Number Prop Question Bank

[Gen]

Hi All,

Can you please explain the set up on this (a little confused on how to set up the equation with the remainder).

Thanks,

Gen

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?

Answer: 0

[Nov1907]

Let us say The quotient when x is divided by 19 is integer k. So we can write 2 equations from the given information:

x = 11*y+3 (eqn 1)

x = 19 *k+3 (eqn 2)

subtracting we have 19k-11y=0. This implies y=19 *(k/11) .

Since y and k are integers and 19 is a prime number and not divisible by 11 we have y = 19* integer. So y is a multiple of 19 and the remainder when y is divided by 19 is 0.

[RonPurewal]

Here's a solution that might make a little more sense than the one posted (which is entirely correct, but will most likely be inaccessible to most readers of this forum because it's way too number-theoretical).

You can translate the two remainder statements as follows:
* X is 3 more than a multiple of 11.
* X is 3 more than a multiple of 19.
- If we take these two statements together, we know that the number (X - 3) is a multiple of 11 AND a multiple of 19. Also, THIS number, X - 3, is Y times 11 (because that's what's left over when you get rid of the remainder).
So, 11 times Y is a multiple of 11, but also a multiple of 19. This means that Y itself must be a multiple of 19 (because 11 and 19 have no common factors that could threaten to ruin the show).

Hope that helps.

[Guest]

Is this also an accurate assessment?

x = 11*y+3 (eqn 1)

x = 19 *k+3 (eqn 2)

11y + 3 = 19k + 3
11y = 19k

11y/19 = k


Since k is a integer there is no remainder?

[StaceyKoprince]

Almost! For this to work, you'd have to have some representation of x in the numerator and show that dividing by 19 still gets us an integer. k is an integer, so that part's okay, but what you have in the numerator is not a representation of x. x = 11y + 3. x does not equal 11y.