Difference between revisions of "2001 AMC 12 Problems/Problem 21"

Latest revision as of 12:01, 18 September 2023

Problem

Four positive integers $a$ , $b$ , $c$ , and $d$ have a product of $8!$ and satisfy:

$\begin{array}{rl} ab + a + b & = 524 \\ bc + b + c & = 146 \\ cd + c + d & = 104 \end{array}$

What is $a-d$ ?

$\text{(A) }4 \qquad \text{(B) }6 \qquad \text{(C) }8 \qquad \text{(D) }10 \qquad \text{(E) }12$

Solution

Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows:

$\begin{align*} (a+1)(b+1) & = 525 \\ (b+1)(c+1) & = 147 \\ (c+1)(d+1) & = 105 \end{align*}$

Let $(e,f,g,h)=(a+1,b+1,c+1,d+1)$ . We get:

$\begin{align*} ef & = 3\cdot 5\cdot 5\cdot 7 \\ fg & = 3\cdot 7\cdot 7 \\ gh & = 3\cdot 5\cdot 7 \end{align*}$

Clearly $7^2$ divides $fg$ . On the other hand, $7^2$ can not divide $f$ , as it then would divide $ef$ . Similarly, $7^2$ can not divide $g$ . Hence $7$ divides both $f$ and $g$ . This leaves us with only two cases: $(f,g)=(7,21)$ and $(f,g)=(21,7)$ .

The first case solves to $(e,f,g,h)=(75,7,21,5)$ , which gives us $(a,b,c,d)=(74,6,20,4)$ , but then $abcd \not= 8!$ . We do not need to multiply, it is enough to note e.g. that the left hand side is not divisible by $7$ . (Also, a - d equals $70$ in this case, which is way too large to fit the answer choices.)

The second case solves to $(e,f,g,h)=(25,21,7,15)$ , which gives us a valid quadruple $(a,b,c,d)=(24,20,6,14)$ , and we have $a-d=24-14 =\boxed{10}$ .

Video Solution

https://youtu.be/zlzievV5e-U

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.

@@ Line 1: / Line 1: @@
 == Problem ==
-Solve the following system of equations for <math>c</math>:
+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>
-\begin{align*}
+\begin{array}{rl}
-a - b &= 2 (c+d)\\
+ab + a + b & = 524
-b &= a-2 \\
+\\
-d &= c+5
+bc + b + c & = 146
-\end{align*}
+\\
+cd + c + d & = 104
+\end{array}
 </cmath>
-== Solution 1 ==
+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:
@@ Line 42: / Line 59: @@
 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>.
-== Solution 2 ==
+== Video Solution ==
+https://youtu.be/zlzievV5e-U
-As above, we can write the equations as follows:
-<cmath>
-\begin{align*}
-a - b = 2(c+d)
-b = a - 2
-d = c + 5
-\end{align*}
-</cmath>
-From the second equation, we can conclude that <math>a-b=2</math>. Plugging this into the first equation yields that <math>c+d=1</math>. The last equation implies that <math>d-c=5</math>, so by inspection, we know that <math>d=3</math> and <math>c=\boxed{-2}</math>.
-<math></math>
-~AopsUser101
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Difference between revisions of "2001 AMC 12 Problems/Problem 21"

Latest revision as of 12:01, 18 September 2023

Contents

Problem

Solution

Video Solution

See Also