Numeric Quant questions GMAT

Dear gauravsoni,
I'm happy to respond. First of all, this post really shouldn't be in the "ask GMAT experts" section, but in the math section. Furthermore, you really should post each one separately in its own thread. Finally, of course, these are not in standard GMAT form, because they only have four answer choices each. Nevertheless, since I'm here already, I'll answer these.

#1. what is the remainder when \((x^3+2x^2-9x-8)^2\) is divided by \((x-3)\)?
This relies on a Precalculus-level theorem known as the Remainder Theorem. This theorem is 100% beyond anything the GMAT would test or for which it would hold students responsible. The Remainder Theorem says that, when polynomial function P(x) is divided by (x - k), the remainder must be P(k). Thus, to find the remainder, we just have to substitute x = 3 into the expression:
(x^3+2x^2-9x-8)^2 = (3^3+2(3^2)-9*3-8)^2 = (27 + 18 - 27 - 8)^2 = 10^2 = 100
Answer = (D)
Again, the entire content of this question is 100% beyond anything the GMAT would test.

#2. What is the smallest number by which 31250 should be divided to make it a perfect square?

The number here is a little big for what the GMAT would have folks do without a calculator, but the concept is well within the GMAT's purview. This one is about factors. See this post:
https://magoosh.com/gmat/2012/gmat-math-factors/
First, we need to find the prime factorization of this number
31250 = 10*3125 = 10*25*125 = 2*5*5*5*5*5*5 = 2*(5^6)
For a perfect square, we need an even number of prime factors. WE are all set with factors of 5 --- already six of those. The extra 2 is problem, but if we divide by 2, we are left with (5^6), which is the square of (5^3 = 125).
Answer = (B)

#3. (1555 x 1557 x 1559) / 13 and \((102^2 - 101^2) /17\) , what is the sum of the remainders of these divisions ?
This is, in an number of ways, much much harder than anything the GMAT would ask. Here is a blog about remainders, with examples what the GMAT might ask:
https://magoosh.com/gmat/2012/gmat-quant ... emainders/
To be honest, I do quite of bit math in my head, but I am not sure how to tackle some of the aspects of this question without a calculator!! Here's what I see. Let's deal with the second term first, because this is simpler. For this, we need the Difference of Two Squares formula to factor it. See:
https://magoosh.com/gmat/2012/gmat-quant ... o-squares/
https://magoosh.com/gmat/2013/three-alge ... -the-gmat/
Thus,
(102^2 - 101^2) = (102 + 101)*(101 - 101) = 203*1 = 203
Well, 13 goes into 130, leaving 203 - 130 = 73. Then, 13 goes into 65, leaving 73 - 65 = 8. That's the remainder from the second piece.
Now, the first piece. This is very tricky. Notice that a multiple of 13 is 156, so 1560 is also a multiple of 13, which means 1560 - 13 = 1547 is also a multiple of 13. Thus,
1555 divided by 13 has a remainder of 8
1557 divided by 13 has a remainder of 10
1559 divided by 13 has a remainder of 12
Using the rebuilding the dividend equation, given in that blog on remainders, we can say:
1555 = 13*M + 8
1557 = 13*N + 10
1559 = 13*P + 12
where M, N, and P are some positive integers that won't matter in the solution. When we multiply these three numbers,
1555*1557*1559 = (13*M + 8)(13*N + 10)(13*P + 12)
then, any term that has in it 13*M or 13*N or 13*P is automatically divisible by 13, and will leave no remainder. The only term of that product that possibly could have a remainder is the final term, 8*10*12 = 960, so the question simply becomes: what is the remainder when 960 is divided by 13?
Well, 13 goes into 91, so it goes into 910, which leaves 960 - 910 = 50. Now, 13 goes into 39, which leaves 50 - 39 = 11. The remainder from the first piece is 11.
The sum of the two remainders is 11 + 8 = 19
I've checked my math a few times, but I get this answer, 19, which is not one of the listed answers.

#4. Two different numbers when divided by the same divisor leaves 11 and 12 as remainders respectively and when their sum was divided by the same divisor , remainder was 4. what is the divisor?
Let P & Q be the two original numbers, and let r be the divisor. We again will use the rebuilding the dividend equation, a tremendously useful equation. It's discussed in this blog:
https://magoosh.com/gmat/2012/gmat-quant ... emainders/
From the first two statements, we get
(a) P = r*k + 11
(b) Q = r*p + 12
where k & p are some positive integers. Then, the second sentence gives us
(c) P + Q = r*t + 4
where t is some other positive integer. Add the first two statements, (a) and (b)
P + Q = r*k + r*p + 11 + 12
(d) P + Q = r*(k + p) + 23
Compare (d) to (c) ---- first terms of both is some multiple of r, so the only way the two numbers at the end could differ is if they differ by a multiple of r. Well, 23 - 4 = 19, so it must be the case that r = 19.
Answer = (D)

Does all this make sense?
Mike