I am a three digit number. My hundreds digit is five more than my once digit. My tens digit is 8 less than my hundreds digit. What number am I?
Your number:
A B C
where
A = hundreds digit
B = tens digit
C = once digit
with conditions:
A ≤ 9
B ≤ 9
C ≤ 9
Hundreds digit is five more than my once digit mean:
A = C + 5
so
C = A - 5
Tens digit is 8 less than my hundreds digit mean:
B = A - 8
B = C + 5 - 8
B = C - 3
so
C = B + 3
C = C
A - 5 = B + 3
Add 5 to both sides
A = B + 3 + 5
A = B + 8
Subtract B to both sides
A - B = 8
Since A ≤ 9 and B ≤ 9
This is possible only if :
A = 9 , B = 1
so
A = 9
B = 1
C = A - 5
C = 9 - 5
C = 4
OR
C = B + 3
C = 1 + 3
C = 4
Your number A B C is 9 1 4
could also be
803
To find the three-digit number, we need to break down the provided information into equations and solve them step by step.
Let's assign variables to each digit of the number:
Let the ones digit be represented by 'x'
Let the tens digit be represented by 'y'
Let the hundreds digit be represented by 'z'
Now, based on the given information, we can write three equations:
1. The hundreds digit is five more than the ones digit:
z = x + 5
2. The tens digit is 8 less than the hundreds digit:
y = z - 8
3. The number is a three-digit number, so the hundreds digit cannot be zero:
z ≠ 0
To find the value of x, y, and z, we can substitute the value of z from equation 1 into equation 2:
y = (x + 5) - 8
y = x - 3
Now we can substitute the value of y from the above equation into equation 3:
z ≠ 0
Based on these equations, let's consider different values for z:
For z = 1:
If z = 1, then y = 1 - 3 = -2 (which is not possible as tens digit cannot be negative)
For z = 2:
If z = 2, then y = 2 - 3 = -1 (which is not possible as tens digit cannot be negative)
For z = 3:
If z = 3, then y = 3 - 3 = 0 (which is not possible as tens digit cannot be zero)
For z = 4:
If z = 4, then y = 4 - 3 = 1, and x = 4 + 5 = 9
Therefore, the three-digit number is 491.
So, the answer to the riddle is 491.