Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1993 based on Smallest positive Integer.
Find 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.
Integers
Divisibility
Algebra
Answer: is 495.
AIME I, 1993, Question 6
Elementary Number Theory by David Burton
Let us take the first of each of the series of consecutive integers as a,b,c
then n=a+(a+1)+...+(a+8)=9a+36=10b+45=11c+55
or, 9a=10b+9=11c+19
or, b is divisible by 9
10b-10=10(b-1)=11c
or, b-1 is divisible by 11
or, least integer b=45
or, 10b+45=10(45)+(45)=495.
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