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

Check the Answer

Answer: is 126.

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

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

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