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22 Jul 2011, 20:24
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If M and N are postivie integers that have remainders of 1 and 3, respectively, when divided by 6, which of following could Not be a possible value of M + N?
a) 86 b) 52 c)34 d)28 e)10
why is 10 not the answer, since I use plug in: M = 1*6+1 =7 and N = 1*6+3 =9, they are already the smallest integer, thus 10 should off the limit, why is 10 is the answer, pls tell me. Thank you
a) 86 b) 52 c)34 d)28 e)10
why is 10 not the answer, since I use plug in: M = 1*6+1 =7 and N = 1*6+3 =9, they are already the smallest integer, thus 10 should off the limit, why is 10 is the answer, pls tell me. Thank you
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22 Jul 2011, 22:21
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M = 6m + 1
N = 6n + 3
1 and 1 that you plug in is m and n, it is not the smallest number
condition given is "M, N is positive number" ; not the m,n
so m,n could be 0,1 ; which gives M=1 and N=9
N = 6n + 3
1 and 1 that you plug in is m and n, it is not the smallest number
condition given is "M, N is positive number" ; not the m,n
so m,n could be 0,1 ; which gives M=1 and N=9
22 Jul 2011, 23:09
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I think following can be reason why 10 can not be answer
first M & N are positive integers, so we only know reminder for M is 1 & N is 3.
so M= 6x+1 where x is can digit
N = 6y+3
M+N = 6*(x+y)+4
We clearly get reminder for M+N will be 4. Now first put x=0, & y=0, we get M+N=4, then x=1 & y =0, we get M+N=10
Only A is having reminder of 2. So A is right answer not E i.e. 10
first M & N are positive integers, so we only know reminder for M is 1 & N is 3.
so M= 6x+1 where x is can digit
N = 6y+3
M+N = 6*(x+y)+4
We clearly get reminder for M+N will be 4. Now first put x=0, & y=0, we get M+N=4, then x=1 & y =0, we get M+N=10
Only A is having reminder of 2. So A is right answer not E i.e. 10
