PP-T2-DS

[Huang Hsin-Yi]

If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of p^2t?

(1) m has more than 9 positive factors
(2)m is a multiple of p^3

Ans:B

根據前天學到的因子數公式
可得m=p^x*T^y
(1)(X+1)(Y+1)>9
那為什麼這樣推不出p^2t的factor?

還請大大指點一二

[sunnyblue13s]

m有超過9個正因數....
(X+1)(Y+1)>9
那...X或Y的可能值是無限多....
無限多代表著...不充分~~對於is m a multiple of p^2t來說~~

倍數問題→界定因子...確認倍數...

[Huang Hsin-Yi]

m有超過9個正因數....
(X+1)(Y+1)>9 那...X或Y的可能值是無限多....

但要滿足這個條件至少X>2 或Y>2,才能滿足,不是嗎?

無限多代表著...不充分~~對於is m a multiple of p^2t來說~~

[sunnyblue13s]

(X+1)(Y+1)...一個大就一個會變小.....無法定義出"值"
因此，就無法精確的判斷 - Is m a multiple of p^2t?

[Huang Hsin-Yi]

(X+1)(Y+1)...一個大就一個會變小.....無法定義出"值"
因此，就無法精確的判斷 - Is m a multiple of p^2t?

舉例來說以最低值x=2 ,y=3 3*4>9
therefore
m=p^2*t^3,那p^2*t 是它的factor呀?

我到底是那裡錯了