a number rounds off to 4000. the digit in the hundreds place is twice the digit in the tens place. the sum of the digits is 12. find the number uses only two different digits. find the number.
4800
Let n=wxyz
wxyz <= 4000
w= 3 (choose the closest possible number to be rounded off to 4)
x=2y
1<=y<=4 (because x value should not have 2 digits i.e. x=2(5)= 10)
w+x+y+z= 12
3+2y+y+z=12
3y+z= 9
*assume y=4
3(4)+z= 9
z= -3 : NOT POSSIBLE
*assume y=3
3(3)+z=9
z=0 : POSSIBLE
*assume y=2
3(2)+z=9
z=3; BUT, x=2y= 2(2)=4, hence you cannot round it off to a thousand
---
w=3
x=2(3)= 6
y=3
z=0
w+x+y+z=12
3+6+3+0=12
3630<= 4000
Therefore, the number is 3630.
To find the number, let's start by analyzing the given information. We know that the number rounds off to 4000. This means that the digit in the thousands place must be 4, as it is the closest value to make the number round to 4000.
Let's call the digit in the tens place "x". Since the digit in the hundreds place is twice the digit in the tens place, it would be 2x.
We also know that the sum of the digits is 12. So we can write the equation:
4 + 2x + x = 12
Combining like terms, we get:
3x + 4 = 12
Subtracting 4 from both sides, we have:
3x = 8
Dividing both sides by 3, we find:
x = 8/3
Since the question asks for a number using only two different digits, we need to choose either 8 or 3 as the digit for the tens place and the other digit for the hundreds place.
If we choose 8 for the tens place, the hundreds place would be twice that, which is 16. However, this would result in a number greater than 4000, so it is not a valid solution.
If we choose 3 for the tens place, the hundreds place would be twice that, which is 6. So we have the number 4632.
Therefore, the number that meets all the given conditions is 4632.
3500 <= n <= 4499
Let n = abcd
If a=4, b+c+d=12 and h=2d
Assuming c is not zero, c=4, since b≠c
But 4848 is no good
If a=3, c=3, so n = 3633
Note that if c=0, 4008 works as well