Difference between revisions of "1997 AIME Problems/Problem 3"
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− | This is the most important part: Notice <math>(90a+9b-1)</math> is <math> | + | This is the most important part: Notice <math>(90a+9b-1)</math> is <math>-1 \pmod{10a+b}</math> and <math>1000(10a + b)</math> is <math>0\pmod{10a+b}</math>. That means <math>(100x+10y+z)</math> is also <math>0\pmod{10+b}</math>. Rewrite <math>(100x+10y+z)</math> as <math>n\times(10a+b)</math>. |
− | <math>(90a+9b-1) | + | <math>(90a+9b-1)\times n(10a+b)= 1000(10a + b)</math> |
− | <math>(90a+9b-1) | + | <math>(90a+9b-1)\times n= 1000</math> |
− | Now we have to find a number that divides 1000 using prime factors 2 or 5 and is <math> | + | Now we have to find a number that divides 1000 using prime factors 2 or 5 and is <math>8\pmod9</math>. It is quick to find there is only one: 125. That gives 14 as <math>10a+b</math> and 112 as <math>100x+10y+z</math>. Therefore the answer is <math>112 + 14 = \boxed{126}</math>. |
Latest revision as of 19:34, 4 August 2021
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 be the two-digit number,
be the three-digit number. Putting together the given, we have
. Using SFFT, this factorizes to
, and
.
Since , we can use trial and error on factors of 1000. If
, we get a non-integer. If
, we get
and
, which satisifies the conditions. Hence the answer is
.
Solution 2
As shown above, we have , so
.
must be just a little bit smaller than 9, so we find
,
, and the solution is
.
Solution 3
To begin, we rewrite
as
and
This is the most important part: Notice is
and
is
. That means
is also
. Rewrite
as
.
Now we have to find a number that divides 1000 using prime factors 2 or 5 and is . It is quick to find there is only one: 125. That gives 14 as
and 112 as
. Therefore the answer is
.
-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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.