Let m be the smallest positive integer such that m^2+(m+1)^2+ +(m+10)^2 is the square of a positive integer n . Find m+n .

Correct answer: 95 aligned ∑_ r=1 ^ 11 (m+r-1)^2 & =11 m^2+(1^2+2^2+3^2+ +10^2)+2 m(1+2+3+ +10) & =11(m^2+10 m+35) & =11((m+5)^2+10) aligned The least possible value of m=18 The required sum =11(23^2+10) aligned & =11 × 11 × 49 & =77^2 aligned Then, m=18 and n=77 m+n=95

Olympiad workbook

Exam: Olympiad workbook
Subject: IOQM
Chapter: Sequences and Series

Olympiad workbook · IOQM · Sequences and Series

Let m be the smallest positive integer such that m^2+(m+1)^2+...

Let

m

be the smallest positive integer such that

m^2+(m+1)^2+\ldots+(m+10)^2

is the square of a positive integer

n

. Find

m+n

Correct answer

Step-by-step solution

\begin{aligned} \sum_{r=1}^{11}(m+r-1)^2 & =11 m^2+\left(1^2+2^2+3^2+\ldots+10^2\right)+2 m(1+2+3+\ldots+10) \\ & =11\left(m^2+10 m+35\right) \\ & =11\left((m+5)^2+10\right) \end{aligned}

The least possible value of

m=18

The required sum

=11\left(23^2+10\right)

\begin{aligned} & =11 \times 11 \times 49 \\ & =77^2 \end{aligned}

Then,

m=18

and

n=77 \therefore \quad m+n=95

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