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value of N

by anant03 » Tue Sep 01, 2015 8:18 am
If N is a positive three-digit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N?

(1) The tens digit of N is 5.

(2) The units digit of N is 5.

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by DavidG@VeritasPrep » Tue Sep 01, 2015 8:41 am
If N is a positive three-digit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N?

(1) The tens digit of N is 5.

(2) The units digit of N is 5.

Statement 1: if we know the tens digit is 5, we have some number, A5B. So now the question is can we determine A and B? We're told that each digit of N is a factor of N. So we know that 5 is a factor of the number. This means that the number must end in 0 or 5. But it can't end in 0 because each digit must be a factor of N, and 0 is never a factor of anything. So N must end in 5. Now we have A55. We know A must be a factor of the number. If can't be 1, because we're told that N is greater than 200.

- It can't be 2, because 2 is not a factor of 255. (Any odd number will not have any even factors)
- It can't be 3, because 3 is not a factor of 355
- 4? Nope
- 5? Yes, this will work. 5 is a factor of 555. So 555 is a possibility for N.
- 6? Nope
- 7? Not a factor of 755.
- 8? Nope
- 9? Not a factor of 955.

So we know the number must be 555. 1 alone is sufficient.

Statement 2: Now we're told that units digit is 5, so we're starting with XY5. Well, we already know that 555 is one possibility, so the question is, can we find any other value that will work? And we can. 515 will also work. Because there are multiple possibilities, statement 2 alone is not sufficient.

Answer is A
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by [email protected] » Tue Sep 01, 2015 9:07 am
Hi anant03,

This question is thick with Number Property rules and Arithmetic patterns (knowing the Rule of 3 and the Rule of 9 would be helpful when working through this prompt).

We're told that N is a positive 3-digit number greater than 200 AND that each DIGIT of N is a FACTOR of N. We're asked for the value of N.

Fact 1: The tens digit is 5

From this Fact, we have.....

_5_ as our 3-digit number.

Since 5 MUST be a factor of this number, we would be restricted to numbers that END in 5 or 0. However, 0 is NOT a factor of ANY number, so the 3-digit number CANNOT end in 0. This leaves us with...

_55

Since we now have an odd number, the first digit CANNOT be even (since even numbers cannot divide evenly into odd numbers). We're also told that the number must be greater than 200, so there are only 4 options to consider:

355, 555, 755, 955

3 does not divide into 355
7 does not divide into 755
9 does not divide into 955

The only number that fits this description is 555.
Fact 1 is SUFFICIENT

Fact 2: The units digit of N is 5

From this Fact, we have.....

_ _5 as our 3-digit number.

From our prior work in Fact 1, we know that 555 is a possible value and that the other two digits MUST be odd.

315 is evenly divisible by 3 and 1 and 5, so this is another possible value for N.
Fact 2 is INSUFFICIENT

Final Answer: A

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by Max@Math Revolution » Wed Sep 02, 2015 9:25 pm
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.


If N is a positive three-digit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N?

(1) The tens digit of N is 5.

(2) The units digit of N is 5.
==>Since we have 3 digit integer in the original condition, we have 3 variables and we need 3 equations to match the number of variables and equations. Since we have 1 each in 1) and 2), E is likely the answer.

Using 1) & 2), n=555 and is a unique answer. Therefore C may seem like the answer. However, since this is an integer problem, considering common mistake type 4(A) (If it's easily C, consider A, B), we can look at this again. 1) gives us n =555 as a unique answer, but 2) gives us n= 555, 515 which is not unique, and therefore not sufficient.
Therefore A, not C, is the answer.
These problems are 495,051 level problems that are related to variable approach method and common mistake types,


If you know our own innovative logics to find the answer, you don't need to actually solve the problem.
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by Matt@VeritasPrep » Fri Sep 04, 2015 6:29 pm
We could also break it down logically, using a few number properties.

S1:

Since 5 is a factor of our number, our number must end in 0 or a 5. But if it ends in 0, then 0 is a factor of N, and this is impossible: 0 isn't a factor of any number. So our number ends in a 5 as well.

Now let's consider our possible integers: 255, 355, 455, 555, 655, 755, 855, and 955. 2, 4, 6, and 8 can't be factors of anything ending in 55, since they're even and 55 is odd. 3 and 9 aren't factors of 355 and 955, respectively, since 3+5+5 isn't divisible by 3, etc. Hence we only have 555 and 755 left. Since 756 = 700 + 56 = 7*100 + 7*8, 756 divides by 7. So 755 CAN'T divide by 7, and 555 is the only option; SUFFICIENT.

S2::

The units digit is 5. This implies that our number, whatever it is, must divide by 5 ... but we already knew that! So there's not much else we can do. (If you're skeptical, consider a few possibilities: 515 and 555, both of which could be the value of N. Since we have more than one value, this is INSUFFICIENT.)
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