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1997 AIME Problems/Problem 3

Revision as of 11:54, 10 August 2020 by Jackshi2006 (talk | contribs) (→‎Solution 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 $(-1) mod (10a+b)$ and $1000(10a + b)$ is $(0) mod (10a+b)$. That means $(100x+10y+z)$ divides $(10+b)$. Rewrite $(100x+10y+z)$ as $n*(10a+b)$.


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


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


Now we have to find a number that divides 1000 using prime factors 2 or 5 and is $(8)mod9$. It is quick to find there is only one: 125. That gives 14 as $10(a+b)$ and 112 as $x+y+z$.


-jackshi2006

See also

1997 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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