AMC10 2025 A

AMC10 2025 A · Q11

AMC10 2025 A · Q11. It mainly tests Sequences & recursion (algebra), Divisibility & factors.

The sequence $1,x,y,z$ is arithmetic. The sequence $1,p,q,z$ is geometric. Both sequences are strictly increasing and contain only integers, and $z$ is as small as possible. What is the value of $x+y+z+p+q$?

数列 $1,x,y,z$ 是等差数列。数列 $1,p,q,z$ 是等比数列。两个数列都是严格递增的且仅包含整数，且 $z$ 尽可能小。$x+y+z+p+q$ 的值是多少？

(A) 66 66

(B) 91 91

(D) 132 132

(E) 149 149

Answer

Correct choice: (E)

正确答案：（E）

Solution

Since the geometric sequence is more restrictive, we can test values for the common ratio until we find one that works. After a few tests, we find that a common ratio of $4$ results in the geometric sequence $1,4,16,64,$ so the arithmetic sequence is $1,22,43,64.$ The answer is $4+16+64+22+43=\boxed{\text{(E) }149}.$ A more generalized solution is as follows. Let the common difference of the arithmetic sequence be $d$, and the common ratio of the geometric sequence be $r.$ Hence, the two sequences are $1,1+d,1+2d,1+3d$ and $1,r,r^2,r^3.$ Since $z=1+3d=r^3,$ the arithmetic sequence is $1,1+d,1+2d,r^3.$ Since $d=\dfrac{1+3d-1}{3}=\dfrac{r^3-1}{3}$ is a positive integer, we seek the smallest $r\neq1$ such that $r^3-1=(r-1)(r^2+r+1)$ is divisble by $3,$ so the smallest $r$ is $4$. The rest follows like above.

由于等比数列限制更多，我们可以测试公比的值，直到找到合适的。经过几次测试后，发现公比为 $4$ 时，等比数列为 $1,4,16,64$，于是等差数列为 $1,22,43,64$。答案是 $4+16+64+22+43=\boxed{\text{(E) }149}$。更一般的解法如下。设等差数列的公差为 $d$，等比数列的公比为 $r$。于是两个数列分别为 $1,1+d,1+2d,1+3d$ 和 $1,r,r^2,r^3$。由于 $z=1+3d=r^3$，等差数列为 $1,1+d,1+2d,r^3$。由于 $d=\dfrac{1+3d-1}{3}=\dfrac{r^3-1}{3}$ 是正整数，我们寻找最小的 $r\neq1$ 使得 $r^3-1=(r-1)(r^2+r+1)$ 能被 $3$ 整除，因此最小的 $r$ 是 $4$。其余如上。

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