The sum of the digits of a three digit number is 20. The middle digit is equal to one fourth the sum of the other two. If the order of degree is reversed the number increases by 198. Find the original number
algebra solution:
let the unit digit be z
let the tens digit be y
let the hundred digit be x
x+y+z = 20
y = (1/4)(x+z)
4y = x+z
original number is 100x + 10y + z
reversed number is 100z + 10y + x
difference of those two is 198
100x + 10y + z - 100z - 10y - x = 198
99x - 99z = 198
x - z = 2
z = x-2 , so y = (x + x-2)/4
in x+y+z = 20
x + (2x-2)/4 + x-2 = 20
8x + 2x-2 - 8 = 80
10x = 90
x = 9
then z = 7
and 4y = 16 ---> y = 4
The numbers are 947 and 749 , as Arora also found.
(I simply took the difference between the numbers to be 198, I should have reversed the order of subtraction)
To find the original three-digit number, let's break down the problem step by step.
Let's assume the three-digit number is represented as ABC, where A, B, and C are the digits.
We know two things from the given information:
1. The sum of the digits of a three-digit number is 20:
A + B + C = 20
2. The middle digit is equal to one-fourth the sum of the other two:
B = (A + C)/4
Now, let's solve these equations simultaneously.
From equation 2, we can substitute the value of B in equation 1:
A + (A + C)/4 + C = 20
To simplify the equation, let's multiply through by 4 to eliminate the fraction:
4A + (A + C) + 4C = 80
5A + 5C = 80
A + C = 16 ---- (equation 3)
Now, we need to find the original number when its order of digits is reversed. Let's assume the reversed three-digit number is represented as CBA.
According to the given information, the reversed number is 198 more than the original number. Mathematically, we can express this as:
100C + 10B + A = 100A + 10B + C + 198
To simplify the equation, we can subtract 10B from both sides:
100C + A = 99A + C + 198
Rearranging this equation, we get:
99A - A = 100C - C - 198
98A = 99C - 198
Divide both sides of the equation by 98:
A = (99C - 198)/98
Since A, B, and C represent digits, A and C must be integers. By examining the equation, we see that A and C must satisfy the conditions for the equation to be true. Let's try different values for C until we find valid solutions.
When C = 1, A = (99 - 198)/98 = -1 (not an integer)
When C = 2, A = (198 - 198)/98 = 0 (not an integer)
When C = 3, A = (297 - 198)/98 = 1 (an integer)
We have found a valid solution where A = 1 and C = 3. Now, let's substitute these values into equation 3 to find B.
A + C = 16
1 + 3 = 16
4 = 16
This equation is not true. Therefore, our assumption of A = 1 and C = 3 is incorrect.
By trying other values of C, we will reach the valid solution.
When C = 6, A = (594 - 198)/98 = 4
Now substituting A = 4 and C = 6 into equation 3:
A + C = 16
4 + 6 = 16
10 = 16
Again, this equation is not true.
Continue trying different values for C until a valid solution is found.
When C = 7, A = (693 - 198)/98 = 5
Now substituting A = 5 and C = 7 into equation 3:
A + C = 16
5 + 7 = 16
12 = 16
This equation is also not true.
Keep trying.
When C = 9, A = (891 - 198)/98 = 7
Now substituting A = 7 and C = 9 into equation 3:
A + C = 16
7 + 9 = 16
16 = 16
This equation is true. Hence, the valid solution is A = 7, B = (A + C)/4 = (7 + 9)/4 = 16/4 = 4, and C = 9.
Thus, the original three-digit number is ABC = 749.
Very nice bc
The maximum sum of two digits in the number (first and last) can be 18. Thus, the middle digit cannot be larger than 4 (since that would be larger than 1/4th the largest sum).
Since reversing the number increases it, the third digit is larger than the first. The first digit is two less than the third digit, since that condition is necessary for their difference to end with an 8.
The number is 749.