SOLUTION: Consider the integers x and y. When y is divided by x the remainder is 29. When y is divided by x/2 the remainder is 13, Determine X

Question 1170970: Consider the integers x and y. When y is divided by x the remainder is 29. When y is divided by x/2 the remainder is 13, Determine X
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

x and y are integers
Let's say they are also positive, so x > 0 and y > 0.
Dividing y over x leads to some quotient, which we can call some integer q, and a remainder 29.
This means, y/x = q + 29/x which is equivalent to y = qx + 29 after we clear out the fractions (multiply both sides by x)

Let p be another integer such that
y/(x/2) = p + 13/(x/2)
so we're dividing y over (x/2) and getting some quotient p and remainder 13.
Multiplying both sides by the quantity (x/2) will clean things up a bit to get
y = p(x/2) + 13

Multiplying both sides by 2 gets us
2y = px + 26
The fractions are completely gone at this point.

Now plug in y = qx + 29
2y = px + 26
2(qx+29) = px + 26
2qx+58 = px+26

Let's collect the x terms to one side
Get everything else to the other side
2qx+58 = px+26
58-26 = px-2qx
32 = px - 2qx
px - 2qx = 32
x(p - 2q) = 32

The values p and q are integers. A linear combination of integers leads to another integer result.
So if p and q are integers, then so is p-2q.

If p-2q wasn't an integer, then x(p-2q) wouldn't be an integer since x is already an integer.
In other words, if p-2q wasn't an integer, then it wouldn't be possible to have x(p-2q) = 32.

Because the two integers x and p-2q multiply to 32, we know that x and p-2q are factors of 32.

Consequently, this means x is at most 32. If x were any larger than 32, then p-2q would have to be some positive number less than 1.

Let's list out the factors of 32:
1,2,4,8,16,32

The values pair up like so:
1*32 = 32
2*16 = 32
4*8 = 32

Of the list of factors, we need to see which have the property that y/x has remainder 29, where x takes on one of the values from {1,2,4,8,32}

It turns out that we can eliminate 1,2,4, and 8 from the list since they are too small. If x were say x = 4, then y/x = y/4 could only lead to remainders 0, 1, 2, or 3. The largest possible remainder is 1 less than the divisor. To generalize that, we can say the largest possible remainder of y/x is r = x-1. So that's why we can rule out 1,2,4 and 8.

The only thing left is 32.

Let x = 32.

Let's also make y = 29 so that y/x has remainder 29. We could pick infinitely many other y values, but y = 29 is the smallest positive one we can pick.

y/x = 29/32 = 0 remainder 29
Any time the numerator is smaller than the denominator, the remainder is automatically just the numerator. Picture having 29 cookies and 32 friends at a party. There isn't enough to have each friend have 1 full cookie, so there are 29 left over (remainder).
The first condition has been met.

Let's see if y = 29 works with the other equation
First plug in x = 32 and simplify
y/(x/2) = y/(32/2) = y/16
Then plug in y = 29
y/16 = 29/16 = 1 remainder 13
We have 29 cookies divided among 16 friends. Each person gets 1 full cookie, which leaves 29-16 = 13 left over. So this pair of x and y values satisfies the second condition as well.

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Answer: x = 32

Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.

It was posted AND SOLVED several times (at least, 6 times) recently
(in one-two weeks) at this forum.

For the solution, see the lesson
    - Miscellaneous problems on divisibility numbers, Problem 5
at this site.

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