1997 AIME Problems/Problem 3

Problem

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. What is the sum of the two-digit number and the three-digit number?

Solution

Let $x$ be the two-digit number, $y$ be the three-digit number. Putting together the given, we have $1000x+y=9xy \Longrightarrow 9xy-1000x-y=0$ . Using SFFT, this factorizes to $(9x-1)\left(y-\dfrac{1000}{9}\right)=\dfrac{1000}{9}$ , and $(9x-1)(9y-1000)=1000$ .

Since $89 < 9x-1 < 890$ , we can use trial and error on factors of 1000. If $9x - 1 = 100$ , we get a non-integer. If $9x - 1 = 125$ , we get $x=14$ and $y=112$ , which satisifies the conditions. Hence the answer is $112 + 14 = \boxed{126}$ .

Solution 2

As shown above, we have $1000x+y=9xy$ , so $1000/y=9-1/x$ . $1000/y$ must be just a little bit smaller than 9, so we find $y=112$ , $x=14$ , and the solution is $\boxed{126}$ .

Solution 3

To begin, we rewrite $(10a+b)*(100x+10y+z)*9 = 10000a + 1000b + 100x + 10y + z$

as

$(90a+9b-1)(100x+10y+z) = 10000a + 1000b$

and

$(90a+9b-1)(100x+10y+z) = 1000(10a + b)$

This is the most important part: Notice $(90a+9b-1)$ is $-1 \pmod{10a+b}$ and $1000(10a + b)$ is $0\pmod{10a+b}$ . That means $(100x+10y+z)$ is also $0\pmod{10+b}$ . Rewrite $(100x+10y+z)$ as $n\times(10a+b)$ .

$(90a+9b-1)\times n(10a+b)= 1000(10a + b)$

$(90a+9b-1)\times n= 1000$

Now we have to find a number that divides 1000 using prime factors 2 or 5 and is $8\pmod9$ . It is quick to find there is only one: 125. That gives 14 as $10a+b$ and 112 as $100x+10y+z$ . Therefore the answer is $112 + 14 = \boxed{126}$ .

-jackshi2006

1997 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1997 AIME Problems/Problem 3

Contents

Problem

Solution

Solution 2

Solution 3

See also