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11 Oct 2022, 02:01
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
21% (02:25) correct
79%
(02:28)
wrong
based on 173
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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 1998, is a = b?
(1) c = 4
(2) a = 3
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(1) c = 4
(2) a = 3
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11 Oct 2022, 20:30
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Bunuel
Proper method would be to check what the question means.
Let us see the case when a=b.
(i) a=b=c,
then the sum would be the number itself as only one number can be formed.
(ii) a=b and c is different
Total 3-digit number that can be formed is 3!/2!=3. aac, aca, caa
When you add them each digit gets added once => (100a+10a+c)+ (100a+10c+a)+ (100c+10a+a)
=> 222a+111c=1998…….…111(2a+c)=1998……….2a+c=18
Case when a is not equal to b
Total 3-digit numbers possible is 3! or 6. Thus, 2(a+b+c)*111=1998……..a+b+c=9
What do we know now.
If a=b, then 2a+c=2b+c=18
If all digits are different, then a+b+c=9
(1) c=4
a) a=b => \(2a+4=18………a=b=7……….774+747+477=1998\)
b) a is not equal to b => \(a+b+c=9……...a+b+4=9……..a+b=5……...a=2 / / and / / b=3…….234+243+324+342+423+432=1998\)
Insufficient
(2) a=3
a) a=b…..2*3+c=18……..c=12……..Not possible as c<10.
We can mark B as the answer, but let us check the values.
b) a is not equal to b……a+b+c=9…….3+b+c=9…….b+c=6…..a,b,c can be 3,2,4 or 3,4,2 or 3,1,5 or 3,5,1.
Sufficient
B
Bunuel, please check the OA.
gmatprepper94111
Joined: 17 Jul 2022
Last visit: 23 Nov 2022
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Schools: Stanford '25
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WE:Corporate Finance (Finance: Venture Capital)
11 Oct 2022, 22:13
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----
What do we know to start?
----
We have the following six possible configurations, where each of a,b,c each are a digit of a three-digit number.
ABC
ACB
BAC
BCA
CAB
CBA
Then we know the sum of these numbers is 1998, i.e. ABC + ACB + BAC + BCA + CAB + CBA = 1998
So, knowing a number of the format XYZ = 100x + 10y + z, we can convert the above statement into something more workable:
100(a + a + b + b + c + c) + 10(b + c + a + c + a + b) + (c + b + c + a + b + a) = 1998
100(2a + 2b + 2c) + 10(2a + 2b + 2c) + (2a + 2b + 2c) = 1998
111(2a + 2b + 2c) = 1998
222(a + b + c) = 1998
a + b + c = 1998 / 222 = (999 * 2) / (2 * 111) = 999 / 111 = 9
So we know a + b + c = 9
Off the bat, we should recognize that knowing two of these numbers necessarily tells us the third number. So the answer can't be E. But let's see if we can do better.
---------------
What does A tell us?
---------------
c = 4
a + b + c = 9
a + b + 4 = 5
a + b = 5
Note that a + b is an odd number. There is no way we could get 5 from a + b if a = b. You could also figure this out by plugging in the possible combinations of a and b.
So Statement A is sufficient
---------------
What does B tell us?
---------------
a = 3
Let's pull back the statement we used earlier: a + b + c = 9
If a = 3, then 3 + b + c = 9
This means b + c = 6
Is it possible for b to be 3? Of course! Suppose a = b = c = 3. Then this checks out.
But it's not necessary for b to be 3. Consider the following counterexample: b = 2, c = 4. This works, but a != b.
So statement B is not sufficient
-------
The answer is A
What do we know to start?
----
We have the following six possible configurations, where each of a,b,c each are a digit of a three-digit number.
ABC
ACB
BAC
BCA
CAB
CBA
Then we know the sum of these numbers is 1998, i.e. ABC + ACB + BAC + BCA + CAB + CBA = 1998
So, knowing a number of the format XYZ = 100x + 10y + z, we can convert the above statement into something more workable:
100(a + a + b + b + c + c) + 10(b + c + a + c + a + b) + (c + b + c + a + b + a) = 1998
100(2a + 2b + 2c) + 10(2a + 2b + 2c) + (2a + 2b + 2c) = 1998
111(2a + 2b + 2c) = 1998
222(a + b + c) = 1998
a + b + c = 1998 / 222 = (999 * 2) / (2 * 111) = 999 / 111 = 9
So we know a + b + c = 9
Off the bat, we should recognize that knowing two of these numbers necessarily tells us the third number. So the answer can't be E. But let's see if we can do better.
---------------
What does A tell us?
---------------
c = 4
a + b + c = 9
a + b + 4 = 5
a + b = 5
Note that a + b is an odd number. There is no way we could get 5 from a + b if a = b. You could also figure this out by plugging in the possible combinations of a and b.
So Statement A is sufficient
---------------
What does B tell us?
---------------
a = 3
Let's pull back the statement we used earlier: a + b + c = 9
If a = 3, then 3 + b + c = 9
This means b + c = 6
Is it possible for b to be 3? Of course! Suppose a = b = c = 3. Then this checks out.
But it's not necessary for b to be 3. Consider the following counterexample: b = 2, c = 4. This works, but a != b.
So statement B is not sufficient
-------
The answer is A
