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BrentGMATPrepNow
When x is divided by a, the quotient is b and the remainder is c. If a, b, c and x are different positive integers, what is the smallest possible value of x?

A) 4
B) 5
C) 6
D) 7
E) 8

------ASIDE---------------
There's a nice rule that says, "If N divided by D equals Q with remainder R, then N = DQ + R"
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3
---------------------------

From the given information, we can write x = ab + c
We want to minimize the value of x.
Also keep in mind that a, b, c and x are different positive integers.

So let's test the three smallest positive integers, 1, 2 and 3.

If a = 1, b = 2 and c = 3, then x = ab + c = (1)(2) + 3 = 5
If a = 2, b = 1 and c = 3, then x = ab + c = (2)(1) + 3 = 5 [notice that reversing the values of a and b resulted in the same value for x, since ab = ba]
If a = 1, b = 3 and c = 2, then x = ab + c = (1)(3) + 2 = 5
If a = 2, b = 3 and c = 1, then x = ab + c = (2)(3) + 1 = 7

So, 5 is the smallest possible value of x.

Answer: B
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BrentGMATPrepNow
BrentGMATPrepNow
When x is divided by a, the quotient is b and the remainder is c. If a, b, c and x are different positive integers, what is the smallest possible value of x?

A) 4
B) 5
C) 6
D) 7
E) 8

------ASIDE---------------
There's a nice rule that says, "If N divided by D equals Q with remainder R, then N = DQ + R"
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3
---------------------------

From the given information, we can write x = ab + c
We want to minimize the value of x.
Also keep in mind that a, b, c and x are different positive integers.

So let's test the three smallest positive integers, 1, 2 and 3.

If a = 1, b = 2 and c = 3, then x = ab + c = (1)(2) + 3 = 5
If a = 2, b = 1 and c = 3, then x = ab + c = (2)(1) + 3 = 5 [notice that reversing the values of a and b resulted in the same value for x, since ab = ba]
If a = 1, b = 3 and c = 2, then x = ab + c = (1)(3) + 2 = 5
If a = 2, b = 3 and c = 1, then x = ab + c = (2)(3) + 1 = 7

So, 6 is the smallest possible value of x.

Answer: C

BrentGMATPrepNow don't you mean (B) 5 is the correct answer

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Hoozan
BrentGMATPrepNow don't you mean (B) 5 is the correct answer

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Oops, not sure where 6 came from :)
All fixed!

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