How many factors does positive integer z have?

5. Divisibility/Multiples/Factors

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Statement 1:

First, z must have a factor of 5 and 7, and z is a multiple of 8 so we should start from z = 5*7*8. Next we should question if there are any other z's that satisfy this criterion.

If we add any additional factors into z, for example z = 5*7*8*3, then the greatest integer that divides both would be 8*3, not 8 anymore. Thus we cannot add any additional factors at all and we are forced to have z = 5*7*8. Since z is a fixed value we can determine the number of factors, sufficient.

Statement 2:

This is saying the least common multiple of z and 14 is 280.

Method 1: \(280 = 14*20 = 2*7*2*2*5 = 2^3 * 5 * 7\).
14 = 2*7 only contributes the 7 for the LCM, we need a \(2^3\) and a \(5\) from z. Then we can have \(z = 2^3 * 5\) or \(z = 2^3*5*7\). Insufficient.

Method 2: Using the formula \(LCM = \frac{a*b}{GCF}\), we have \(280 = \frac{z*14 }{ GCF}\). We can try a couple GCF's that make sense in the context of b = 14, such as GCF = 1, GCF = 2, GCF = 7, and GCF = 14. Then we could receive values such as z = 20 (reject becasue the GCF of 20 and 14 would be 2), z = 40 (accept), z = 140 (reject since the GCF of 140 and 14 is 14), z = 280 (accept).
Thus z = 40 or z = 280 are both acceptable, insufficient.

Ans: A

Hi,

From the first statement you surely know that 2,5,7,4,8 are factors of z. Z can be=8X5X7X3 & still z/5 or z/7 will be integers. We cannot say that from 1st statement that we know all the factors. If 8 is the largest factor then we are surely missing out on 3. What about 3 ? What if it is too a factor? Statement 1 and 2 combine will give us a firm answer that 5 X 7 X 8 are the only factors.

Please advise if the thought process is correct.

Yes, I have the same problem here. Why don't we consider 3. If statement 1 & 2 are combined then only 3 is out. Please suggest.

Posted from my mobile device

i couldn't get it under 3 min however with the following in mind
A - provides that it's 5*7*8 through this we can make out the factors sufficient
B- no it could be 20 and 14 and any other combination clearly insufficient
therefore IMO A

(1) z/5 and z/7 are integers and the greatest integer that divides them both is 8.
Whatever be the integers we try to substitute with 8 will yield a integer bigger than 8 dividing the integers Z/5 , Z/7

Clearly sufficient and the factors are 1,2,5,7,4 ruling out all negative terms

(2) The smallest integer that is divisible by both z and 14 is 280.
the number could be multiplied by 3*4*5*6 and so on the , there is no outside limit

Clearly insufficient

Therefore IMO A

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