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Difficulty:
35%
(medium)
Question Stats:
74% (01:28) correct
26%
(01:55)
wrong
based on 887
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How many positive integer divisors does 8! have?
A. 14
B. 15
C. 24
D. 56
E. 96
How many positive integer.png [ 15.11 KiB | Viewed 22639 times ]
A. 14
B. 15
C. 24
D. 56
E. 96
Attachment:
How many positive integer.png [ 15.11 KiB | Viewed 22639 times ]
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nick13
Positive integers mean 1 and above. So the smallest is 1 and largest is 8!.
Now, the method to find positive integer divisors is as shown by gmatophobia
Get the number in its prime factors and take product of all exponents after adding one to each exponent.
Number of 2s: \(\frac{8}{2}+\frac{8}{2^2}+\frac{8}{2^3}\)=4+2+1=7
Number of 3s: \(\frac{8}{3}+\frac{8}{3^2}\)=2+0=2
Number of 5s and 7s will be 1 each.
Thus, 8!=\(2^7*3^2*5*7\) and factors are \((7+1)(2+1)(1+1)(1+1)=8*3*2*2=96\)
Clearly, Option E has been wrongly written as 95. Moreover, an odd number of positive factors means the number is Perfect Square and 8! is clearly not a perfect square.
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gmatophobia
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nick13
\(8! = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 = 1 * 2 * 3 * 2^2 * 5 * (3 * 2) * 7 * 2^3 \)
\(= 2^7 * 3^2 * 5^1 * 7^1\)
Number of factors \(= 8 * 3 * 2 * 2 = 96\)
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Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
General Discussion
gmatophobia
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Joined: 22 Dec 2016
Last visit: 09 Sep 2026
Posts: 3,174
Location: India
Concentration: Strategy, Leadership
Posts: 3,174
pablovaldesvega
The prime number 8! consists are 2, 3, 5, and 7
- Power of 2 in 8! = 7
- Power of 3 in 8! = 2
- Power of 5 in 8! = 1
- Power of 7 in 8! = 1
\(8! = 2^7 * 3^2 * 5^1 * 7^1\)
Number of factors = \((7+1)*(2+1)*(1+1)*(1+1) = 8 * 3 * 2 * 2 = 12 * 8 = 96\)
Option E
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Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
can someone please explain how this to 2^7∗3^2∗5^1∗7^1 gets reduced to Number of factors being 8∗3∗2∗2=96
is there a specific formula or method to this ?
is there a specific formula or method to this ?
diya7
Finding the Number of Factors of an Integer
First, make the prime factorization of an integer \(n = a^p * b^q * c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\), and \(p\), \(q\), and \(r\) are their respective powers.
The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and \(n\) itself.
Example: Finding the number of all factors of 450: \(450 = 2^1 * 3^2 * 5^2\)
The total number of factors of 450, including 1 and 450 itself, is \((1+1)(2+1)(2+1) = 2*3*3 = 18\) factors.
2. Properties of Integers
- Theory
Math: Number Theory
Number Properties: Tips and hints
Advanced Number Properties on GMAT (All 5 parts)
Do you make these 3 mistakes in GMAT Even-Odd Questions?
Everything about Factorials on the GMAT
Solving Complex Problems Using Number Line
For other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread
Hope it helps.
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Bunuel
Is this actually a formula we just have to keep in mind? like b^2+4ac
JoeAa
Yes, knowing and understanding the formula for finding the number of factors can be useful for the GMAT.
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can you help explain the intuition behind this rule?> i had no idea to use this.
(7+1)(2+1)(1+1)(1+1)=8∗3∗2∗2=96
(7+1)(2+1)(1+1)(1+1)=8∗3∗2∗2=96
chetan2u
So, the idea is that we can have seven two's in a factor or we can have 0 2's ie one. Likewise for all numbers mentioned. Consider the number 12.
12's factors are: 1,2,3,4,6,12
It's prime factorization is: 2^2*3. So notice each factor can be represented in the following manner 3*2^0, 3*2^1, 3*2^2, 3^0*2, 3^0*2^0. I hope this is clear.
12's factors are: 1,2,3,4,6,12
It's prime factorization is: 2^2*3. So notice each factor can be represented in the following manner 3*2^0, 3*2^1, 3*2^2, 3^0*2, 3^0*2^0. I hope this is clear.
Dbrunik
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