徐永強數學-p-set2-16

[邁向高等學府]

16. Is the tens digit of two-digit positive integer p divisible by 3?
(1) p – 5 is a multiple of 3
(2) p – 11 is a multiple of 3

answer is E
but I think it is D , When considering the two conditions respectively, I think each can answer the quesition, that is, none of the tens digit of two-digit positive integer can be divided by 3
can anybody explain how to solve this problem?
or there are some flaws in my resoning?

[汪汪北鼻]

1)p-5=3a, p=3a+5(a>=2)->insufficient
2)p-11=3b,p=3b+11(b>=1)->insufficient

considering both 1) and 2), also insufficient
let's assume p=20, p-5=15, p-11=9, the tens digit of positive integer p can't be divided by 3

but if p=32, p-5=27,p-11=21, the tens digit of positive integer p can be divided by 3

thus, the answer is E

[邁向高等學府]

1)p-5=3a, p=3a+5(a>=3)->insufficient
2)p-11=3b,p=3b+11(b>=1)->insufficient

considering both 1) and 2), also insufficient
let's assume p=20, p-5=15, p-11=9, the tens digit of positive integer p can't be divided by 3

but if p=32, p-5=27,p-11=21, the tens digit of positive integer p can be divided by 3

thus, the answer is E

我的想法
p=3a+5
a=1-> p=8
2 -> 11
3 -> 14
4 -> 17

p=3a+11
b=1-> p=14
2 -> 17
3 -> 20
以此類推…
分別根據二個條件…
不就可以回答本題之問題了嗎?(十位數p是否可被3整除?)
p是不可被3整除的
因此為D...

不知道那裡想錯了…

[nicoleliu]

我的想法
p=3a+5
a=1-> p=8
2 -> 11
3 -> 14
4 -> 17

p=3a+11
b=1-> p=14
2 -> 17
3 -> 20
以此類推…
分別根據二個條件…
不就可以回答本題之問題了嗎?(十位數p是否可被3整除?)
p是不可被3整除的 :^)
因此為D...

不知道那裡想錯了…

你是不是把題目想錯了....
題目是問十位數p 中的"the tens digit " 是否可被3整除?
並不是問十位數p 是否可被3整除喔~~

[邁向高等學府]

我的想法
p=3a+5
a=1-> p=8
2 -> 11
3 -> 14
4 -> 17

p=3a+11
b=1-> p=14
2 -> 17
3 -> 20
以此類推…
分別根據二個條件…
不就可以回答本題之問題了嗎?(十位數p是否可被3整除?)
p是不可被3整除的 :^)
因此為D...

不知道那裡想錯了…

你是不是把題目想錯了....
題目是問十位數p 中的"the tens digit " 是否可被3整除?
並不是問十位數p 是否可被3整除喔~~

謝謝
原來是關鍵字沒看清…
所以在兩個條件下，皆有可與不可被3整除的十位數…所以合在一起看也無法有答案…因此是E…