GMAT Practice Question: The integer 120 has many factorizations. For example, 120=(2)(60), 120=(3)(4)(10), and...
Question
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?
- 2
- 3
- 4
- 5
- 6
Topics: number theory, factoring
Solution
Step 1
We name the first integer n and the count of consecutive factors k.
\text{First integer} = n, \text{Number of factors} = k
Step 2
We translate \textbf{\text{"consecutive integers"}} into unknowns and set their product to 120.
n(n+1)...(n+k-1) = 120
Step 3
For 2 factors there is no integer solution.
n(n+1) = 120
Step 4
For 3 factors there is one solution.
n(n+1)(n+2) = 120 4 \times 5 \times 6 = 120
Step 5
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
Step 6
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
Answer
4
