115k views
3 votes
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.​

1 Answer

4 votes

Answer: x = 32

===============================================

Step-by-step explanation:

When we divide y over x to get a remainder of 29, this means we are dividing over a value greater than 29. This means x > 29.

If we divided over a number 29 or less, then the remainder would be smaller than the divisor.

------------------------------------------

The variable x is some integer. When we divide y over (x/2) we get some other integer. This implies that x/2 itself is also an integer. That tells us that x is some even integer.

So far we know that x > 29 and x is some even number.

------------------------------------------

y divided by x leads to a remainder of 29. This means

y/x = q + 29/x

for some integer q, which is the quotient.

Multiply both sides by x

x*(y/x) = x*(q + 29/x)

y = qx + 29

We'll come back to this later.

------------------------------------------

Now divide y over (x/2) to get a remainder 13. We'll let n be the quotient this time

y/(x/2) = n + 13/(x/2)

which is equivalent to

y = n(x/2) + 13

We can multiply both sides by 2 further getting

2y = 2*( n(x/2) + 13 )

2y = nx + 26

Plug in y = qx+29 from earlier

2(qx+29) = nx + 26

2qx + 58 = nx + 26

Let's group the x terms to one side and everything else on the other side

2qx - nx = 26-58

2qx - nx = -32

-nx + 2qx = -32

nx - 2qx = 32

I multiplied both sides by -1 to turn that -32 into 32

Factoring x from the left side gives

nx - 2qx = 32

x(n - 2q) = 32

We have x as an integer, which is stated in the instructions.

In order to have x(n-2q) = 32 be true, this must mean n-2q is an integer.

We can say n-2q = 32/x, so 32/x is some integer because x is an integer.

------------------------------------------

In short, x and n-2q are both integers. In order for the two integers to multiply to 32, the factors must be smaller than 32 or equal to 32.

So
x \le 32 and
n-2q \le 32

We don't need to worry about the second inequality since all we care about is x.

Recall that earlier we stated that x > 29 and x is even.

Combine this with
x \le 32 and we form the compound inequality
29 < x \le 32

This basically says x is a value from the set {30, 32}

But x = 30 is not a factor of 32. There's no way to have x(n-2q) = 32 be a true equation if x = 30. We must rule x = 30 out.

The only thing left is x = 32. This is the final answer.

------------------------------------------

If x = 32 and y = 29, then we fit the requirements of the problem.

y/x = 29/32 = 0 remainder 29

y/(x/2) = 29/16 = 1 remainder 13

The solution has been confirmed.

User Lei Yang
by
7.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories