1993 AIME Problems/Problem 6
Contents
Problem
What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?
Solution
Solution 1
Denote the first of each of the series of consecutive integers as . Therefore, . Simplifying, . The relationship between suggests that is divisible by . Also, , so is divisible by . We find that the least possible value of , so the answer is \boxed{495}$.
=== Solution 2 ===
Let the desired integer be$ (Error compiling LaTeX. ! Missing $ inserted.)na, \ b, \ cn = 9a + 36 = 10b + 45 = 11c + 55n \equiv 0 \pmod{9}$$ (Error compiling LaTeX. ! Missing $ inserted.)n \equiv 5 \pmod{10}$$ (Error compiling LaTeX. ! Missing $ inserted.)n \equiv 0 \pmod{11}\boxed{495}$=== Solution 3 ===
Let$ (Error compiling LaTeX. ! Missing $ inserted.)n$be the desired integer. From the given information, we have <cmath> \begin{align*}9x &= a \\ 11y &= a \\ 10z + 5 &= a, \end{align*}</cmath> here,$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)x,yza911,\text{lcm}[9,11]=99.a=99m,m.10z + 5 = 99m.99(5) = \boxed{495}$is the smallest integer that can be represented in such a way.
=== Solution 4 === By the method in Solution 1, we find that the number$ (Error compiling LaTeX. ! Missing $ inserted.)n9a+36=10b+45=11c+55a,b,cnnnn\boxed{495}$. Solution by Zeroman.
See also
1993 AIME (Problems • Answer Key • Resources) | ||
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