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AMC10 2024 A

AMC10 2024 A · Q19

AMC10 2024 A · Q19. It mainly tests Sequences & recursion (algebra), Primes & prime factorization.

The first three terms of a geometric sequence are the integers $a,\,720,$ and $b,$ where $a<720<b.$ What is the sum of the digits of the least possible value of $b?$
一个等比数列的前三项为整数$a,720,b$,其中$a < 720 < b$。求$b$的最小可能值的各位数字之和?
(A) 9 9
(B) 12 12
(C) 16 16
(D) 18 18
(E) 21 21
Answer
Correct choice: (E)
正确答案:(E)
Solution
For a geometric sequence, we have $ab=720^2=2^8 3^4 5^2$, and we can test values for $b$. We find that $b=768$ and $a=675$ works, and we can test multiples of $5$ in between the two values. Finding that none of the multiples of 5 divide $720^2$ besides $720$ itself, we know that the answer is $7+6+8=\boxed{\textbf{(E) } 21}$. (Note: To find the value of $b$ without bashing, we can observe that $2^8=256$, and that multiplying it by $3$ gives us $768$, which is really close to $720$. ~ YTH) Note: The reason why $ab=720^2$ is because $b/720 = 720/a$. Rearranging this gives $ab = 720^2$ Note: Another reason that $ab=720^2$ is because the $\sqrt{ab}=720$ (as the middle term in a geometric series is always the geometric mean [the geometric mean is the square root of the product of the first and last terms of the series]) and squaring on both sides results in $ab=720^2$.
对于等比数列,$ab = 720^2 = 2^8 3^4 5^2$,可测试$b$的值。发现$b=768$,$a=675$可行,并在两者间测试$5$的倍数,发现除$720$本身外无其他$5$的倍数整除$720^2$,故答案为$7+6+8=\boxed{\textbf{(E) } 21}$。 (注:不需暴力求解$b$,可观察$2^8=256$,乘以$3$得$768$,非常接近$720$。~ YTH) 注:$ab=720^2$的原因是$b/720 = 720/a$,重排得$ab=720^2$。 注:另一原因是等比数列中项为首尾乘积的几何平均,即$\sqrt{ab}=720$,两边平方得$ab=720^2$。
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