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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

But try the problem first…

Answer: is 495.

Source

Suggested Reading

AIME I, 1993, Question 6

Elementary Number Theory by David Burton

## Try with Hints

First hint

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

Second Hint

or, 9a=10b+9=11c+19

or, b is divisible by 9

Final Step

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

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