The answer states that if y^3 is divisible by 9, y must be divisible by 3.
How can you intuitively arrive at this conclusion (I agree this is true, but would never have arrived at this conclusion)? And more importantly, are there any other extensions of these divisibility/exponent rules?
Ans. B
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
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
Is y^3 divisible by 9?
Source: Beat The GMAT — Data Sufficiency |
If y^3 is divisible by 9 then multiples of y will also satisfy the formula.
4^3 is 64 and therefore has a cross sum of 10 -> not divisible by 9
6^3 is 216 and therefore has a cross sum of 9 -> divisible by 9 -> B is correct
4^3 is 64 and therefore has a cross sum of 10 -> not divisible by 9
6^3 is 216 and therefore has a cross sum of 9 -> divisible by 9 -> B is correct
















