Difference between revisions of "1997 AIME Problems/Problem 3"
(cleanup) |
m (→Solution 3: Replacing the mods with pmods) |
||
(15 intermediate revisions by 6 users not shown) | |||
Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
− | Let <math>x</math> be the two-digit number, <math>y</math> be the three-digit number. Putting together the given, we have <math>1000x+y=9xy \Longrightarrow | + | Let <math>x</math> be the two-digit number, <math>y</math> be the three-digit number. Putting together the given, we have <math>1000x+y=9xy \Longrightarrow 9xy-1000x-y=0</math>. Using [[Simon's Favorite Factoring Trick|SFFT]], this factorizes to <math>(9x-1)\left(y-\dfrac{1000}{9}\right)=\dfrac{1000}{9}</math>, and <math>(9x-1)(9y-1000)=1000</math>. |
− | Since < | + | Since <math>89 < 9x-1 < 890</math>, we can use trial and error on factors of 1000. If <math>9x - 1 = 100</math>, we get a non-integer. If <math>9x - 1 = 125</math>, we get <math>x=14</math> and <math>y=112</math>, which satisifies the conditions. Hence the answer is <math>112 + 14 = \boxed{126}</math>. |
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | As shown above, we have <math>1000x+y=9xy</math>, so <math>1000/y=9-1/x</math>. <math>1000/y</math> must be just a little bit smaller than 9, so we find <math>y=112</math>, <math>x=14</math>, and the solution is <math>\boxed{126}</math>. | ||
+ | |||
+ | |||
+ | == Solution 3== | ||
+ | |||
+ | To begin, we rewrite <math>(10a+b)*(100x+10y+z)*9 = 10000a + 1000b + 100x + 10y + z</math> | ||
+ | |||
+ | as | ||
+ | |||
+ | <math>(90a+9b-1)(100x+10y+z) = 10000a + 1000b</math> | ||
+ | |||
+ | and | ||
+ | |||
+ | <math>(90a+9b-1)(100x+10y+z) = 1000(10a + b)</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)\times n(10a+b)= 1000(10a + b)</math> | ||
+ | |||
+ | |||
+ | <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>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>. | ||
+ | |||
+ | |||
+ | |||
+ | -jackshi2006 | ||
== See also == | == See also == | ||
Line 11: | Line 44: | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
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.