MathDB

25

Part of 2021 AMC 12/AHSME Fall

Problems(2)

Polynomial Problem

Source: 2021 AMC12A #25

11/11/2021
Let m≥5m\ge 5 be an odd integer, and let D(m)D(m) denote the number of quadruples (a1,a2,a3,a4)\big(a_1, a_2, a_3, a_4\big) of distinct integers with 1≤ai≤m1\le a_i \le m for all ii such that mm divides a1+a2+a3+a4a_1+a_2+a_3+a_4. There is a polynomial q(x)=c3x3+c2x2+c1x+c0q(x) = c_3x^3+c_2x^2+c_1x+c_0such that D(m)=q(m)D(m) = q(m) for all odd integers m≥5m\ge 5. What is c1?c_1?
(<spanclass=′latex−bold′>A</span>) −6(<spanclass=′latex−bold′>B</span>) −1(<spanclass=′latex−bold′>C</span>) 4(<spanclass=′latex−bold′>D</span>) 6(<spanclass=′latex−bold′>E</span>) 11(<span class='latex-bold'>A</span>)\: {-}6\qquad(<span class='latex-bold'>B</span>) \: {-}1\qquad(<span class='latex-bold'>C</span>) \: 4\qquad(<span class='latex-bold'>D</span>) \: 6\qquad(<span class='latex-bold'>E</span>) \: 11
algebrapolynomialAMCAMC 12AMC 12 A
Sum of Remainders

Source: 2021 Fall AMC 12B 25

11/17/2021
For nn a positive integer, let R(n)R(n) be the sum of the remainders when nn is divided by 22, 33, 44, 55, 66, 77, 88, 99, and 1010. For example, R(15)=1+0+3+0+3+1+7+6+5=26R(15) = 1+0+3+0+3+1+7+6+5=26. How many two-digit positive integers nn satisfy R(n)=R(n+1) ?R(n) = R(n+1)\,?
<spanclass=′latex−bold′>(A)</span>0<spanclass=′latex−bold′>(B)</span>1<spanclass=′latex−bold′>(C)</span>2<spanclass=′latex−bold′>(D)</span>3<spanclass=′latex−bold′>(E)</span>4<span class='latex-bold'>(A) </span>0\qquad<span class='latex-bold'>(B) </span>1\qquad<span class='latex-bold'>(C) </span>2\qquad<span class='latex-bold'>(D) </span>3\qquad<span class='latex-bold'>(E) </span>4
AMCAMC 12AMC 12 B2021 AMC Fallnumber theorycombinatoricsCasework