A number is increased by 3 when it is rounded to the
nearest ten thousand or to the nearest thousand.
What could the number be?
300
To solve this problem, we need to find a number that satisfies the given condition: when it is rounded to the nearest ten thousand or to the nearest thousand, it is increased by 3.
Let's assume the number is "x".
When x is rounded to the nearest ten thousand, the last four digits should become zero. Therefore, we can express this rounding as (x - x % 10000).
Similarly, when x is rounded to the nearest thousand, the last three digits should become zero. So we can represent this rounding as (x - x % 1000).
According to the problem, both of these rounded values are 3 more than the original number, so we can write two equations:
x - x % 10000 = x + 3
x - x % 1000 = x + 3
To find the values of x that satisfy these equations, we can solve them:
Simplifying the first equation:
x - x % 10000 = x + 30000
- x % 10000 = 30000
x % 10000 = -30000
Simplifying the second equation:
x - x % 1000 = x + 3000
- x % 1000 = 3000
x % 1000 = -3000
Here, we can see that the modulus (remainder) is a negative number, which means there are no whole number solutions for x that will satisfy these equations.
Therefore, there is no possible value for the number x that satisfies the given condition.