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## Two and Three-digit numbers | AIME I, 1997 | Question 3

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

AIME I, 1997, Question 3

Elementary Number Theory by David Burton

## Try with Hints

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

here 1000p+q=9pq

$\Rightarrow 9pq-1000p-q=0$

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

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

from factors of 1000 gives 9p-1=125

$\Rightarrow p=14,q=112$

$\Rightarrow 112+14=126$.