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

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</math>
 
</math>
  
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== Solution ==
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== Solution 1 ==
  
 
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:
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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>.
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== Solution 2 ==
 +
 +
As above, we can write the equations as follows:
 +
 +
<cmath>
 +
\begin{align*}
 +
(a+1)(b+1) & = 525
 +
\\
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(b+1)(c+1) & = 147
 +
\\
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(c+1)(d+1) & = 105
 +
\end{align*}
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</cmath>
 +
 +
Looking at the first two equations, we know that <math>a+1</math> but not <math>b+1</math> is a multiple of 5, and looking at the last two equations, we know that <math>(d+1)</math> but not <math>(c+1)</math> must be a multiple of 5 (since if <math>b+1</math> or <math>c+1</math> was a multiple of 5, then <math>(b+1)(c+1)</math> would also be a multiple of 5).
 +
 +
Thus, <math>a \equiv d \equiv 0 \hspace{1 mm} \text{(mod 5)}</math>, and <math>a - d \equiv 0 \hspace{1 mm}  \text{(mod 5)}</math>. The only answer choice where this is true is <math>\boxed{\text{D) 10}}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 21:49, 10 July 2017

Problem

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

\begin{align*} ab + a + b & = 524 \\ bc + b + c & = 146 \\ cd + c + d & = 104 \end{align*}

What is $a-d$?

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

Solution 1

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 $a-d=24-14 =\boxed{10}$.

Solution 2

As above, we can write the equations as follows:

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

Looking at the first two equations, we know that $a+1$ but not $b+1$ is a multiple of 5, and looking at the last two equations, we know that $(d+1)$ but not $(c+1)$ must be a multiple of 5 (since if $b+1$ or $c+1$ was a multiple of 5, then $(b+1)(c+1)$ would also be a multiple of 5).

Thus, $a \equiv d \equiv 0 \hspace{1 mm} \text{(mod 5)}$, and $a - d \equiv 0 \hspace{1 mm} \text{(mod 5)}$. The only answer choice where this is true is $\boxed{\text{D) 10}}$.

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

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