# GMAT Question of the Day - Data Sufficiency - Number Properties/Arithmetic

If p and q are positive integers and p/q is a terminating decimal is q/p an integer?

(1) (q^1/5)/10 is an integer

(2) p = 1024

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If p and q are positive integers and p/q is a terminating decimal is q/p an integer?

(1) (q^1/5)/10 is an integer

(2) p = 1024

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If v is greater than or equal to 1 is r/v a terminating decimal?

(1) v - 1 is divisible by 3

(2) r = 99

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A terminating decimal broken down to its simplest form only has 2's and 5's in the denominator. That's because fractions are based on 10's and 2's and 5's are the only prime factors of 10. Some examples of terminating decimals are 1/2, 2/5, 1/4, 1/8, 3/20, 7/100. Examples of non-terminating decimals are 1/3, 4/6, 1/9, 2/45, 5/70.

Statement (1) tells us that v could 4, 7, 10, 13, 16, 19... Any number that has a remainder of 1 when divided by 3. 10 works but 7 doesn't so this is insufficient.

Statement (2) only tells us about the numerator so it must be insufficient.

Putting them together, a value of 10 for v will make a terminating decimal but a value of 7 for v will still result in a non-terminating decimal. Insufficient.

When the positive integer x is divided by the positive integer y the result can be expressed as a terminating decimal with J, K, and M representing the tenths, hundredths, and thousandths digits respectively. What is the value of J + K + M?

(1) x < 10

(2) y = 56

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Let's have a quick review on GMAT terminating decimal questions. The idea is that in order for a decimal to terminate, the fractional equivalent can only have 5's and 2's in the denominator. 1/2, 1/4, 1/5, 1/8, and 1/10 are examples of this. So we know that the denominator here (after any simplifying with the numerator) must have only 5's and 2's. Many GMAT tutoring students wonder how one is supposed divine this in advance. Good question.

Statement (1) x can be any of the digits 1-9. This doesn't tell us what the decimal will be when x is divided by y. y could be 8 and x could 1 or 3 each yielding a terminating decimal with different answers for the sum of J, K, and M. Insufficient.

Statement (2) If y is 56 then we know that x must have at least one 7 so that the 7 in 56 can be eliminating leaving only and 8 in the denominator (so that the decimal will terminate). However we don't know anything else about x. x could be 21 or 7. Each number yields different values for J, K, and M. Insufficient.

Statement (1) and (2) Putting bother statements together leaves only one possible value for x: 7 So the fraction is 1/8 or .125. J + K + M = 8.

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