Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1997 based on Two and Three-digit numbers.
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.
Twodigit Number
Threedigit Number
Factors
Answer: is 126.
AIME I, 1997, Question 3
Elementary Number Theory by David Burton
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
\(\Rightarrow p=14,q=112\)
\(\Rightarrow 112+14=126\).
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