IanStewart wrote:
firas92 wrote:
Three positive integers a, b and c leave the same remainder y when divided by x, where x is a prime number greater than 3 and y is an integer greater than 1. What is the remainder when x is divided by y?
(1) c + 3 when divided by x yields an integer
(2) a + x - 2 when divided by x yields an integer
Statement 1 tells us c is 3 less than some multiple of x. To find a remainder, we'd want to know how much greater c is than a multiple of x, so Statement 1 is not very useful.
If, as Statement 2 tells us, a+x-2 is divisible by x, then a-2 must be divisible by x as well (since subtracting x from a multiple of x will always give us another multiple of x). And if a-2 is divisible by x, then a must be 2 greater than an exact multiple of x, which is another way of saying (since x > 2) "the remainder is 2 when we divide a by x". So y = 2, and when we divide the odd prime x by 2, we must get a remainder of 1, and Statement 2 is sufficient alone.
I have no idea what the unknown "b" has to do with anything.
I will just add to this brilliant explanation by Ian.
Why statement 2 is sufficient is as following-
Given, a= k1*x + y --------- (1)
Statement 2 gives:
a + x - 2 = k2*x
Rearranging,
a - 2 = (k2 - 1)*x
a - 2 = Kx
Substitute value of a from one above,
k1*x + y - 2 = Kx
Rearrange,
y = x(K - k1) + 2
y = x*C + 2. \Here, K, k1, k2, C are all constants.
Hence, this is the form:
Dividend = Divisor * Quotient + Remainder
Where, Remainder = 2. Your answer.