Difference between revisions of "2001 AMC 12 Problems/Problem 21"
Aopsuser101 (talk | contribs) (→Solution 2) |
(→Problem) |
||
(3 intermediate revisions by 3 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | + | Four positive integers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> have a product of <math>8!</math> and satisfy: | |
+ | |||
<cmath> | <cmath> | ||
− | \begin{ | + | \begin{array}{rl} |
− | a | + | ab + a + b & = 524 |
− | b &= | + | \\ |
− | d &= | + | bc + b + c & = 146 |
− | \end{ | + | \\ |
+ | cd + c + d & = 104 | ||
+ | \end{array} | ||
</cmath> | </cmath> | ||
− | == Solution | + | What is <math>a-d</math>? |
+ | |||
+ | <math> | ||
+ | \text{(A) }4 | ||
+ | \qquad | ||
+ | \text{(B) }6 | ||
+ | \qquad | ||
+ | \text{(C) }8 | ||
+ | \qquad | ||
+ | \text{(D) }10 | ||
+ | \qquad | ||
+ | \text{(E) }12 | ||
+ | </math> | ||
+ | |||
+ | == Solution == | ||
Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows: | Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows: | ||
Line 41: | Line 58: | ||
The second case solves to <math>(e,f,g,h)=(25,21,7,15)</math>, which gives us a valid quadruple <math>(a,b,c,d)=(24,20,6,14)</math>, and we have <math>a-d=24-14 =\boxed{10}</math>. | The second case solves to <math>(e,f,g,h)=(25,21,7,15)</math>, which gives us a valid quadruple <math>(a,b,c,d)=(24,20,6,14)</math>, and we have <math>a-d=24-14 =\boxed{10}</math>. | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/zlzievV5e-U | ||
== See Also == | == See Also == |
Latest revision as of 13:01, 18 September 2023
Contents
Problem
Four positive integers ,
,
, and
have a product of
and satisfy:
What is ?
Solution
Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows:
Let . We get:
Clearly divides
. On the other hand,
can not divide
, as it then would divide
. Similarly,
can not divide
. Hence
divides both
and
. This leaves us with only two cases:
and
.
The first case solves to , which gives us
, but then
. We do not need to multiply, it is enough to note e.g. that the left hand side is not divisible by
. (Also, a - d equals
in this case, which is way too large to fit the answer choices.)
The second case solves to , which gives us a valid quadruple
, and we have
.
Video Solution
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.