GWD12-3, 12-13

[Petra]

Q3:
What is the remainder when the two-digit, positive integer x is divided by 3 ?
(1) The sum of the digits of x is 5.
(2) The remainder when x is divided by 9 is 5.

A: A. However, I think that (2)is sufficient because x=9n+5....


Q13:
4^3 + 4^3 + 4^3 + 4^3 =

A. 4^3
B. 4^6
C. 4^8
D. 4^9
E. 4^12

A: A. Is there anything wrong in this question?


Thanks for your help...
I hate to become stupid while the exam is coming....

[homaru]

No.12
M: DCDDE, CBBAD, EC4(4)CA, DDDCE, DAAAA, CACAA, AABBD, EB

答案是D 跟4^4 沒錯!

[Petra]

Oh...thank you...
I might have checked the answer first....

[訪客]

Q3:
What is the remainder when the two-digit, positive integer x is divided by 3 ?
(1) The sum of the digits of x is 5.
(2) The remainder when x is divided by 9 is 5.




..

請教這題 8-|

[homaru]

Q3:
What is the remainder when the two-digit, positive integer x is divided by 3 ?
(1) The sum of the digits of x is 5.
(2) The remainder when x is divided by 9 is 5.
..

請教這題 8-|

1)x的每位數加起來是5,如果每位數加起來是3的倍數可以整除3
加起來是5,那就是餘2, ex 14/3=4...2; 23/3=7...2

2)x=9n+5=9n+3+2=3(3n+1)+2-->餘數2

[stan(屎蛋)]

12-3 應該是這樣吧
由題目可知，它是要問餘數
第一個可知，x可能為14.23.32.41.50
你把這五個都除以3餘數皆為2....故(1)suff
第二可知x=9n+5 n=1 n=2.......n=5其實是14.23.32.41.50
和上述條件1一樣，皆可求出餘數是2..故(2)也suff
所以答案是d
應該是這樣吧，不知道小弟有沒有說錯.......
ㄏㄏ

原來川叔...已經解了.....網頁沒有重新整哩，所以沒看到....ㄎㄎ.. ;)

[訪客]

恩...恩..
謝謝囉

[Grace]

Q13:
4^3 + 4^3 + 4^3 + 4^3 =

A. 4^3
B. 4^6
C. 4^8
D. 4^9
E. 4^12

A: A. Is there anything wrong in this question?

這題答案不是應該是E (4^3)^4=4^12
我那裡 弄錯了嗎??? :'(

[吳志道]

4^3+4^3+4^3+4^3=4*4^3=4^4(這是我的答案說^^~)
Grace指數有問題的話..看OG前面整理(math那)整理的粉好喔^^

[blacklilu]

4^3 + 4^3 + 4^3 + 4^3 =
可看成： 4^3 ( 1+1+1+1) = 4^3 X 4 ^ 1 = 4 ^ (3+1) = 4^4

[撥撥]

Q3:
What is the remainder when the two-digit, positive integer x is divided by 3 ?
(1) The sum of the digits of x is 5.
(2) The remainder when x is divided by 9 is 5.
..

請教這題 8-|

1)x的每位數加起來是5,如果每位數加起來是3的倍數可以整除3
加起來是5,那就是餘2, ex 14/3=4...2; 23/3=7...2

2)x=9n+5=9n+3+2=3(3n+1)+2-->餘數2

想請問第一個條件之下
是不是只有被(1)3除的時候(2)只有2digits，會有這個情況？

[SugarChunda]

第一個條件在"被3除" & "非2位數" 的時候應該都成立 :
我的通式解如下 :
當2位數時, 令該2位數為 10X+(5-X), 則可以得到 [10X+(5-X)]/3 =9X/3 + 5/3 ..... 所以得知餘數一定會是 2

當3位數時, 令該3位數為 100X+10Y+(5-X-Y), 則可以得到[100X+10Y+(5-X-Y)]/3 = (99X/3)+(9Y/3)+5/3 ..... 所以得知餘數一定會是 2

以此類推,不管幾位數只要是數字總和是5,除以3的餘數都可以得知是2...
希望對大家有幫助 !!

[撥撥]

第一個條件在"被3除" & "非2位數" 的時候應該都成立 :
我的通式解如下 :
當2位數時, 令該2位數為 10X+(5-X), 則可以得到 [10X+(5-X)]/3 =9X/3 + 5/3 ..... 所以得知餘數一定會是 2

當3位數時, 令該3位數為 100X+10Y+(5-X-Y), 則可以得到[100X+10Y+(5-X-Y)]/3 = (99X/3)+(9Y/3)+5/3 ..... 所以得知餘數一定會是 2

以此類推,不管幾位數只要是數字總和是5,除以3的餘數都可以得知是2...
希望對大家有幫助 !!

謝謝Sugar大美女喔∼∼

[NEWUNIVERSE]

謝謝 通式解 很棒唷^^

[chris8888]

Q13:
4^3 + 4^3 + 4^3 + 4^3 =

A. 4^4
B. 4^6
C. 4^8
D. 4^9
E. 4^12

A: A

A應該是這樣吧, 上面都沒改, 我以為沒答案