AMC 10A 2012 (Problem 22)The sum of the first mmm positive odd integers is 212212212 more than the sum of the first nnn positive even integers. What is the sum of all possible values of nnn?(A) 255\text{(A)}\;255(A)255(B) 256\text{(B)}\;256(B)256(C) 257\text{(C)}\;257(C)257(D) 258\text{(D)}\;258(D)258(E) 259\text{(E)}\;259(E)259Related TopicsCoreToolkit 10 — Difference of squaresMajorToolkit 6 — Square of a sumMinorToolkit 25 — Arithmetic Sequence and SeriesHints (12)Hint 11+3+⋯+(2m−1)=212+(2+4+⋯+2n)1+3+\cdots+(2m-1)=212+(2+4+\cdots+2n)1+3+⋯+(2m−1)=212+(2+4+⋯+2n)Hint 2Use Toolkit 25 — Arithmetic Sequence and Series.Hint 3m2=212+n(n+1)m^2=212+n(n+1)m2=212+n(n+1)Hint 4Use Toolkit 6 — Square of a sum and Toolkit 10 — Difference of squares.Hint 5m2=212+(n+12)2−14m^2=212+\left(n+\frac12\right)^2-\frac14m2=212+(n+21)2−41Hint 64m2=848+(2n+1)2−14m^2=848+(2n+1)^2-14m2=848+(2n+1)2−1Hint 74m2−(2n+1)2=8474m^2-(2n+1)^2=8474m2−(2n+1)2=847Hint 8(2m+2n+1)(2m−2n−1)=847(2m+2n+1)(2m-2n-1)=847(2m+2n+1)(2m−2n−1)=847Hint 92m+2n+12m+2n+12m+2n+1 and 2m−2n−12m-2n-12m−2n−1 are factors of 847847847.Hint 10m,n>0⟹2m−2n−1<2m+2n+1m,n>0\Longrightarrow 2m-2n-1<2m+2n+1m,n>0⟹2m−2n−1<2m+2n+1847>0847>0847>0(2m+2n+1)(2m−2n−1)=847(2m+2n+1)(2m-2n-1)=847(2m+2n+1)(2m−2n−1)=847⟹2m−2n−1>0\Longrightarrow 2m-2n-1>0⟹2m−2n−1>0Hint 11847=7⋅112847=7\cdot11^2847=7⋅1122m+2n+12m−2n−1mn84712122111217322877112216\begin{array}{c|c|c|c}2m+2n+1&2m-2n-1&m&n\\ \hline847&1&212&211\\121&7&32&28\\77&11&22&16\end{array}2m+2n+1847121772m−2n−11711m2123222n2112816Hint 12Ans=211+28+16=255\text{Ans}=211+28+16=255Ans=211+28+16=255Final Answer(A) 255255255Related Problems (8)AMC 10B 2023 (Problem 14)BMO1 2016/2017 (Problem 3)AMC 10A 2024 (Problem 11)AMC 10/12A Spring 2021 (Problem 10)AMC 10A/12A 2024 (Problem 15/9)AMC 10A 2023 (Problem 23)AMC 10A 2020 (Problem 14)AMC 10B Spring 2021 (Problem 15)