GMAT Practice Question: What is the greatest positive integer n such that 5^{n} divides 10!-(2)(5!)^{2} ?
Question
What is the greatest positive integer n such that 5^{n} divides 10!-(2)(5!)^{2} ?
- 2
- 3
- 4
- 5
- 6
Topics: factorials, divisibility, number theory, prime factorization, exponents
Solution
Step 1
We first factor out the common 5! from both terms to simplify the expression.
10! - 2(5!)^2 = 5! \times (6 \times 7 \times 8 \times 9 \times 10 - 2 \times 5!)
Step 2
We then multiply 6 and 7 inside the parentheses.
6 \times 7 = 42
Step 3
We then multiply the result by 8.
42 \times 8 = 336
Step 4
We simplify 336 \times 9 by distributing to avoid large-factor multiplication.
336 = 300 + 36 300 \times 9 = 2700 36 \times 9 = 324 2700 + 324 = 3024
Step 5
We then multiply the result by 10.
3024 \times 10 = 30240
Step 6
We then subtract twice 5! from that result to complete the parentheses.
2 \times 5! = 2 \times 120 = 240 30240 - 240 = 30000
Step 7
This means the expression becomes 5! times 30000.
5! \times 30000 = 120 \times 30000
Step 8
We break each factor into its prime factors to find the exponent of 5 in the product.
5! = 2 \times 3 \times 4 \times 5 = 2 \times 3 \times (2 \times 2) \times 5 = 2^3 \times 3 \times 5 30000 = 3 \times 10^4 = 3 \times (2 \times 5)^4 = 3 \times 2^4 \times 5^4
Step 9
We then combine the prime factors and apply \textbf{Divisibility via Prime Exponents} to find the largest power of 5 dividing the product.
2^3 \times 3 \times 5 \times 2^4 \times 3 \times 5^4 = 2^7 \times 3^2 \times 5^5
Answer
5
