

The numerator is $n$ because $n$ is what gets divided. The denominator is $5 \times 7$ because that's what it's being divided by. For the fraction to produce an integer, everything in the denominator has to cancel with something in the numerator, which means $n$ must contain both a 5 and a 7 somewhere in its prime factorization.

That's the whole hand the problem deals: all we know for sure about $n$ is that a 5 and a 7 are in there. It might contain other prime factors too, but nothing else is guaranteed.

Step 2: Test Each Roman Numeral by Prime Factors

Each Roman numeral asks the same question: does this number's prime factorization have to fit inside $n$'s?

Take Roman numeral I. The prime factorization of 12 is $2 \times 2 \times 3$, and nothing in it is guaranteed. If $n = 5 \times 7 \times 2 \times 2 \times 3$, then $n$ happens to be divisible by 12. If $n = 35$, it isn't. Both are allowed by the problem, so 12 divides $n$ sometimes but not surely.

Roman numeral II is the opposite case. The prime factorization of 35 is $5 \times 7$, exactly the two factors the problem guarantees. Whether $n$ is just 35 or 35 multiplied by any pile of other primes, a factor of $5 \times 7$ cancels completely, so 35 divides $n$ in every allowed case.

Roman numeral III is close to II but not quite it. Since $70 = 2 \times 5 \times 7$, dividing $n$ by 70 needs a 2 to cancel, and the problem never promised a 2. When $n = 35$, the division by 70 fails. So 70 divides $n$ sometimes, the same status as 12.

Step 3: Answer the Question Asked

The question asks what $n$ must be divisible by. Only Roman numeral II survives that standard, so:

The answer is (C).

The Number-Testing Alternative

This question also solves by trying a concrete value for $n$, and 35 is the natural pick since it's the smallest number the problem allows. Dividing: $\frac{35}{12}$ is not an integer, so numeral I fails. $\frac{35}{35} = 1$, so numeral II holds. $\frac{35}{70} = \frac{1}{2}$, so numeral III fails. That's (C) in a few lines, and a second case like $n = 105$ can confirm it if wanted.

Testing numbers is a fine front-line approach, and on this question it's fast. The caveat is range. A question like "if $n$ is divisible by both 25 and 7, then $n^{17}$ must be divisible by which of the following?" stops being friendly to testing, because the numbers blow up while the prime factor analysis barely changes. The cancellation setup handles small and large cases the same way, which is why it's worth learning even if testing numbers stays your default.

Why This Problem Matters

The Roman numeral format pressures a specific reading: must versus might. A numeral that divides $n$ in some allowed cases is a real temptation, because it's divisible in the cases you happen to picture. The fraction setup converts that judgment into a checkable rule. A numeral divides $n$ surely only when every one of its prime factors is guaranteed to be in $n$, and "guaranteed" means the problem stated it or forced it.

It's also a natural follow-on to Episode 10's divisibility work, which built the prime-cancellation view from the ground up. If the fraction setup above felt foreign, that page is the place to start.

Ready for the next one? "If 4 Is One of the Solutions to x² + 3x + k = 10..." — GMAT® Worked Solution.

Want the full strategy? Read: GMAT® Percents, Divisibility, and Quadratics: Translate First, Solve Second

From Episode 11 of Real GMAT® Problems (The GMAT® Strategy Podcast).