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by karthikpandian19 » Thu Dec 15, 2011 11:50 pm
If M and N are positive integers that
have remainders of 1 and 3, respectively,
when divided by 6, which of the
following could NOT be a possible
value of M+N?
(A) 86
(B) 52
(C) 34
(D) 28
(E) 10
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by ankush123251 » Fri Dec 16, 2011 12:02 am
Quickest way as i believe is to try out the numbers directly...
Numbers that leave 1 and 3 as remainders on dividing by 6 are:
Remainder 1 = 1,7,13,19,25... don't stretch this as maximum sum is 86 in options
Remainder 3 = 3,9,15,21,27 ......
Try you the options with minimum i.e try getting 10,28 34,52.
As you can see 86 which is option A is answer.
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by karthikpandian19 » Fri Dec 16, 2011 12:17 am
Is there a logic way with the possibility of NUMBER PROPERTIES & DIVISIBILITY RULES?????
ankush123251 wrote:Quickest way as i believe is to try out the numbers directly...
Numbers that leave 1 and 3 as remainders on dividing by 6 are:
Remainder 1 = 1,7,13,19,25... don't stretch this as maximum sum is 86 in options
Remainder 3 = 3,9,15,21,27 ......
Try you the options with minimum i.e try getting 10,28 34,52.
As you can see 86 which is option A is answer.
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by GmatMathPro » Fri Dec 16, 2011 2:57 am
karthikpandian19 wrote:If M and N are positive integers that
have remainders of 1 and 3, respectively,
when divided by 6, which of the
following could NOT be a possible
value of M+N?
(A) 86
(B) 52
(C) 34
(D) 28
(E) 10
M leaves a remainder of 1 when divided by 6, so M=6a+1 for some positive integer, a. Likewise, N=6b+3 for some positive integer, b.

M+N = 6a+1 + 6b+3 = 6(a+b) + 4. This tells us that M+N is 4 more than a multiple of 6.

Note that all of the choices are 4 more than a multiple of 6 except for 86.
Pete Ackley
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by ankush123251 » Fri Dec 16, 2011 3:11 am
Yes logically its like this,

M = 6x + 1
N= 6y + 3

M + N = 6(x + y) + 4.
Now check the options to fit in above format.
86 does not fit thus the answer.
Note in the above method also you try and first options thus the first method i posted is quicker.
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by batmannavneet » Fri Dec 16, 2011 6:19 am
M leaves remainder of 1, which means M-1 is divisible by 6
Similarly, N-3 is divisible by 6

Therefore, M+N-4 is divisible by 6.

Trying out the options, A fits the answer
karthikpandian19 wrote:If M and N are positive integers that
have remainders of 1 and 3, respectively,
when divided by 6, which of the
following could NOT be a possible
value of M+N?
(A) 86
(B) 52
(C) 34
(D) 28
(E) 10
Top
Join the discussion
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