Question 36; Page 39; Number properties

[ashishparmar01]

Q; If x and y are positive integers and x/y has a remainder of 5, what is 5, what is the smallest possible value of xy?

The answer explanation says that x should have a smallest value of 5 and I somehow fail to understand the reasoning. Can someone please explain as to why x has the smallest value of 5?
if we x/y with remainder 5, as in the question and replace with 5/6 (x=5 and y=6, as in explanation), the remainder would not be 5.

[pratik.mehta]

Q; If x and y are positive integers and x/y has a remainder of 5, what is 5, what is the smallest possible value of xy?

The answer explanation says that x should have a smallest value of 5 and I somehow fail to understand the reasoning. Can someone please explain as to why x has the smallest value of 5?
if we x/y with remainder 5, as in the question and replace with 5/6 (x=5 and y=6, as in explanation), the remainder would not be 5.

I have the same question, can somebody answer this?

[esledge]

We want to minimize xy, so we'll minimize both x and y. (That's a bit of a simplification, but it works here.)

If x/y has a remainder of 5, we know that:
(1) x has to be 5 more than a multiple of y. Mathematically, x = ny + 5, where n = an integer.
(2) y has to be an integer greater than 5.

Why rule (2)? Consider some examples.
29/6 leaves a remainder of 5 because 29 = 4*6 + 5
17/6 leaves a remainder of 5 because 17 = 3*6 + 5
25/5 leaves a remainder of 5 because 25 = 4*5 + 5. NO! That's wrong.

The remainder is what is left over after the divisor (denominator) has gone into the dividend (numerator) as many times as possible. If the divisor is 5, and the "remainder" is 5 or more, that implies that the divisor could go into the dividend at least once more. Thus, the remainder must ALWAYS be less than the divisor. For a divisor of 5, possible remainders are 0, 1, 2, 3, and 4 (not 5!). For a divisor of 6, possible remainders are 0, 1, 2, 3, 4, and 5. The minimum possible divisor in this problem is 6.

So, we know that:
(2) minimum y is 6.
(1) minimum x is 5 more than a multiple of y = 6. Thus, x = 5, 11, 16, 21, 26. The minimum x must be 5.

Minimum xy = (5)(6) = 30.

[ashishparmar01]

Thanks for your response Emily. Appreciate the same.

[JonathanSchneider]

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