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Divisibility & Remainders

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Divisibility & Remainders

by nishant1309 » Tue Nov 08, 2011 4:06 am
Q: A natural number N, when divided by D leaves a remainder of 17. When 3N is divided by D if the remainder is 7, then what is the remainder when 8N is divided by D?

1. 1

2. 18

3. 4

4. Cannot be determined.

I would appreciate explanations. OA after some discussion.
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by user123321 » Tue Nov 08, 2011 4:13 am
N = Dx + 17
3N = Dy + 7

multiply first eq by 3
3N = 3Dx + 51
this should be equivalent to what is given in problem.
So D should be 51-7 = 44 or factors of 44
And D should be 22 or 44 because D > 17.

now N = 44x+17 or N = 22x+17
when 8N div by 44, the remainder is remainder of 17*8 div by 44 = 136 by 44 = 4
when 8N div by 22, the remainder is still the same which is 4

IMO C

edited once.

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Last edited by user123321 on Tue Nov 08, 2011 4:50 am, edited 1 time in total.
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by rijul007 » Tue Nov 08, 2011 4:15 am
N = xD + 17
3N = 3xD + 51 = 3xD + 44 +7

44 is divisible by D
D = either 4,11 or 44
we know that D > 17
D = 44


8N = 8xd + 17*8 = yD + 136



136 divided by 44 gives remainder 4

Option 3
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by nishant1309 » Tue Nov 08, 2011 6:41 am
Thanks folks!

It may look silly but I simplified 136/44 to 34/11 to get the remainder = 1. I shouldn't have simplified... right?
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by amit2k9 » Tue Nov 08, 2011 7:53 am
N= DQ1+17 -----1
3N= 3DQ1+ 51 ----2

but 3N= DQ2+ 7 ---3
equating 2 and 3

3DQ1+51 = DQ2+7 => D(Q2-3Q1)= 44 as D > 17 => D= 22|44

Now, 8N = DQ3 + K ---4
also, 8N= 8DQ1 + 17*8 ---5

equating 4 and 5

DQ3+ K = 8DQ1+ 17*8 => K = D(8Q1-Q3) + 17*8 => D[(8Q1-Q3)+4] for D= 22|44

thus remainder = 4.
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