i am a four digit number. all of my digits are different. my first digit is twice my fourth digit. my second digit is twice my first digit.my last digit is twice my third digit. the sum of my digits is 15. what number am i?
a four digit number equals the sum of twice of its reversed and half of it. also its first digit is 4 times the last digit and second digits is 7 times the 3rd digit .the sum of four digits of the number is ?
I am a four digit number with 6 hundreds and 5 ones .the sum of all my digits is 15 what am I.
To solve this problem, start by assigning variables to each digit of the four-digit number. Let's use the variables A, B, C, and D, where A represents the first digit, B represents the second digit, C represents the third digit, and D represents the fourth digit.
We are given the following information:
1. A is twice D.
2. B is twice A.
3. D is twice C.
4. The sum of the digits (A + B + C + D) is 15.
Now, let's translate each of the given statements into equations:
1. A = 2D
2. B = 2A = 2(2D) = 4D
3. D = 2C
4. A + B + C + D = 15
Now we can start solving the equations:
Substitute A = 2D into the equation for B:
B = 4D
Substitute D = 2C into the equation for A:
A = 2(2C) = 4C
Now, substitute A = 4C and B = 4D into the equation for the sum of the digits:
4C + 4D + C + D = 15
5C + 5D = 15
Divide both sides of the equation by 5:
C + D = 3
Now, we have a system of two equations:
C + D = 3 (Equation 1)
4C + 4D = 15 (Equation 2)
Solve Equation 1 for C:
C = 3 - D
Now substitute this value of C into Equation 2:
4(3 - D) + 4D = 15
12 - 4D + 4D = 15
12 = 15
Since this equation is not true, it means that there is no solution to the system of equations. Therefore, there is no four-digit number that satisfies all the given conditions.
Let the fourth digit be x
First digit = 2x
Second digit = 2(2x) = 4x
Third digit = x/2
Thus x+2x+4x+x/2 = 15
15x/2=15
x=2
Thus the number is (2*2)(4*2)(2/2)(2)
= 4812