if n is the product of the integers from 1 to 8, inclusive, how many different primer factors greater than 1 does n have?
so n = 1*2*3*4*5*6*7*8
prime factors of those are: 2,3,5,7
so 4 prime factors?
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Featured Course
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
Prime Factors
Source: Beat The GMAT — Problem Solving |
sfreeman wrote:if n is the product of the integers from 1 to 8, inclusive, how many different primer factors greater than 1 does n have?
product of n: 1*2*3*4*5*6*7*8
We can re-write this as:
1*2*3*(2^2)*5*(3*2)*7*(2^3)
See what I did? I reduced everything integer in the product to its factors.
Now its easy, we count all the factors greater than 1
# of 2s: 7
# of 3s: 2
# of 5s: 1
# of 7s: 1
total # of prime factors: 7+2+1+1 = 11 different prime factors. (Also note that since the question did not ask for DISTINCT or UNIQUE factors, we count all prime numbers, even as the prime numbers repeat themselves.
I hope that helped! Let me know if I am correct!
-
PussInBoots
- Master | Next Rank: 500 Posts
- Posts: 157
- Thanked: 3 times
pathaniaus you gotta stop smoking man 
The answer is 4. There are only 4 different prime factors: 2, 3, 5, and 7.
The answer is 4. There are only 4 different prime factors: 2, 3, 5, and 7.
1 is not a prime numberpathaniaus wrote:sfreeman wrote:if n is the product of the integers from 1 to 8, inclusive, how many different primer factors greater than 1 does n have?
product of n: 1*2*3*4*5*6*7*8
We can re-write this as:
1*2*3*(2^2)*5*(3*2)*7*(2^3)
See what I did? I reduced everything integer in the product to its factors.
Now its easy, we count all the factors greater than 1
# of 2s: 7
# of 3s: 2
# of 5s: 1
# of 7s: 1
total # of prime factors: 7+2+1+1 = 11 different prime factors. (Also note that since the question did not ask for DISTINCT or UNIQUE factors, we count all prime numbers, even as the prime numbers repeat themselves.
I hope that helped! Let me know if I am correct!
