the sum of the in a three digit number is 9. The tens digit is half the sum of the sum of the other two and the hundreds digit is half the units digit. Find the total value of the number.
ABC
B = 1/2 (A+C)
A = (1/2) C
A + B + C = 9
so
B= (1/2)(3/2) C
(1/2)C +(3/4) C + C = 9
(9/4) C = 9
C = 4
then
A = 2
B = 3
so
234
Oh, boy! It's time for some number fun! Let's solve this puzzle step by step:
We know that the sum of the digits is 9. So, let's call the hundreds digit "H," the tens digit "T," and the units digit "U."
According to the problem, H + T + U = 9. That's our first equation!
Now, the problem says that the tens digit is half the sum of the other two digits. So we can write T = (H + U)/2. This is our second equation!
Lastly, the hundreds digit is half the units digit. In other words, H = U/2. This is our third equation!
Now we have a system of three equations. Let's solve it!
From the third equation, we can substitute H = U/2 into the first equation: U/2 + T + U = 9. Simplifying this, we have 3U + 2T = 18.
We also have the second equation, T = (H + U)/2. Substituting H = U/2, we have T = (U/2 + U)/2. Simplifying this, we have T = 3U/4.
Now we substitute the value of T from the second equation into the new equation we found: 3U + 2(3U/4) = 18. Simplifying, we get U = 4.
Substituting U = 4 into H = U/2, we get H = 2.
Finally, substituting U = 4 into T = 3U/4, we get T = 3.
So the digits of the number are 2, 3, and 4. Putting them together, the total value of the number is 234.
Ta-da! We did it! Hope you enjoyed this numerical clownery!
Let's break down the problem step by step.
Step 1: Let's assume the three-digit number is represented by ABC, where A represents the hundreds digit, B represents the tens digit, and C represents the units digit.
Step 2: According to the problem, the sum of the digits in the three-digit number is 9. This can be written as A + B + C = 9.
Step 3: The problem states that the tens digit (B) is half the sum of the other two (A and C). Therefore, B = (A + C)/2.
Step 4: The problem also states that the hundreds digit (A) is half the units digit (C). So, A = C/2.
Step 5: Now we can substitute A and B in terms of C into the equation from step 2. We have (C/2) + (C/2) + C = 9.
Step 6: Simplify the equation: 2C + 2C + 4C = 36. This becomes 8C = 36.
Step 7: Divide both sides of the equation by 8: C = 4. This means that the units digit (C) is 4.
Step 8: Substitute the value of C into the equation from step 4 to find A: A = 4/2 = 2. This means that the hundreds digit (A) is 2.
Step 9: Substitute the values of A and C back into the equation from step 3 to find B: B = (2 + 4)/2 = 6/2 = 3. This means that the tens digit (B) is 3.
Step 10: Now we have A = 2, B = 3, and C = 4, so the three-digit number ABC is 234.
Therefore, the total value of the number is 234.
To find the total value of the three-digit number, let's break down the information given step by step:
1. The sum of the digits is 9.
Let's assume the digits of the three-digit number are represented as ABC, where A represents the hundreds digit, B represents the tens digit, and C represents the units digit. According to the information, we know that A + B + C = 9.
2. The tens digit is half the sum of the other two digits.
This information tells us that B = (A + C) / 2.
3. The hundreds digit is half the units digit.
This information states that A = C / 2.
Now, let's solve the problem by substituting the equations:
From equation 2, we can replace B in equation 1:
A + (A + C) / 2 + C = 9.
From equation 3, we can replace A in equation 1:
C / 2 + (C + C) / 2 + C = 9.
Simplifying the equation:
C/2 + (2C + C) / 2 + C = 9.
C/2 + 3C/2 + C = 9.
(6C + 3C + 2C) / 2 = 9.
11C/2 = 9.
Now, let's solve for C:
Multiply both sides by 2:
11C = 18.
C = 18/11.
Since the units digit should be a whole number, we round the value to the nearest whole number:
C ≈ 1.
Now, substitute the value of C back into equation 2 to find B:
B = (A + C) / 2.
B = (A + 1) / 2.
Since B is also a whole number, and B is half of the sum of A and C, it means A + C must be even. From our previous result, we know that C is 1, so A must be even.
The possibilities for A and C that fulfill these conditions are: A = 2 and C = 1, or A = 4 and C = 2. However, since A should be half of C (from equation 3), the only valid solution is A = 2 and C = 1.
Now, substitute A = 2 and C = 1 into equation 1 to find B:
2 + B + 1 = 9.
B + 3 = 9.
B = 6.
Therefore, the three-digit number is 261.