Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Tough and tricky 4: addition problem [#permalink]
11 Oct 2009, 16:42

Expert's post

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

84% (02:04) correct
16% (01:50) wrong based on 84 sessions

Tough and tricky 4: addition problem

AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

Re: Tough and tricky 4: addition problem [#permalink]
12 May 2013, 02:04

Expert's post

khosru wrote:

AKProdigy87 wrote:

The answer is D: 9.

Logical deduction.

1) Two 2-digit numbers sum to a 3 digit number in the form AAA. Since AB < 100 and CD < 100, AB + CD must be less than 200. So AAA = 111, and A = 1.

2) Now we know, since A is equal to 1, that AB < 20. The only way that AB + CD = 111, if AB < 20, is that if CD > 90. Therefore, C = 9.

sorry I disagree. as 99 +12 =111 0r 12+99=111 C can be 9 or 1, again 76+25=111 or 25+76=111, C can be 2 or 7. so my answer is E.

That's not correct.

AB and CD are two digit integers, their sum can give us only one 3-digit integer of a kind of AAA: 111. So, A=1 --> 1B+CD=111.

C can not be less than 9, because no 2-digit integer with first digit 1 (mean that it's <20) can be added to 2-digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111) --> C=9.

Re: Tough and tricky 4: addition problem [#permalink]
12 May 2013, 10:11

Different approach...

A + C + x (Carryover) = AA.. Since all integers are positive C cant be zero...

=> C + x (Carryover) = 10

Now the only carryover that can come from B + D is 1

Hence C = 9 _________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

Re: Tough and tricky 4: addition problem [#permalink]
13 May 2013, 00:46

1

This post received KUDOS

Bunuel wrote:

Tough and tricky 4: addition problem

AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determined

A + C gives A . So c must be either 0 or 10. I the answer choices there is no option with zero. so c must be 10.for c to be 10 we must get a carry over and c MUSt be 9 as the maximum carry over we can get is 1 . so D _________________

"Kudos" will help me a lot!!!!!!Please donate some!!!

Completed Official Quant Review OG - Quant

In Progress Official Verbal Review OG 13th ed MGMAT IR AWA Structure

Yet to do 100 700+ SC questions MR Verbal MR Quant

Re: Tough and tricky 4: addition problem [#permalink]
13 May 2013, 03:11

Bunuel wrote:

Tough and tricky 4: addition problem

AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determined

AB<100 and CD<100.So , 20<=AB+CD<200.From 20 to 199 which is in the form of AAA is only one number i.e 111.So A=1,and AB=1B<20 because the tens place has 1.

1B+CD=111,let us consider 1.) B=2 => CD = 99 AB+CD=12+99=111, C=9

2)B=3 => AB=13 =>CD = 98 . C=9

take B=4,5,6,7,8,9 ,we get C= 9 _________________

......................................................................................... Please give me kudos if my posts help.

Re: Tough and tricky 4: addition problem [#permalink]
26 Sep 2013, 05:17

SrinathVangala wrote:

Bunuel wrote:

Tough and tricky 4: addition problem

AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determined

A + C gives A . So c must be either 0 or 10. I the answer choices there is no option with zero. so c must be 10.for c to be 10 we must get a carry over and c MUSt be 9 as the maximum carry over we can get is 1 . so D

Correct, also C cannot be zero because the question says they are all positive integers

gmatclubot

Re: Tough and tricky 4: addition problem
[#permalink]
26 Sep 2013, 05:17

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

As part of our focus on MBA applications next week, which includes a live QA for readers on Thursday with admissions expert Chioma Isiadinso, we asked our bloggers to...