The problem is i am a four digit number 3 of the digits are the same the sum of the digits are 28 if rounded to 10, 100, 1000 i am still the same number....
2006-07-26 17:29:14 · 6 answers · asked by mexmami79 1 in Science & Mathematics ➔ Mathematics
ever notice how some people are so picky. like the only way they have to prove how smart they are is to point out your mistakes.
2006-07-26 17:38:10 · answer #1 · answered by biggun4570 4 · 1⤊ 0⤋
The answer:
S000 (A number of base 29)
where S(base 29) = 28 (base 10)
Round it to the nearest 10, 100 or 1000 it is still S000
^_^
2006-07-27 00:55:08 · answer #2 · answered by kevin! 5 · 0⤊ 0⤋
The only answer is 1999, which rounds to 2000 in all three cases.
(4888 rounds to 4890, 4900 and 5000 respectively)
(7777 rounds to 7780, 7800 and 8000 respectively)
And the alternate arrangements are invalid too:
9199 --> 9200, 9200, 9000
9919 --> 9920, 9900, 10000
9991 ---> 9990, 10000, 10000
So the only answer is: 1999 (1+9+9+9 = 28, and rounds to 2000 in all 3 cases)
2006-07-26 18:04:49 · answer #3 · answered by Puzzling 7 · 0⤊ 0⤋
Here are four answers & 3 ways!
A; 1,999 - 9199, 9919 & 9991
B, 4,888 - 8488, 8848 & 8884.
PS: 7,777 would be 3rd answer.
2006-07-26 17:45:48 · answer #4 · answered by Math_Maestro 7 · 0⤊ 0⤋
Can you break your paragraph into sentences?
2006-07-26 17:34:51 · answer #5 · answered by EE 2 · 0⤊ 0⤋
8884?
2006-07-26 17:33:40 · answer #6 · answered by J.D. 6 · 0⤊ 0⤋