<p>I’m too lazy to check anything else but don’t forget that n=1 works as well.</p>
<p>Oh yeah, you’re right. Maybe it’s 6625. That would make a good AMC trick question.</p>
<p>@rspence: How did you do sigma(2016) = 6552?</p>
<p>Here’s another Number Theory problem. 
Let a, b be natural numbers.
- Show that a^2012 | b^2011 implies that a | b.
- Find two natural numbers a, b such that a^2011 | b^2012 and a does not divide b.
- Show that if a has less than 2011 divisors and a^2011 | b^2012 then a | b.</p>
<p>@yellowcat you can factor 2016 = (2^5)(3^2)(7). The sum of divisors is (1 + 2 + … + 2^5)(1 + 3 + 9)(1 + 7).</p>
<p>I’ll leave out the formalities of the proof, and sort-of-obvious stuff will be assumed (I proved them, but I don’t feel like writing out lots of (p<em>i)^(a</em>i)'s and stuff).</p>
<ol>
<li><p>All primes in the prime factorization of a must be in the prime factorization of b. Let’s focus on one prime and call that p. Say the exponent of p in the prime factorization of a is A, the exponent in the pfactorization of b is B. We wish to prove that p^A | p^B, or A <= B. Then we have p^(2011B) = kp^(2012A). Let k = p^x. Then we have 2011B = 2012A + x. Divide both sides by 2012; it’s clear from there that A < B and thus p^A | p^B and hence a | b.</p></li>
<li><p>a = 2^2012, b = 2^2011</p></li>
<li><p>We follow the proof of (1) except now we have 2011A + x = 2012B. Divide both sides by 2012 for 2011A/2012 + x/2012 = B; x is positive so 2011A/2012 <= B. Using a well-known fact and the fact that a has less than 2011 divisors, we know that A <=2010. If A=2010, the inequality and the fact that B is an integer forces B >=2010 and thus A <= B. If A < 2010, we have the same result (the proof is pretty intuitive; basically multiplying a number less than 2010, say X, by 2011/2012 can’t get you less than or equal to X-1). A <= B implies, as we well know, a | b.</p></li>
</ol>
<p>And regarding the sigma formula, it’s online somewhere, I don’t want to type the long thing out.</p>
<p>Sorry if I am too lazy to do stuff ;)</p>
<p>I think you mixed up your “divides” symbol. a|b means “a divides b,” or “b is divisible by a.” For example, 5|20. So #1 should be:</p>
<p>Show that b^2011 | a^2012 implies that b | a.</p>
<p>Here’s a problem that could have been a Level 6 on … the old SAT.</p>
<p>Col A 100,210 x 90,021
Col B 100,021 x 90,210</p>
<p>Is A < B, or A > B, or A = B, or is it undefined?</p>
<p>Try this problem without a calculator in less than 40 seconds.</p>
<p>idk what Col means, but if it means compute, then A<B can be figured out in about 10 seconds. and how can that be undefined ._.</p>
<p>@rspence then that would mean #1 is not true. Take a = 2^2011 and b = 2^2012, for example.</p>
<p>^ It’s odd that anyone might pick D – undefined. These are just numeric operations. It would be a pretty fuzzy math that let one column be bigger sometimes and other times not. </p>
<p>Still, as a tutor, I miss the old two-column format…easy to prep for.</p>
<p>But as Kyrix1’s post shows, current CCers don’t necessarily know what we older folks are talking about with the two-column format. That’s ok – a lot of them don’t know why we call it “dialing” when you push the buttons on a phone.</p>
<p>@Kyrix1, if a = 2^2011 and b = 2^2012, then a^2012 | b^2011 (they’re equal) and 2^2011 | 2^2012, that is true. a|b implies b/a is an integer, that’s how | is defined.</p>
<p>A < B, pretty easy.</p>
<p>A similar problem: Arrange the following integers from least to greatest: 15^100, 30^80, 60^60.</p>
<p>
If a = 2^2011 and b = 2^2012, then b^2011 | a^2012 is satisfied by b | a is not…</p>
<p>@pckeller: cus you used to twist a dial to each number in turn :D</p>
<p>
</p>
<p>The ETS once had such high hopes for the 4-choice QC problems that they thought the entire SAT math section could be QC format.</p>
<p>Their internal research later showed the format to be quite a bit more coachable than the usual 5-choice MC questions, which eventually led to the removal of QCs in 2005.</p>
<p>@rspence: From least to greatest would be 60^60, 15^100, 30^80. Correct?</p>
<p>^dang 2:56 am? that’s hardcore…
anyways you’re right, cancel out 15^60 and then you’re comparing 4^60, 15^40, 2^80<em>15^20. 4^60=2^120=8^40 so 15^40>4^60, and 2^8>15^2 so 2^80>15^20, 2^80</em>15^20>15^40. or you can be a pro like sonnhard and use wolfram alpha</p>
<p>ok b/c my previous problem was too hard 
the first problem of NIMO:
Let f(x) = (x^4+2x^3+4x^2+2x+1)^5. Compute the prime satisfying f(p) = 418195493.</p>
<p>@yellowcat429, yep.</p>
<p>@funsummer 418195493 ≡ 3 (mod 10) and out of the fifth powers 1^5, 2^5, …, 10^5, only 3^5 ≡ 3 (mod 10) so x^4 + … + 1 is 3 mod 10.</p>
<p>41819543 is roughly 4000*10^5, a little more than 50^5, but clearly less than 60^5. 53^5 ≡ 3 (mod 10) so set p^4 + 2p^3 + 4p^2 + 2p + 1 = 53. p = 2, we’re done.</p>
<p>Here’s one from the Caltech - Harvey Mudd Math Competition (CHMMC):</p>
<p>a and b are positive real numbers with a > b and ab = 8. Find the minimum value of (a^2 + b^2)/(a - b). </p>
<p>Darn, I noticed we keep deviating away from the “SAT Level 6” questions…but those aren’t challenging enough for us. Maybe we should move to the math challenge thread, or start a new one…</p>
<p>yeah, you’re right 
what you could also have done is see that if something to the 5th is odd, then it’s odd, and x^4+2x^3+4x^2+2x+1 is only odd if x^4 is even.</p>
<p>here’s what i have for your problem:
= (a-b) + (2ab)/(a-b)
= (a-b) + 16/(a-b)</p>
<br>
<br>
<p>np:
solve (x+1)(x+2)(x+3)(x+4)=-1 in reals</p>
<p>or we can revise this to be “v_Enhance SAT Level 1 questions” XD</p>
<p>Some algebraic manipulation => end up with (a-b) + 16/(a-b). Now let a-b = h, so we minimize h + 16/h. Taking the derivative (or by intuition), we find a critical point at h = 4, so the minimal value is 8.</p>
<p>edit: darn got sniped. also DARN how did i forget amgm</p>
<p>OR you can set h+16/h=k, and then h^2-kh+16=0; since h=a-b=real, the min value of k is 8 to make the discriminant 0. xp</p>