Data Sufficiency

ostrowskiamy wrote:"If N is a positive integer, is the units digit = 0?"
(1) Both 14 and 25 are factors of N
(2) N = (2^5)(3^2)(5^7)(7^6)

My answer = B.

Correct answer = D.

I understand that when multiplying together the prime factors of 14 and 25, I get a number with a units digit of 0. But, what if the numbers was, for instance, NOT the number that I got from multiplying the prime factors together, but something way larger that also is divisible by 14 and 25? Do I have to test multiple variations, or can I just "know" that as long as 14 and 25 are factors, any multiple of a product of their factors will always result in number with a units digit of 0? Does that make sense? :/

Thank you!

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IMPORTANT: A lot of integer property questions can be solved using prime factorization.
For questions involving factors, we can say:
If k is a factor of N, then k is "hiding" within the prime factorization of N

Examples:
3 is a factor of 24 <--> 24 = 2x2x2x3
5 is a factor of 70 <--> 70 = 2x5x7
6 is a factor of 330 <--> 330 = 2x3x5x11
8 is a factor of 56 <--> 56 = 2x2x2x7

Okay, now let's solve the question:

Target question: If N is a positive integer, is the units digit = 0?

Statement 1: Both 14 and 25 are factors of N
If 14 is a factor of N, then 14 is hiding in the prime factorization of N.
That is, N = (2)(7)(?)(?))(?)(?)...
Also, if 25 is a factor of N, then 25 is hiding in the prime factorization of N.
So, N = (2)(7)(5)(5))(?)(?)...
At this point, if we combine a (2) and a (5) in the prime factorization of N, we get 10, which means N is a multiple of 10, which means the units digit of N must be 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: N = (2^5)(3^2)(5^7)(7^6)
Here, we could find the exact value of N, in which case we could definitively determine whether or not the units digit of N is 0
Since we could answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

Cheers,
Brent

First recognize that all multiples of 10 have a zero as their units digit.
So, rather than ask, "Is the units digit of N equal 0?" we could ask, "Is N a multiple of 10?"
Another way to rephrase the question is to ask "Is 10 a factor of N?"
We can even involve prime factorizations here and ask, "Is 10 hiding in the prime factorization of N?"
Since 10 = (2)(5), we could even ask, "Is there are 2 and a 5 hiding in the prime factorization of N?"

At this point, all that matters is whether or not 2 and 5 are both in the prime factorization of N.

If N = (2)(3)(5)(11), then we can be certain that N is a multiple of 10, since we have a 2 and a 5 in the prime factorization.

If N = (2)(2)(2)(5)(5)(7)(13)(29), then we can be certain that N is a multiple of 10, since we have a 2 and a 5 in the prime factorization.

Conversely, if N = (2)(3)(3)(7)(7)(41), then we can be certain that N is not a multiple of 10, since we do not have a 2 and a 5 in the prime factorization.

Does that help?

Cheers,
Brent

ostrowskiamy wrote:"If N is a positive integer, is the units digit = 0?"
(1) Both 14 and 25 are factors of N
(2) N = (2^5)(3^2)(5^7)(7^6)

If the unit's digit of any number is 0, then the number must be a multiple of 10, i.e. multiple of both 2 and 5.

Statement 1: N is multiple of 14 and 25. Hence, N is a multiple of both 2 and 5.

Sufficient

Statement 2: Clearly N is multiple of both 2 and 5.

Sufficient

The correct answer is D.

ostrowskiamy wrote:I understand that when multiplying together the prime factors of 14 and 25, I get a number with a units digit of 0. But, what if the numbers was, for instance, NOT the number that I got from multiplying the prime factors together, but something way larger that also is divisible by 14 and 25? Do I have to test multiple variations, or can I just "know" that as long as 14 and 25 are factors, any multiple of a product of their factors will always result in number with a units digit of 0? Does that make sense? :/

You almost answered yourself.
The number may not be the number you got by multiplying 14 and 25, but something way larger - still it will be multiple 14 and 25. Hence, it'll be multiple of LCM of 14 and 25 = 14*25 = 350. Now any multiple of 350 will always end in 0.

It's important on any DS question that you try to rephrase the question before you jump to the statements. My fellow experts have both explained well why the REAL question here is "does N have a 2 and a 5 as factors?" Starting with that rephrased question makes it much easier to analyze the statements!

I also found something telling in your question:

I understand that when multiplying together the prime factors of 14 and 25, I get a number with a units digit of 0.

If you were multiplying these numbers together, but not analyzing why, you were doing too much math! Whenever you see the language of divisibility (factor, multiple, divisor, etc), it never helps to make those numbers bigger! You should always instead make them smaller - break them down into prime factors, and ask yourself what those prime factors tell you.

Do I have to test multiple variations, or can I just "know" that as long as 14 and 25 are factors, any multiple of a product of their factors will always result in number with a units digit of 0?

With Number Properties, testing numbers is usually an inefficient approach; think about the properties of the numbers instead. You should be asking yourself "how can I know if a number ends in a 0?" If you start with that question, you'll realize "if I have a 2 and a 5, I'll know."