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Two different numbers when divided by the same divisor left

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Two different numbers when divided by the same divisor left [#permalink] New post 17 May 2003, 07:30
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Two different numbers when divided by the same divisor left a
remainder of 11 and 21 respectively and when the sum of the numbers
was divided the remainder was 4. what is the divisor??
a) 36
b) 28
c) 14
d) 9
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 [#permalink] New post 18 May 2003, 04:10
Given that:

R=kX+11
Q=mX+21
R+Q=nX+4
**************************************
Add P+Q=(k+m)X+32, which is equal to nX+4

32-4=28
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 [#permalink] New post 20 May 2003, 06:53
just an extenion to your solution...
it could be either of 28 or 14 but since the remainder of 21 implied the divisor should be larger than 21 the anwer is 28
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 [#permalink] New post 30 Jun 2004, 10:01
Can someone please explain the solution to this problem ?
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 [#permalink] New post 30 Jun 2004, 11:49
Hi Dookie,

let the first no be x.

x/n = (an integer) + 11 = u + 11 - equation 1

similarly

y/n = (an integer) + 21 = v + 21 - equation 2


(x+y)/n = (an integer) + 4 = w + 4 - equation 3

eq1+eq2 = eq3 - equation 4


simplifies
n(u+v) + 32 = nw + 4 - here u,v,w are integers.

this means that n is a factor of 28.........bcoz n<=32-4 or n<=28

the factors of 28 are 14,7 etc.

since we have seen a remainder of 21 this means that n>21 , therefore n=28.

- ash
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ash
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 [#permalink] New post 30 Jun 2004, 12:09
I substituted the numbers and got the ans as 28...

the easiest way with substitution...take a number that s more than 21 cos the least number that can divide 21 perfect =21 ...

take /37/28= reminder 11 and add 10 to this reminder and that wud get U 47 which gives a reminder of 21...

add both 37+47 = 88
now 88/28 leaves a reminder 4...

hope that helps!

Have fun :-D
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  [#permalink] 30 Jun 2004, 12:09
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