I guess the question should be like 27, 54 or both. Pick some numbers to solve:
3,6 and 9 are positive consecutive multiples of 3. They are divisible by 27 (3*9) and by 54 (6*9). Hence both.
- Deepak
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
Number properties
Source: Beat The GMAT — Problem Solving |
Because they are three multiples of 3, their product will definitely be able to divide 27, which is just 3^3. Consecutive multiples of three rotate in evenness and oddness; because they are consecutive, and because there are three of them, at least one of them is even, and so their product will definitely be able to divide 54, whose prime factorization is just 2*3^3.jnbimmer wrote:If r,s,and t are consecutive positive multiples of 3, is rst divisible by 27, 54, or both?
Kaplan Teacher in Toronto
Let s=r+3 and t=r+6
Then rst = r(r+3)(r+6)
Since r,s & t are each multiples of 3, the resulting number rst has three 3s as one of its factors. Therefore rst is divisible by 27 since 27=3x3x3.
Also rst must be even as at least one of rst is even. Therefore rst is divisible by 54 since an even number contains the factor 2 that is also present in 54.
Then rst = r(r+3)(r+6)
Since r,s & t are each multiples of 3, the resulting number rst has three 3s as one of its factors. Therefore rst is divisible by 27 since 27=3x3x3.
Also rst must be even as at least one of rst is even. Therefore rst is divisible by 54 since an even number contains the factor 2 that is also present in 54.
Rule: A number is divisible by 3 if the sum of the digits is divisble by 3.
27: 2+7 = 9
54: 5+4 = 9
If r,s,and t are consecutive positive multiples of 3, rst is divisible by both.
27: 2+7 = 9
54: 5+4 = 9
If r,s,and t are consecutive positive multiples of 3, rst is divisible by both.
Hi there,nittanylion530 wrote:Rule: A number is divisible by 3 if the sum of the digits is divisble by 3.
27: 2+7 = 9
54: 5+4 = 9
If r,s,and t are consecutive positive multiples of 3, rst is divisible by both.
actually, using the rule the way you did doesn't quite lead us to the answer: It is a given that 27 and 54 are divisible by 3...We need to know whether rst is divisible by 27 and 54.
Let's think about 27 first. What would a number have to do in order to divide by 27 successfully? Well, because 27 is "made up of" three 3s (ie, its prime factorization is 3^3), a number would have to have three 3s in it in order to be successfully divided by 27. Well, because r, s, and t are all multiples of 3, we know rst has three 3s in it, and so of course rst can be divided by 27.
54 is just 2*27. So, in order to successfully divide by 54, a number would, in addition to having three 3s, also have to have a 2. Because these are consecutive multiples of 3, we know that at least one of them is even, and so rst has at least one 2 in it. So, rst would also be a multiple of 54.
Kaplan Teacher in Toronto
