"What Is the Lowest Positive Integer That Is Divisible by Each of the Integers 1 Through 7..." — GMAT® Worked Solution

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

What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive?

(A) 420

(B) 840

(C) 1,260

(D) 2,520

(E) 5,040

Try it before reading on.

Step 1: Write What's Given and Asked

Given: the integers 1, 2, 3, 4, 5, 6, and 7. Asked: the lowest positive integer divisible by each of them.

Both halves of the ask matter. The integer has to be divisible by all seven numbers, and it has to be the lowest such integer. Satisfying the first condition while ignoring the second is the trap this problem is built around.

Step 2: Why the Obvious Answer Is a Trap

The quick instinct is to multiply all seven numbers:

$1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 = 5040$

That number is divisible by each integer from 1 to 7, and it's sitting in the answer choices as (E). It just isn't the lowest. Multiplying everything together guarantees divisibility but over-delivers it, collecting more prime factors than the conditions require.

Step 3: Think of Divisibility as a Fraction

A number is divisible by $y$ when the division produces an integer. In fraction form, $x$ divided by $y$ comes out an integer only when every prime factor of $y$ cancels with something in the numerator.

A small example: 6 divided by 3. Prime factor the top, $\frac{2 \times 3}{3}$, watch the 3 cancel, and the integer 2 is what remains. The same check works for bigger numbers. Since $1500 = 2^2 \times 3 \times 5^3$ and $20 = 2^2 \times 5$, the fraction $\frac{1500}{20}$ has a denominator that cancels completely, so 1,500 is divisible by 20 without any long division.

So the question becomes a construction problem: build the smallest numerator that lets every denominator from 1 to 7 cancel.

Step 4: Build the Numerator One Divisor at a Time

Walk through the divisors in order, and for each one add the prime factors it needs that the numerator doesn't already have:

Rows 4 and 6 show the savings. The number 4 needs $2 \times 2$, but a single 2 is already in the numerator from handling 2, so one more 2 finishes the job. The number 6 needs $2 \times 3$, and both factors are already there from handling 2 and 3 separately. Nothing gets added. The factorial would have paid for those factors twice; the "lowest" condition pays once.

Step 5: Multiply It Out

The finished numerator is $1 \times 2 \times 3 \times 2 \times 5 \times 7$:

$2 \times 3 = 6$

$6 \times 2 = 12$

$12 \times 5 = 60$

$60 \times 7 = 420$

The answer is (A).

Why This Problem Matters

The trap is answering the question you didn't get asked. 5,040 satisfies the divisibility condition and ignores "lowest," and it's the natural landing spot for reading the stem as "the product of 1 through 7" instead of "divisible by each of 1 through 7." Reading the question twice, with attention on the adjectives, is a cheap defense.

There's also a faster path on this particular question worth knowing about: test answer choice (A) by dividing 420 by each integer from 1 to 7. Each division comes out even, and 420 is the smallest choice on the list. Difficulty ratings on the GMAT® come from test-taker data, and this question rates easy partly because plenty of people test the first choice and finish quickly.

Testing the choices wins this particular question because 420 happens to be the first option and the divisors are small. As the numbers grow, dividing by seven divisors under time pressure gets slower, while the cancellation system treats bigger constraints the same way. Neither route is universally right: this problem is a good one for solving both ways during review, then noting which one you'd trust under time pressure.

Want the full strategy? Read: GMAT® Rates and Divisibility: Two Systems That Scale

From Episode 10 of Real GMAT® Problems (The GMAT® Strategy Podcast).