Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1993 based on Smallest positive Integer.
Smallest positive Integer – AIME I, 1993
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.
- is 107
- is 495
- is 840
- cannot be determined from the given information
Key Concepts
Integers
Divisibility
Algebra
Check the Answer
Answer: is 495.
AIME I, 1993, Question 6
Elementary Number Theory by David Burton
Try with Hints
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.
Other useful links
- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA