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 500,000 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:
Given that a, b, c, and d are different nonzero digits and [#permalink]
17 Sep 2004, 06:33
00:00
A
B
C
D
E
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 - a, which of the following could not be a solution to the addition problem below?
10d + 11c < 100 - a means that the max value for a=9, b=9, c=7, d=8
IMO, Odd man out is (C) 8581 since
- to get a 5 in the hundredth digit, b+b has to be 7+7 and get a carry over from tenth digit
- to give a carry over of 1, d+c = 18, but the maximum possible value for d+c = 15
a<100-10d-11c
or, a+c<100-10d-10c or a+c<10(10-d-c) ... A
Now a+c has to be >0 10-d-c>0 or d+c<10 .... hence we know for sure that adding d and c wont have a carry over
Also a+c can either be <10 or <20
if less than 10 a+c has no carry over and c+d has no carry over. Hence in the sum - b+b has to be even
if less than 20 and greater than 10 a+c has carryover of 1
10-d-c=2 ... from A
or d+c=8 .....
d+c+carryover_from_a+c=8+1=9
Hence b+b has no carryover to add ..... and hence b+b has to be even
10d + 11c < 100 - a <=> 10(d+c)+(c+a) <100
and this represents tens+units digits of the sum
So there is no impact on the hundreds digit of the sum (=2.b)
(A) 3689
(B) 6887
(C) 8581
(D) 9459
(E) 16091
The sole solution with an odd number possibly equal to 2b is C with 5 or 15 which is impossible so answer si C
10d + 11c < 100 - a <=> 10(d+c)+(c+a) <100 and this represents tens+units digits of the sum So there is no impact on the hundreds digit of the sum (=2.b)
(A) 3689 (B) 6887 (C) 8581 (D) 9459 (E) 16091
The sole solution with an odd number possibly equal to 2b is C with 5 or 15 which is impossible so answer si C
I follow your point up to where you have the bold statement. could you elaborate how you selected C as the odd one out from there, Why 5 or 15
The hundreds number of the sum could have several origins :
we know that it is at least equal to the units digit of 2.b (2.b could be above or below 10...)
you could have to add +1 if 10(d+c)+(c+a) >99 but we know from the stem that it is < 100 so only 2.b is contributing to the hundreds digit of the sum ; 2.b beeing even, the sole not compliant number is C