AMC12 2001 A

AMC12 2001 A · Q21

AMC12 2001 A · Q21. It mainly tests Manipulating equations, Primes & prime factorization.

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 $a-d$?

四个正整数 $a$, $b$, $c$, 和 $d$ 的乘积为 $8!$，且满足： \[\begin{array}{rl} ab + a + b & = 524 \\ bc + b + c & = 146 \\ cd + c + d & = 104 \end{array}\] 求 $a-d$。

(A) 4 4

(B) 6 6

(D) 10 10

(E) 12 12

Answer

Correct choice: (D)

正确答案：（D）

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

使用 Simon's Favorite Factoring Trick，可将三式改写为： \begin{align*} (a+1)(b+1) & = 525 \\ (b+1)(c+1) & = 147 \\ (c+1)(d+1) & = 105 \end{align*} 令 $(e,f,g,h)=(a+1,b+1,c+1,d+1)$，则有： \begin{align*} ef & = 3\cdot 5\cdot 5\cdot 7 \\ fg & = 3\cdot 7\cdot 7 \\ gh & = 3\cdot 5\cdot 7 \end{align*} 显然 $7^2$ 整除 $fg$。另一方面，$7^2$ 不可能整除 $f$，否则它也会整除 $ef$。同理，$7^2$ 也不可能整除 $g$。因此 $7$ 同时整除 $f$ 与 $g$。这只剩两种情况：$(f,g)=(7,21)$ 和 $(f,g)=(21,7)$。第一种情况解得 $(e,f,g,h)=(75,7,21,5)$，从而 $(a,b,c,d)=(74,6,20,4)$，但此时 $abcd \not= 8!$。无需实际相乘，只需注意例如左边不被 $7$ 整除即可。（并且此时 $a-d=70$，也明显过大，不符合选项。）第二种情况解得 $(e,f,g,h)=(25,21,7,15)$，从而得到有效的四元组 $(a,b,c,d)=(24,20,6,14)$，于是 $a-d=24-14 =\boxed{10}$。

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