The integer 120 has many factorizations. For examp... | Free GMAT Practice Question | GMAT Panda

👋 Hi! I'm GMAT Panda. Ask me anything about this question, or click a button below to get started!

The integer 120 has many factorizations. For example, 120 = (2) (60), 120 = (3) (4) (10), and 120 = (- 1) (- 3) (4) (10) . In how many of the factorizations of 120 are the factors consecutive integers in ascending order?

Explanation	Calculations	Help
$We name the first integer n and the count of consecutive factors k .$	$First integer = n, Number of factors = k$	Theory & Tactics Method Card TRANS1-A Click to view full details
$We translate "consecutive integers" into unknowns and set their product to 120.$	$n (n + 1) ... (n + k - 1) = 120$	Theory & Tactics Method Card TRANS5-B Click to view full details
$For 2 factors there is no integer solution.$	$n (n + 1) = 120$
$For 3 factors there is one solution.$	$n (n + 1) (n + 2) = 120 4 \times 5 \times 6 = 120$
$For 4 factors there are two solutions.$	$n (n + 1) (n + 2) (n + 3) = 120 2 \times 3 \times 4 \times 5 = 120 (- 5) \times (- 4) \times (- 3) \times (- 2) = 120$
$For 5 factors there is one solution.$	$n (n + 1) (n + 2) (n + 3) (n + 4) = 120 1 \times 2 \times 3 \times 4 \times 5 = 120$

Scroll horizontally to view all columns

4

The integer 120 has many factorizations. For example, 120 = (2) (60), 120 = (3) (4) (10), and 120 = (- 1) (- 3) (4) (10) . In how many of the factorizations of 120 are the factors consecutive integers in ascending order?

Explanation	Calculations	Help
$We name the first integer n and the count of consecutive factors k .$	$First integer = n, Number of factors = k$	Theory & Tactics Method Card TRANS1-A Click to view full details
$We translate "consecutive integers" into unknowns and set their product to 120.$	$n (n + 1) ... (n + k - 1) = 120$	Theory & Tactics Method Card TRANS5-B Click to view full details
$For 2 factors there is no integer solution.$	$n (n + 1) = 120$
$For 3 factors there is one solution.$	$n (n + 1) (n + 2) = 120 4 \times 5 \times 6 = 120$
$For 4 factors there are two solutions.$	$n (n + 1) (n + 2) (n + 3) = 120 2 \times 3 \times 4 \times 5 = 120 (- 5) \times (- 4) \times (- 3) \times (- 2) = 120$
$For 5 factors there is one solution.$	$n (n + 1) (n + 2) (n + 3) (n + 4) = 120 1 \times 2 \times 3 \times 4 \times 5 = 120$

Scroll horizontally to view all columns

4