Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1997 based on Two and Three-digit numbers.

## Two and Three-digit numbers – AIME I, 1997

Sarah intended to multiply a two digit number and a three digit number, but she left out the multiplication sign and simply placed the two digit number to the left of the three digit number, thereby forming a five digit number. This number is exactly nine times the product Sarah should have obtained, find the sum of the two digit number and the three digit number.

- is 107
- is 126
- is 840
- cannot be determined from the given information

**Key Concepts**

Twodigit Number

Threedigit Number

Factors

## Check the Answer

But try the problem first…

Answer: is 126.

AIME I, 1997, Question 3

Elementary Number Theory by David Burton

## Try with Hints

First hint

Let p be a two digit number and q be a three digit number

here 1000p+q=9pq

\(\Rightarrow 9pq-1000p-q=0\)

Second Hint

\((9p-1)(q-\frac{1000}{9})\)=\(\frac{1000}{9}\)

\(\Rightarrow(9p-1)(9q-1000)\)=1000

Final Step

from factors of 1000 gives 9p-1=125

\(\Rightarrow p=14,q=112\)

\(\Rightarrow 112+14=126\).

## 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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