AMC 10A/12A 2026 (Problem 10/6)For what odd positive integer nnn is the mean of 1,4,9,16,…,n21,4,9,16,\ldots,n^21,4,9,16,…,n2 exactly 32%32\%32% greater than its median?(A) 19\text{(A)}\;19(A)19(B) 39\text{(B)}\;39(B)39(C) 49\text{(C)}\;49(C)49(D) 89\text{(D)}\;89(D)89(E) 169\text{(E)}\;169(E)169Related TopicsCoreSums of PowersMajorMean, Median, Mode, and RangeCheck AnswerYour answer:ABCDECheckHints (3)Hint 11+4+9+16+⋯+n2n=1.32(n+12)2\frac{1+4+9+16+\cdots+n^2}{n}=1.32\left(\frac{n+1}{2}\right)^2n1+4+9+16+⋯+n2=1.32(2n+1)2Hint 2n(n+1)(2n+1)6n=1.32(n+12)2\frac{n(n+1)(2n+1)}{6n}=1.32\left(\frac{n+1}{2}\right)^26nn(n+1)(2n+1)=1.32(2n+1)2Hint 32n+16=1.32⋅n+14\frac{2n+1}{6}=1.32\cdot\frac{n+1}{4}62n+1=1.32⋅4n+12n+13=1.32(n+1)2\frac{2n+1}{3}=\frac{1.32(n+1)}{2}32n+1=21.32(n+1)4n+2=3.96n+3.964n+2=3.96n+3.964n+2=3.96n+3.960.04n=1.960.04n=1.960.04n=1.96n=1.960.04=1964=49n=\frac{1.96}{0.04}=\frac{196}{4}=49n=0.041.96=4196=49Final Answer(C) 49Related Problems (7)AMC 10B 2025 (Problem 18)AMC 10A/12A 2025 (Problem 4/3)AMC 10B/12B 2025 (Problem 17/14)2024 AMC 10A Problem 122024 AMC 12A Problem 52024 AMC 12A Problem 21AMC 10B/12B 2024 (Problem 15/10)