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If M and N are positive integers that have remainders of 1 [#permalink]

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19 Nov 2009, 13:43

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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?

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

Guys i am going mad for this question according to me the answer should be E but OA is A please explain how come it cannot be 10

M=6p+1, where p is integer >=0, so M can be 1, 7, 13, etc. N=6q+3. where q is integer >=0, so N can be 3, 9, 15, etc.

M+N=6(p+q)+4, hence M+N is multiple of 6 plus 4 = 10, 16, 22, 28, 34, etc.

Only answer which is not of this type is 86, 86=13*6+2

but still can you give me the the value of M and N which will add up to 10. If you check the other way round the least numbers are 7 and 9 which adds up to 16 so how come 10 is possible?

but still can you give me the the value of M and N which will add up to 10. If you check the other way round the least numbers are 7 and 9 which adds up to 16 so how come 10 is possible?

The problem in your reasoning is that you are not considering the values of M and N when, p and q are 0. If you do you'll get the least values: M=1 and N=3. So if you take the value of M when p=1 and value of N when q=0 you'll get:

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?

This question is fairly easy, but I don't get it, why 10 is not false. When divided by 6, the smallest numbers that have a remainder of 1 and 3, respectively, are 7 and 9. 7+9=16. So I don't get it, why 10 is correct.

Can someone please explain. I hope I don't have a error in thinking about this.
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Consider giving kudos, if you like my thread!

FYI: Questions are from the workbook of my prep course.

The smallest numbers are 1 and 3: 1 div by 6 = 0 with remainder if 1 3 divided 6 =0 with remainder of 3

so 10 could be 7+3 => 7 gives a remainder of 1 and 3 gives a remainder of 3 when divided by 6 or 1+9 => 9 gives a remainder of 3 and 1 gives a remainder of 1 when divided by 6

When m and n are divided by 6, their remainders are 1 and 3.

In order to have a remainder of 1, M must be 7,13,19, 25 and so on. We can set up an equation: 6*n + 1 for every number M has to be. The same is with N. Their we have 9,15,21,27 etc. = 6*n+9.

If we sum it up, we have: 16+12n.

Now we can test every solution and the one that doesen't work is solution A.

Re: If M and N are positive integers that have remainders of 1 [#permalink]

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24 Apr 2012, 04:54

I did it this way M=6p+1 N=6q+1

We need M+N=6p+1+6q+3=6(p+q)+4

Pick numbers for p & q Since it is an addition of p & q and the answer to this expression should be an integer (because all the numbers being added are integers), we just need to choose values so that we get integer multiples of 6 so p+q=0 ; M+n = 4 p+q=1 ; M+N = 10 P+q=2 ; M+N = 16

and so on, so basically you get something like - 4,10,16,22,28,34,..... all the other options were turning up.

Then I directly tried p+q=12 because it was closer to 86 for the first option, i got 76, then tried with 13 - got 84, 14 got 90.. no 86.

Your approach was not wrong but a little cumbersome. Instead of picking values for p and q and trying to get to the options, pick the options and find out whether they suit this format.

52 = 6*8 + 4 so p+q = 8. Hence 52 can be the value of M+N Similarly check for other options. At every step, you get closer to the solution else you could end up waiting for a long time before you get 4 of the 5 options.
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Re: If M and N are positive integers that have remainders of 1 [#permalink]

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25 Apr 2012, 10:18

1+3=4 Using six times table to recognize the the nearest integer uner each option i.e.- 6*4 =24. Whichever answer does not have four as a remainder wins the selection.

Re: If M and N are positive integers that have remainders of 1 [#permalink]

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19 Oct 2012, 00:22

sagarsabnis 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 & N are divided by 6 : R =1 & 3 resp

M+ N divided by 6 : R 1+ 3 = 4

So divide the options by 6 .. & R should be 4.

Only ..86 has a remiander of 2 .....instead of 4..

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

@bunuel..why are we considering the integers 1 and 3 in the list of integers when the stem question says that when each of the numbers is divided by six,the remainder is 1 and 3 respectively..when 1 is divided by six am sure we would not get 1 as a remainder neither would we get 3 as a remainder when the integer 3 is divided by 6

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

@bunuel..why are we considering the integers 1 and 3 in the list of integers when the stem question says that when each of the numbers is divided by six,the remainder is 1 and 3 respectively..when 1 is divided by six am sure we would not get 1 as a remainder neither would we get 3 as a remainder when the integer 3 is divided by 6

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That's not correct.

When a smaller integer is divided by a larger integer, the quotient is 0 and the remainder is the smaller integer. For example, 7 divided by 11 has the quotient 0 and the remainder 7 since \(7=11*0+7\)

Hence, 1 divided by 6 yields the remainder of 1 and 3 divided by 6 yields the remainder of 3.

Re: If M and N are positive integers that have remainders of 1 [#permalink]

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17 Aug 2014, 03:09

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Re: If M and N are positive integers that have remainders of 1 [#permalink]

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18 Aug 2015, 18:06

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Re: If M and N are positive integers that have remainders of 1 [#permalink]

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21 Sep 2015, 02:13

Assume M = 6a+1 and N = 6b+3 M+N = 6(a+b)+4 Hence answer should be such that when subtracted by 4, it should be divisible by 6. 86-4 = 82 is not divisible by 6. Hence A is the answer.

Re: If M and N are positive integers that have remainders of 1 [#permalink]

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26 Sep 2016, 18:24

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