2006 iTest Problems/Problem 25

Revision as of 19:20, 1 December 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 25 -- Simplify Quotient of Sums)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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
Invalid username
Login to AoPS