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02 Mar 2022, 03:46
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D
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Difficulty:
95%
(hard)
Question Stats:
36% (02:19) correct
64%
(02:04)
wrong
based on 101
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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 3108, is a=b?
(1) c = 9
(2) a = 2
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(1) c = 9
(2) a = 2
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
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02 Mar 2022, 06:36
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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
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
03 Mar 2022, 00:44
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Bunuel
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
