You are here:

Advanced Math/AMC Questions


What is the answer to question one and four?

Problem 1
Problem 1  
Besides the three problems with typos, your approach to these problems isn't really on the right track. I would recommend you read Pólya's How to Solve It for some general exposition about solving problems in mathematics, and you can even refer to this page for more information about the AMC in general.

Problem solving is really more about understanding the fundamentals of mathematics and learning to solve the problems in logical, reasonable, effective ways. There is no formula for every problem -- that would produce winners for the AMC based on memorization and plug-and-chug, not on mathematical ability.

1. If 2(x^2)-2xy+(y^2)=289, where x and y are integers and x is greater or equal to 0, what is the number of different ordered pairs (x,y) which satisfy this equation?

I've thought about completing the square and such methods but I didn't seem to get anywhere with them.

As far as I can tell, completing the square won't work here.

First, you can look to the computer. Obviously at the AMC you can't do this, but you can check by computer to see what things look like. Plotting the curve and looking at the grid of integer points in the plane, you can identify 12 such points (see included image). The question, then, is how do you do this if not by computer?

Frankly, the approach I'd recommend is first to rewrite the equation:

2x^2 - 2xy + y^2 = x^2 + (x^2 - 2xy + y^2) = x^2 + (x-y)^2 = 17^2

These are Pythagorean triples with hypotenuse 17 and legs x and |x-y|. The standard parametrization of Pythagorean triples gives:

|x| = k (m^2 - n^2)
|x-y| = 2kmn
17 = k (m^2 + n^2)

or reverse the legs:

|x| = 2kmn
|x-y| = k (m^2 - n^2)
17 = k (m^2 + n^2)

Now, there are a few cases you need to analyze separately here. Start by noticing that since 17 is prime, either k=17 or m^2 + n^2 = 17 (and whichever is not 17 has to be 1).

Case 1: k = 17, m^2 + n^2 = 1

      Note: m=1, n=0 in this case.

      Case 1a.

            |x| = 17 (1^2 - 0^2) = 17
            |x-y| = 2 * 17 * 1 * 0 = 0

            Case 1a-i. x=17

                  Then y=17 so that x-y=0.

            Case 1a-ii. x=-17

                  Then y=-17 so that x-y=0.

      Case 1b.

            |x-y| = 17 (1^2 - 0^2) = 17
            |x| = 2 * 17 * 1 * 0 = 0

            x=0 immediately means y=17 or -17.

Case 2: m^2 + n^2 = 17, k=1

      Note: m=4, n=1 in this case.

      Case 2a.

            |x| = 1 (16 - 1) = 15
            |x-y| = 2 * 1 * 4 * 1 = 8

            Case 2a-i: x = 15

                  |15-y|=8 has solutions y = 23 and y=7.

            Case 2a-ii: x = -15

                  |-15-y| = |y+15| = 8 has solutions y = -7 and y = -23.

      Case 2b:

            |x-y| = 1 (16 - 1) = 15
            |x| = 2 * 1 * 4 * 1 = 8

            Case 2b-i: x=8

                  |8-y|=15 has solutions y = -7 and y = 23.

            Case 2b-ii: x=-8

                  |-8-y|=|8+y|=15 has solutions y = -23 and y = 7.

Summary of solutions:
Case 1a-i: (17,17)
Case 1a-ii: (-17,-17)
Case 1b: (0,17), (0,-17)
Case 2a-i: (15,7), (15,23)
Case 2a-ii: (-15,-7), (-15,-23)
Case 2b-i: (8,-7), (8,23)
Case 2b-ii: (-8,-23), (-8,7)

This gives you all twelve solutions.

4. If 'k' is the smallest positive integer such that (2^k)(5^300) has 303 digits when expanded, then what is the sum of the digits of the expanded number?

I'm guessing that these types of questions have some type of formula associated with them.

I'm not sure what you mean by "some type of formula" but the number of digits in a number is the log of that number (if you are using base 10 numbers, then it's log base 10). But you won't necessarily need this formula.

So first you should notice that if k=300, then N=2^k 5^300 is exactly 10^300, which is 301 digits long of course. Now of course, you just need to increase k a little bit to make it a power of 2 that makes N 303 digits long. But clearly, if k = 300 + k', then:

N = 2^k 5^300 = 2^k' 2^300 5^300 = 2^k' 10^300

So what power of 2 gives 3 more digits? k'=7 because 2^7 = 128. So digits 303, 302, and 301 are the 1, 2, and 8.

Given k'=7, the answer is really k = 307, so that N = 2^310 5^300. But knowing how to split this up really tells us N = 128 10^300, and the only nonzero digits of that are 1+2+8=11.

Note: I received a.. "complaint" from this user that I was insulting to her by noting that there are typos. The three other problems she submitted literally do not make mathematical sense. "If ((3/1)*(5/3)*(7/5)*...*(2n+1/2n-1))^1/2 , then what is the value of n?" This question makes no sense, I may as well ask "if n+2, what is the value of n?" I am saddened that this user not only is angry for me not pointing out these typos, but is so even when I have provided resources to help her learn how to tell if a mathematical statement is sensible or not (a skill I believe one should have long before one attempts to solve these sorts of critical thinking / mathematical reasoning problems).

She also refers to her instructor, and is mad that the typos must be his/her fault and that she has been blamed (not sure where "blame" was ever an issue). Regardless, it doesn't change the fact that they are there, and by submitting questions and insisting on having carefully checked that they are copied down -- but without really knowing if they make mathematical sense -- I think these types of problems may be too advanced for this user. Her comments about how to solve them also give this impression. This is the concern I expressed, and I tried to provide resources and materials to aid her in her study, but she instead provided a rude and angry rebuttal. So I would preemptively advise readers to ignore whatever rating she gives this answer and let the answer speak for itself. I took the time to explain thoroughly and completely the two questions which could be interpreted in mathematical meaningful ways. That this elicits indignation is quite astonishing.

Advanced Math

All Answers

Answers by Expert:

Ask Experts


Clyde Oliver


I can answer all questions up to, and including, graduate level mathematics. I am more likely to prefer questions beyond the level of calculus. I can answer any questions, from basic elementary number theory like how to prove the first three digits of powers of 2 repeat (they do, with period 100, starting at 8), all the way to advanced mathematics like proving Egorov's theorem or finding phase transitions in random networks.


I am a PhD educated mathematician working in research at a major university.


Various research journals of mathematics. Various talks & presentations (some short, some long), about either interesting classical material or about research work.

BA mathematics & physics, PhD mathematics from a top 20 US school.

Awards and Honors
Various honors related to grades, various fellowships & scholarships, awards for contributions to mathematics and education at my schools, etc.

Past/Present Clients
In the past, and as my career progresses, I have worked and continue to work as an educator and mentor to students of varying age levels, skill levels, and educational levels.

©2016 All rights reserved.