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A number when divided by a divisor leaves 22 as the remainder. Twice the number when divided by same divisor leaves 7 as the remainder. Find the divisor?

      (a) 23  (b) 27

      (c) 33  (d) 37

2 Answers

Best answer
1 votes
1 votes
37 is the answer.
It is clear that divisor must be greater than 22.
Suppose P mod Q = 22,
Then we have to find Q.
Since Q is greater than 22, we can also write it as: P mod Q = 22 mod Q
Now multiplying 2 on both sides.
(2P) mod Q = (2 x 22) mod Q = 7.
Since 44 mod Q = 7.
Then (44 - 7) mod Q = 7 - 7 = 0.
I.e. 37 mod Q = 0,
and since 37 is a prime number, only 1 & 37 can divide 37.
I.e. 37 mod 1 = 0, and 37 mod 37 = 0.
Now since Q can not be smaller than 22, so Q must be 37.
1 votes
1 votes
Let x be the number, d the divisor.

x mod d = 22 which means x = nd + 22 for some integer n. -> (1)

Also, 2x mod d = 7 which means 2x = md + 7 for some integer m. ->(2)

2*(1) - (2) => 0 = (2n-m)d + 37, d = 37 / (m-2n)

d must be an integer and 37 is a prime number. So, only way is m - 2n = 1 (as d cannot be 1 here), which gives d = 37.

We can try n = 1 which gives x = 59

n = 2, gives x = 96

Likewise there are infinite solutions for x, but d is only 37.

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