Suppose n is an integer such that the sum of the digits of n is 2,and 10^10
(b) 10
(c) 9
(d) 8
2006-08-28 20:30:45 · 4 answers · asked by Rohit C 3 in Science & Mathematics ➔ Mathematics
There are 11 different values
10000000001
10000000010
10000000100
10000001000
10000010000
10000100000
10001000000
10010000000
10100000000
11000000000
20000000000
^_^
2006-08-28 20:47:25 · answer #1 · answered by kevin! 5 · 0⤊ 0⤋
The answer is (b).
10^10 is 10000000000
10^11 is 100000000000
You are looking at the numbers
10000000001
10000000010
10000000100
10000001000
10000010000
10000100000
10001000000
10010000000
10100000000
11000000000
2006-08-28 20:46:41 · answer #2 · answered by don.giovanni 3 · 0⤊ 0⤋
I think your looking for these:
1000000001
1000000010
1000000100
1000001000
1000010000
1000100000
1001000000
1010000000
1100000000
Looks like there's only 9.
Those guys below me are right... I didn't add enough zeros.. and the 200000000000 is a clever one....
The answer is 11
2006-08-28 20:38:33 · answer #3 · answered by John H 3 · 0⤊ 0⤋
i can't understand you're question.
any of the choices doesn't satisfy 10^10 < n < 10^11.
2006-08-28 20:35:00 · answer #4 · answered by ←deadstar→ 3 · 0⤊ 0⤋