A number rounds off to 4000. The digit in the hundreds place is twice the digit in tens place. The sum of the digit is 12. The numbers uses only two different digits. Find the number.
4 x y z
x = 2 y but x is <5 so y = 0, 1, or 2
4 + 2y + y + z = 12
so
x + 3y + z = 8
if y = 0, then twice y is also 0 nope
if y = 1, then twice y = 2, nope 421 three digits
if y = 2 , then 4 4 2 z
then if z = 2, the sum is 12
how about 4422 ???
if the number is 3xxx with x>5, then the minimum sum is 3633=15 so, that will not work
If the number is 4xxx, then there must be 2 4's and 2 2's. That leaves only 4422 as a possibility.
To solve this problem, we need to find a number that meets the given criteria:
1. The number rounds off to 4000. This means the actual number is between 3500 and 4499.
2. The digit in the hundreds place is twice the digit in the tens place. Let's assume the digit in the tens place is "x". Therefore, the digit in the hundreds place is "2x".
3. The sum of the digits is 12. Let's suppose the first digit is "y". Therefore, the sum of the digits would be y + 2x + x = 12.
Now, we can solve this as a system of equations:
Equation 1: y + 2x + x = 12
Equation 2: y * 100 + 2x * 10 + x = 4000
Let's solve Equation 1 for "y":
y + 3x = 12
y = 12 - 3x
Substitute this value of "y" into Equation 2:
(12 - 3x) * 100 + 2x * 10 + x = 4000
1200 - 300x + 20x + x = 4000
1200 - 280x = 4000
-280x = 2800
x = -10 (discard this value since the digit can't be negative)
x = 10
Substitute the value of "x" back into Equation 1 to find "y":
y + 3(10) = 12
y + 30 = 12
y = -18 (discard this value since the digit can't be negative)
y = 18
Therefore, the tens digit "x" is 10, and the hundreds digit "2x" is 20. Since the sum of the digits is 12, the units digit is 12 - 10 - 20 = -18 (discard this value). Hence, the only suitable digits are 1 and 8.
Putting it all together, the number is 2188.