I am a number rounding to the nearest thousand makes me 1000. All my digits are even their sum is 12 they are in order from greatest to least and no digits is tepeated
so the number is 1xyz
where x+y+z = 12
and all are even.
can you not guess?
To determine the number that satisfies the given criteria, let's break it down step by step:
1. We are looking for a number that rounds to 1000 when rounded to the nearest thousand. This means the number must be between 500 and 1,499. However, since all digits are even, the number must be at least 1,000.
2. The sum of the digits is 12. Since all digits are even, we need to find two even digits that add up to 12. The pairs of even numbers that satisfy this condition are (6, 6) and (8, 4).
3. The digits must be in order from greatest to least. Considering the pairs of digits, (6, 6) cannot be arranged in descending order since the digits are the same. However, (8, 4) can be arranged as 84.
Therefore, the number that satisfies all the given conditions is 1,084.