[問題]GWD 30 Q2,Q3

[ASTROLOGER]

The function f is defined for each positive three-digit integer n by f(n) = 2x3y5z, where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m) = 9f(v), then m-v = ?

A. 8
B. 9
C. 18
D. 20
E. 80
請問這題如何解?我不知道如何下手,謝謝!我當初很直覺的選A

Q3
At a certain food stand, the price of each apple is ¢ 40 and the price of each orange is ¢ 60. Mary selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean) price of the 10 pieces of fruit is ¢ 56. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is ¢ 52?

A. 1
B. 2
C. 3
D. 4
E. 5
這題我設蘋果x顆,橘子y顆
x+y=10
40x+60y=560
解連立得x=2y=8
因為平均價格要調成52,再設一個方程式
40x+60y=520
得x=4y=6
所以8-6=2橘子要放回兩個,但答案是5個
請問我是哪裏解錯了呢?謝謝!

[Penny Lee]

回答你Q3的問題:

題目說orange要少多少, 表示apple購買數不變, orange會減少, 表示apple+orange的購買數已經不是10個, 所以average price 的分母應該設為(2+Y)個, 也就是:

(40x2+60Y)/(2+Y)=52
可得Y=3

得解8-3=5

[Penny Lee]

Q2 的題目你好像打錯囉... 我查了一下應該是F(n) = 2^x3^y5^z 才對[不是f(n) = 2x3y5z]

解法是:
因為f(m)=9f(v), 所以f(m)=9*2^x3^y5^z=2^x*3^(y+2)*5^z
得m=100x+10(y+2)+z

so m-v=100x+10(y+2)+z-(100x+10y+z)=20

得解

[ASTROLOGER]

謝謝,q2題目關於次方問題,我是直接從gwd文字檔複製出來的,沒想到送出後它的次方沒有顯示出來,不好意思我沒有仔細檢查!

謝謝!