When positive integer x is divided by 11, the quotient is y

[rockrock]

When the positive integer x is divided by 11, the quotient is y and the remainder 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?

a) 0
b) 1
c) 2
d) 3
e) 4

How would you approach this problem by picking numbers?

Also does anyone know if there is a Thursdays with Ron session that deals with advanced number properties/divisibility & primes?

[loving.achin]

Hi rockrock,
We are given : x = 11*y + 3 ---(1)
x = 19*(P) + 3 ---(2)
Where P is some number.

=> That x leaves a remainder of 3 when divided by 11
and
x leaves a remainder of 3 when divided by 19

=> (x-3) is a multiple of 11
and
(x-3) is a multiple of 19

=> (x-3) is a multiple of both 11 and 19.

=> Take any number which is both multiple of 11 and 19. Divide it by 11 to get the value of y (i.e. find y from equation (1) above).

Lets take 11*19*200 to be the number. So (x-3) is 11*19*200
=> from equation (1) .....
x-3 = 11*y
=> y = 19 * 200


Now divide y by 19 => No remainder => 'A' is the solution

[rockrock]

Thanks great explanation! The only difference I would've made would be to just multiply 11*19 to get 209 as (x-3). then plug that into y/19 = 11 with no remainder.

Really appreciate it!

[mschwrtz]

Very clever direct approach.

[catennacio]

A shorter version, we have:

x = 11y + 3 (1)
x = 19k + 3, k int (2)
y = 19p + r, p int (3)
r = ?

from (1) & (2) 11y + 3 = 19k + 3
-> 11y = 19k

11 and 19 are prime numbers, so choose y = 19, k = 11 (you can choose any values as long as the left hand side = right hand side, i.e. y = 38, k = 22)

when y = 19: from (3) -> 19p + r = 19 -> p = 1, r = 0 -> no remainder

[RonPurewal]

that looks workable.