# 2006 iTest Problems/Problem 25

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## Problem

The expression $$\dfrac{(1+2+\cdots + 10)(1^3+2^3+\cdots + 10^3)}{(1^2+2^2+\cdots + 10^2)^2}$$ reduces to $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

## Solution

The sum of integers from $1$ to $n$ is $\tfrac{n(n+1)}{2}$, the sum of squares of integers from $1$ to $n$ is $\tfrac{n(n+1)(2n+1)}{6}$, and the sum of cubes of integers from $1$ to $n$ is $(\tfrac{n(n+1)}{2})^2$. By applying the formulas, we can simplify the expression and reduce common factors. $$\frac{\frac{10 \cdot 11}{2} \cdot (\frac{10 \cdot 11}{2})^2}{(\frac{10 \cdot 11 \cdot 21}{6})^2}$$ $$\frac{(\frac{10 \cdot 11}{2})^3}{(\frac{10 \cdot 11 \cdot 21}{6})^2}$$ $$\frac{10^3 \cdot 11^3}{8} \cdot \frac{36}{10^2 \cdot 11^2 \cdot 21^2}$$ $$\frac{10 \cdot 11 \cdot 9}{2 \cdot 21^2}$$ $$\frac{5 \cdot 11}{49}$$ $$\frac{55}{49}$$ Thus, $m+n = \boxed{104}$.

## See Also

 2006 iTest (Problems) Preceded by:Problem 24 Followed by:Problem 26 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10
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