1997 AIME Problems/Problem 3

Revision as of 13:13, 21 November 2007 by Inscrutableroot (talk | contribs) (Solution)

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 the two-digit number be ab, and the three-digit number be cde.

$ab=10a+b=x$

$cde=100c+10d+e=y$

$abcde=10000a+1000b+100c+10d+e=9(10a+b)(100c+10d+e}$ (Error compiling LaTeX. ! Extra }, or forgotten $.)

$1000x+y=9xy$

$9xy-1000x-y=0$

$(9x-1)(y-\dfrac{1000}{9})=\dfrac{1000}{9}$

$(9x-1)(9y-1000)=1000$

9y-1000 must be positive, so y>111.

Since x is greater than 89, we can try certain factors of 1000:

100: 9x=101, nope.

125: 9x=126, x=14

Then 9y-1000=8, 1008=9y, y=112.

112+14=126

See also

1997 AIME (ProblemsAnswer KeyResources)
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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