MBA.Aspirant wrote:What's the greatest prime factor of 3^38 - 2^28?
What's the approach to this one?
Thanks
There's no way to factor that expression (beyond applying a Difference of Squares once), so there is no way to answer the question without a computer, unless you can somehow use process of elimination with the answer choices. Where is the question from, and what answer choices are given?
There are two related questions that could be answered, in case there's a typo in the post above. If you were asked to find the largest prime factor of
2^38 - 2^28, then you can factor (using the Difference of Squares in the second line):
2^38 - 2^28 = 2^28 (2^10 - 1)
= 2^28 (2^5 + 1)(2^5 - 1)
= 2^28 * 33 * 31
= (2^28)(3)(11)(31)
and the answer is 31.
Or, if you were asked to find the
smallest prime divisor of 3^38 - 2^28, that question can be answered, though you need to use modular arithmetic, something the GMAT doesn't test:
* 3^38 - 2^28 is not divisible by 2 (it's odd, since 3^38 is odd, and 2^28 is even).
* Nor is it divisible by 3, since 3^38 *is* divisible by 3. We'd need to subtract a multiple of 3 from 3^38 to get another multiple of 3, but 2^28 is not divisible by 3.
* Nor is it divisible by 5, since the units digit of 3^38 is 9 and the units digit of 2^28 is 6, so the units digit of 3^38 - 2^28 is 3.
* It *is* divisible by 7. Here we need to use what is known as 'modular arithmetic', which just says that when calculating remainders, we can replace one number with any other that gives the correct remainder. I've never needed this on an actual GMAT question, so the details here are not going to be important to test takers, but since 2^28 = (2^3)^9 (2) = (8^9)(2), and since the remainder is 1 when 8 is divided by 7, the remainder when (8^9)(2) is divided by 7 will be the same as the remainder when (1^9)(2) = 2 is divided by 7, so will be 2. Since 3^38 = (3^2)^19 = 9^19, and since the remainder is 2 when 9 is divided by 7, the remainder when 3^38 is divided by 7 will be equal to the remainder when 2^19 is divided by 7. Now as before, 2^19 = (2^3)^6 (2) = (8^6)(2), which has a remainder of 2 when divided by 7. Thus both 3^38 and 2^28 have a remainder of 2 when divided by 7, and when we subtract we get a remainder of 0, so 3^38 - 2^28 is divisible by 7.
