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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.
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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.
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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.
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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.
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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 [tex]x \le 32[/tex] and [tex]n-2q \le 32[/tex]
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 [tex]x \le 32[/tex] and we form the compound inequality [tex]29 < x \le 32[/tex]
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.
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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.
