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Please explain

by radhika1306 » Fri Aug 03, 2007 3:16 pm
19. 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


ans A

could some one please explain
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by davidnodine » Fri Aug 03, 2007 7:48 pm
19. 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

The easiest way is to pick two numbers that fit the description of M & N. So M=7 and N=9. 7+9= 16 which is divisible by 6 with a remainder of 4. So the answer has to fit that same rule. 10/6=1r4, 28/6=4r4, 34/6=5r4, 52/6=8r4. 86/6= 14r2. So it doesn't fit the rule A is your answer. Try it with other numbers: 599+597= 1196/6 = 182r4
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by givemeanid » Sat Aug 04, 2007 8:48 am
M = 6x + 1
N = 6y + 3

M + N = 6(x+y) + 4
(M+N) - 4 = 6(x+y) = Multiple of 6.

Look at answer choices:
A. 86 - 4 = 82. Not a multiple of 6. CORRECT.
B. 52 - 4 = 48 = 6*8
C. 34 - 4 = 30 = 6*5
D. 28 - 4 = 24 = 6*4
C. 10 - 4 = 6
So It Goes
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by radhika1306 » Tue Aug 07, 2007 4:31 am
thanks the reply really helped
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