Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Tough prime factorisation problem

Expert replies
by selango » Mon Jul 05, 2010 6:38 am
For any integer k > 1, the term “length of an integer” refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 × 2 × 2 × 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

A.5
B.6
C.15
D.16
E.18

OA D

can anyone explain in detail?
--Anand--
Join the discussion
Source: — Problem Solving |

by kvcpk » Mon Jul 05, 2010 7:00 am
Hi Anand,

The wording of the problem is very difficult.

I will expalin How I approached the problem. Others can suggest even better ways.

Given x + 3y < 1000
Now the number of prime factors need to be maximum for both x and y.

Logically, the maximum number of prime multiples that can be accommodated is when the number has more 2 multiples.

So, I took x to be some power of 2.

Let us say 2^10 ->1024 INSUFFICIENT.
2^9 = 512 -> We still need to find y
now 512 + 3(2^p) <1000
3(2^p)<488
2^p<162

So I chose p to be 7

now we have 2^9 + 3(2^7) = 896
so length of x=9, length of y = 7
total is 16.

How do we knowthat it is the maximum?

we need not know.. because, the only other answer that is available is 18.
at the worst case, we can split it into power of x + power of y = 18
atleast one power should be greater than 9.
which is not possible even with the least prime number 2.
2^9 +3*2^9 ???

Hence pick D.
Last edited by kvcpk on Mon Jul 05, 2010 10:24 am, edited 1 time in total.
Join the discussion

by selango » Mon Jul 05, 2010 7:05 am
the maximum number of prime multiples that can be accommodated is when the number has more 2 multiples.

I didn't think this way.Got it.Thanks.
--Anand--
Join the discussion

by kvcpk » Mon Jul 05, 2010 7:12 am
selango wrote:the maximum number of prime multiples that can be accommodated is when the number has more 2 multiples.

I didn't think this way.Got it.Thanks.
Glad that it was helpful!! :)
Join the discussion

by raunakrajan » Mon Jul 05, 2010 9:56 am
kvcpk wrote:Hi Anand,

The wording of the problem is very difficult.

I will expalin How I approached the problem. Others can suggest even better ways.

Given x + 3y < 1000
Now the number of prime factors need to be maximum for both x and y.

Logically, the maximum number of prime multiples that can be accommodated is when the number has more 2 multiples.

So, I took x to be some power of 2.

Let us say 2^10 ->1024 INSUFFICIENT.
2^9 = 512 -> We still need to find y
now 512 + 3(2^p) <1000
3(2^p)<488
2^p<162

So I chose p to be 7

now we have 2^9 * 3(2^7) = 896
so length of x=9, length of y = 7
total is 16.

How do we knowthat it is the maximum?

we need not know.. because, the only other answer that is available is 18.
at the worst case, we can split it into power of x + power of y = 18
atleast one power should be greater than 9.
which is not possible even with the least prime number 2.
2^9 +3*2^9 ???

Hence pick D.
i still din get itt :(
can you explain it probably in a simpler way if possible!>?
Join the discussion

by singhpreet1 » Mon Jul 05, 2010 10:22 am
kvcpk wrote:Hi Anand,

The wording of the problem is very difficult.

I will expalin How I approached the problem. Others can suggest even better ways.

Given x + 3y < 1000
Now the number of prime factors need to be maximum for both x and y.

Logically, the maximum number of prime multiples that can be accommodated is when the number has more 2 multiples.

So, I took x to be some power of 2.

Let us say 2^10 ->1024 INSUFFICIENT.
2^9 = 512 -> We still need to find y
now 512 + 3(2^p) <1000
3(2^p)<488
2^p<162

So I chose p to be 7

now we have 2^9 * 3(2^7) = 896
so length of x=9, length of y = 7
total is 16.

How do we knowthat it is the maximum?

we need not know.. because, the only other answer that is available is 18.
at the worst case, we can split it into power of x + power of y = 18
atleast one power should be greater than 9.
which is not possible even with the least prime number 2.
2^9 +3*2^9 ???

Hence pick D.
i think u need to make a correction:

now we have 2^9 * 3(2^7) = 896

u mean 2^9 +3(2^7)= 896

i hope this was not causing any confusion to anyone.

Preet
Join the discussion

by kvcpk » Mon Jul 05, 2010 10:26 am
@preet - Thanks for the correction.. That was a typo..Edited it now..

@raunakrajan - Can you please tell me till what point you understood it.. I can help you out.
Join the discussion

by singhpreet1 » Mon Jul 05, 2010 10:37 am
raunakrajan wrote: i still din get itt :(
can you explain it probably in a simpler way if possible!>?
Raunak..let me try and have a go and rephrase what has already been stated by kvcpk

lets look at the question again:

For any integer k > 1, the term “length of an integer� refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 X 2 X 2 X 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

A.5
B.6
C.15
D.16
E.18

so we know that both x and y are greater than 1. lets assume x=2, as we have to find the maximum possible length of x and y.
also 2 is the first prime number right? so if we take 2^10 then we get 1024 which does not satisfy the condition x + 3y < 1000.

ok, so lets assume x= 2^9= 512 yea?
now, again as per the condition to find y; x + 3y < 1000; y must be multiplied by 3 yea? so 3y= 3(2^7)= 384, why did i choose 2^7, if i take 2^8= 256, my condition of x + 3y < 1000 will not be satisfied, x= 512+3(256)= 1280 which is not less than 1000, so if i take x=2^9, y= 3(2^7)

i get 512+ 384= 896 which is less than 1000.

so, in 896 the prime factors used to determine x and y are: 9 2's (2^9) and 7 2's (2^7). lets add them up 9+7= 16.

i hope this was helpful to you.

Preet
Join the discussion

by kvcpk » Mon Jul 05, 2010 10:43 am
singhpreet1 wrote: ok, so lets assume x= 2^9= 512 yea?
now, again as per the condition to find y; x + 3y < 1000; y must be multiplied by 3 yea? so 3y= 3(2^6)= 384, why did i choose 2^6, if i take 2^7= 256, my condition of x + 3y < 1000 will not be satisfied, x= 512+3(256)= 1280 which is not less than 1000, so if i take x=2^9, y= 3(2^6)
OOps.. typo..

2^7 = 128
so 512 + 3(128) = 896

2^8 = 256
Join the discussion

by raunakrajan » Mon Jul 05, 2010 1:01 pm
singhpreet1 wrote:
raunakrajan wrote: i still din get itt :(
can you explain it probably in a simpler way if possible!>?
Raunak..let me try and have a go and rephrase what has already been stated by kvcpk

lets look at the question again:

For any integer k > 1, the term “length of an integer� refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 X 2 X 2 X 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

A.5
B.6
C.15
D.16
E.18

so we know that both x and y are greater than 1. lets assume x=2, as we have to find the maximum possible length of x and y.
also 2 is the first prime number right? so if we take 2^10 then we get 1024 which does not satisfy the condition x + 3y < 1000.

ok, so lets assume x= 2^9= 512 yea?
now, again as per the condition to find y; x + 3y < 1000; y must be multiplied by 3 yea? so 3y= 3(2^7)= 384, why did i choose 2^7, if i take 2^8= 256, my condition of x + 3y < 1000 will not be satisfied, x= 512+3(256)= 1280 which is not less than 1000, so if i take x=2^9, y= 3(2^7)

i get 512+ 384= 896 which is less than 1000.

so, in 896 the prime factors used to determine x and y are: 9 2's (2^9) and 7 2's (2^7). lets add them up 9+7= 16.

i hope this was helpful to you.

Preet
thanks a ton boss! really helpful
Join the discussion