Math question.....?

Question

Okay....so...... i really need an answer to this.
I'm a four digit number.
the sum of my digits is 28.
three of my digits are the same.
whether i am rounded to the nearest tens, hundreds, or thousands, i am STILL the same number

Answer 1

Since 3 of the digits are same and total 4 digits add up to 28.
the 3 digits can be 1-9. Given that they round up to the same digit every tens hundreds thousands. My guess is 1999.

Answer 2

The two middle digits must be "99". The way you work this out is from the hint that rounding to ten, hundreds or thousands always gives the same number.

If rounding to the nearest thousand, the answer would be X000 (where X is some digit). Rounding to the nearest 10 should produce X000 too, and this could only happen if the two middle digits are both 9 and the ones-place digit is greater than or equal to 5, so that rounding trickles up all the way to the thousands place.

Subtract 9+9 from 28, to figure out the sum of the first and last digits (10). You know the ones-place digit must be greater than or equal to 5, so combinations of (T,O) <- that's T="thousands digit", O="ones digit" that satisfy these two conditions are: (5,5), (4,6), (3,7), (2,8) and (1,9)

Furthermore, three digits must be the same. Well, you've already got "99" in the middle, so the three-digits-all-same must be all 9's.

That's how you come to the conclusion of "1999".

Answer 3

9 9 9 1 or 8 8 8 4 or 7 7 7 7

Answer 4

1999

1+9+9+9=28
1999 rounded to 10's 100's or 1000's =2000

hope this helps

Answer 5

9991
Ya, I think so.

Answer 6

1.999 or 4.888 or 7.777

Answer 7

7778 ?