Positive integers a, b and c are all less than 10. If the sum of all t

Question

Positive integers a, b and c are all less than 10. If the sum of all the distinguishable three digit numbers that can be formed by juxtaposing these integers is 3108, is a=b?

(1) c = 9
(2) a = 2

Are You Up For the Challenge: 700 Level Questions: 700 Level Questions

Nidzo · Answer

Summing the numbers will look like this:
abc
acb 
bac
bca 
cab 
cba 
------
3108

1) 
c=9 gives us: 
Units digits summed:  2a+2b+18  
Tens digits summed:  20a+20b+180 
Hundreds digits summed: 200a + 200b + 1800

Summing the above three gives us:  1998+222a+222b = 3108 
 222a+222b = 1110

If a=b, then  222a = 555  and  222b = 555 . However,  \frac{555}{222}=2.5  which is not an integer. Therefore a≠b.

SUFFICIENT

2) 
a = 2 gives us,

Units digits summed:  4+2b+2c 
Tens digits summed:  40+20b+20c 
Hundreds digits summed: 400+200b+200c

Summing the above three gives:  444+222b+222c = 3108 
 222b+222c = 2664

If a=b, then b=2 which would give:  444+222c = 2664 
 222c = 2220 
 c = 10

As no number can equal 10, it is impossible for a=b

SUFFICIENT

Answer D

parth04red · Answer

Given: a,b,c are positive integers less than 10.

To find: a = b?

possible 3 digit numbers(distinguishable) are:
abc = 100a + 10b + c
acb = 100a + 10c + b
bca = 100b + 10c + a
bac = 100b + 10a + c
cba = 100c + 10b + a
cab = 100c + 10a + b

It is given that the sum of all possible 3 digit numbers = 3108
thus, on adding the 6 numbers, we have 222a + 222b + 222c = 3108
which gives us 222(a+b+c) = 222*14
thus, a + b + c = 14

St (1): c = 9
which gives us a + b = 5
 If a = b, then we get a = b = 2.5, which is not possible as a,b,c are integers. 
Hence,  a is not equal to b .
St 1 is  Sufficient .

St (2): a = 2
 if a = b = 2, we get c = 10, which is not possible as a,b,c are all less than 10. 
thus,  a is not equal to b .
St 2 is  Sufficient .

Answer:  option D

rsrighosh · Answer

what is the meaning of "juxtapose" ?

Carcass · Answer

two things (different) side by side. to compare or contrast

av1901 · Answer

I went with an unconventional approach, please let me know if I am mistaken  Bunuel

We have three positive integers a, b, and c: All less than 10

We join the three single digit integers to form all possible 3 digit integers:

Now, if each digit is unique : We have a total of 6 possible three digit integers: 3! = 6
 But if any two digits are the same : Then total number of distinguishable three digit integers = 3!/2! = 3

Now: If we have any two integers the same, we only have 3 possible three digit integers and the sum of integers will ALWAYS be LESS THAN 3000. How?
Take for example: a=b=9 and c=8: Possible integers = 998, 989, 899. Sum < 3000. Sum will never ever be greater than 3000 
So, if sum > 3000, then no two integers can be the same, all have to be unique

So, this question is false (as in a never equal to b) irrespective of the statements

Answer: D

p.s. if my approach is correct, can we come across such questions in the actual exam? where statements do not add any value?

Bunuel · Answer

You are absolutely right! The stem is enough to deduce that the answer is NO. You cannot have such question on an actual exam. Thank you for noticing that! Here is an edited version of this question: positive-integers-a-b-and-c-are-all-less-than-10-if-the-sum-of-all-400001.html P.S. Having said that, there is one official question from GMAT Prep with the same flaw: if-zy-xy-0-is-x-z-x-z-101210.html The stem is enough to answer the question there.

Positive integers a, b and c are all less than 10. If the sum of all t

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia