BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

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.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 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
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
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
GMATOnDemand Start 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

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Pl help

Expert replies
by Nijo » Thu Jul 10, 2014 5:55 am
Hi
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?
Choices are 5, 6, 15, 16 and 18

I am unable to understand the explanation
Thanks!
Join the discussion
Source: — Problem Solving |

by Brent@GMATPrepNow » Thu Jul 10, 2014 6:17 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
Notice that powers of 2 will MAXIMIZE the length of an integer.
For example, 16 (aka 2^4) has length 4, whereas 21 (which is greater than 16) has a length of only 2.

Let's recall a few of the higher powers of 2.
2^7 = 128 (length is 7)
2^8 = 256 (length is 8)
2^9 = 512 (length is 9)

At this point, we might fiddle with some possible values of x and y and test them.
If x = 512 and y = 128, then x + 3y < 1000
Here, the length of x is 9, and the length of y is 7.
So, the sum of the length of x and the length of y = 9 + 7 = 16

So, the correct answer might be D, or it might be E

The correct answer can't be E. ,
We know this because 2^10 is greater than 1000, which means neither x nor y can be 2^10 or greater.
Also, 2^9 = 512, so x and y cannot both equal 2^9 (since x + 3y would be GREATER THAN 1000)
These two facts rule out the possibility that the lengths of x and y add to 18.

This means the correct answer must be D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

by GMATGuruNY » Thu Jul 10, 2014 6:49 am
Nijo wrote:Hi
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?
Choices are 5, 6, 15, 16 and 18

I am unable to understand the explanation
Thanks!
Length = the number of prime factors.
Thus, the MAXIMUM length = the MAXIMUM number of prime factors.
To maximize the number of prime factors of x and y such that x+3y < 1000, we must include as many 2's as possible without exceeding the threshold of 1000.

Let x = 2^9 = 512.
Since x is composed of nine 2's, the length of x = 9.

Since x=9 and x+3y < 1000, we get:
512 + 3y < 1000
3y < 488
y < 163 (approx).
Thus, y = 2^7 = 128.
Since y is composed of seven 2's, the length of y = 7.

Thus, the maximum possible sum of x and y = 9+7 = 16.

The correct answer is D.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

by Nijo » Fri Jul 11, 2014 2:38 am
Hi,
Why as many 2's as possible? Why not as many 5's as possible?
Join the discussion

by GMATGuruNY » Fri Jul 11, 2014 4:52 am
Nijo wrote:Hi,
Why as many 2's as possible? Why not as many 5's as possible?
Let's say the constraint is that x < 40.
Thus, the maximum possible value of x is 39.

If the prime-factorization of x is composed solely of 2's, then the greatest possible value of x is 32:
2*2*2*2*2 = 32.
No more prime factors can be included without exceeding the threshold of 39.
Here, since x is composed of 5 prime factors, the length of x is 5.

If the prime-factorization of x is composed solely of 5's, then the greatest possible value of x is 25:
5*5 = 25.
No more prime factors can be included without exceeding the threshold of 39.
Here, since x is composed of 2 prime factors, the length of x is 2.

To MAXIMIZE the length of x, we must MINIMIZE the value of each prime factor.
Thus, we want the prime-factorization of x to include as many 2's as possible.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion