Maths Olympiad Prep

Library / /304 of 520

Algebra Difficulty 3.2 AMC 10/12 Find the answer

If 3p+34=903^p + 3^4 = 90, 2r+44=762^r + 44 = 76, and 53+6s=14215^3 + 6^s = 1421, what is the product of pp, rr, and ss?

Pick one

Solution

Solution 1: Solving
First, we're going to solve for pp. Start with 3p+34=903^p+3^4=90. Then, change 343^4 to 8181. Subtract 8181 from both sides to get 3p=93^p=9 and see that pp is 22. Now, solve for rr. Since 2r+44=762^r+44=76, 2r2^r must equal 3232, so r=5r=5. Now, solve for ss. 53+6s=14215^3+6^s=1421 can be simplified to 125+6s=1421125+6^s=1421 which simplifies further to 6s=12966^s=1296. Therefore, s=4s=4. prsprs equals 2542*5*4 which equals 4040. So, the answer is (B) 40\boxed{\textbf{(B)}\ 40}.

Solution 2: Process of Elimination
First, we solve for ss. As Solution 1 perfectly states, 53+6s=14215^3+6^s=1421 can be simplified to 125+6s=1421125+6^s=1421 which simplifies further to 6s=12966^s=1296. Therefore, s=4s=4. We know that you cannot take a root of any of the numbers raised to pp, rr, or ss and get a rational answer, and none of the answer choices are irrational, so that rules out the possibility that pp, rr, or ss is a fraction. The only answer choice that is divisible by 44 is (B) 40\boxed{\textbf{(B)}\ 40}.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.