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30 May 2012, 13:36
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C
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Difficulty:
65%
(hard)
Question Stats:
59% (01:56) correct
41%
(02:02)
wrong
based on 653
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If m and n are both two digit numbers and m - n = 11x, is x an integer?
(1) The tens digit and the units digit of m are the same
(2) m + n is a multiple of 11
(1) The tens digit and the units digit of m are the same
(2) m + n is a multiple of 11
31 May 2012, 02:24
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If m and n are both two digit numbers and m-n = 11x, is x an integer?
The question basically asks whether m-n is a multiple of 11.
(1) The tens digit and the units digit of m are same --> m could be: 11, 22, 33, ..., 99 --> m is a multiple of 11. Not sufficiient since no info about n.
(2) m+n is a multiple of 11 --> if m=n=11 then the m-n is a multiple of 11 but if m=12 and n=10 then m-n is NOT a multiple of 11. Not sufficient.
(1)+(2) From (1) we have that m={multiple of 11} and from (2) we have that m+n={multiple of 11} --> {multiple of 11}+n={multiple of 11} --> n={multiple of 11} --> m-n={multiple of 11}-{multiple of 11}={multiple of 11}. Sufficient.
Answer: C.
Below might help to understand this concept better.
If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.
If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.
If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.
Hope it's clear.
The question basically asks whether m-n is a multiple of 11.
(1) The tens digit and the units digit of m are same --> m could be: 11, 22, 33, ..., 99 --> m is a multiple of 11. Not sufficiient since no info about n.
(2) m+n is a multiple of 11 --> if m=n=11 then the m-n is a multiple of 11 but if m=12 and n=10 then m-n is NOT a multiple of 11. Not sufficient.
(1)+(2) From (1) we have that m={multiple of 11} and from (2) we have that m+n={multiple of 11} --> {multiple of 11}+n={multiple of 11} --> n={multiple of 11} --> m-n={multiple of 11}-{multiple of 11}={multiple of 11}. Sufficient.
Answer: C.
Below might help to understand this concept better.
If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.
If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.
If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.
Hope it's clear.
General Discussion
10 Sep 2013, 22:25
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Hi Banuel,
one confusion here.
from (1) we know that m is a multiple of 11. we also know that m-n= multiple of 11.
now, if we consider m to be 99 than , 99-n=multiple of 11. can we have any other 2 digit no. which is NOT a multiple of 11 for n in this case ? I think no. so effectively shouldn't the answer be A ? Have I missed something here ?
one confusion here.
from (1) we know that m is a multiple of 11. we also know that m-n= multiple of 11.
now, if we consider m to be 99 than , 99-n=multiple of 11. can we have any other 2 digit no. which is NOT a multiple of 11 for n in this case ? I think no. so effectively shouldn't the answer be A ? Have I missed something here ?
