Prove that , if the difference between a three digit number in the form of a0b that is digit at ten's place is 0 and the sum of it's digits is divided by 99, we always get the number equal to a?
Uncompleted
To prove the given statement, we need to show that for any three-digit number in the form of "a0b," where the digit at the tens place is 0, if the sum of its digits is divided by 99, the result will always be equal to 'a.'
Let's break down the problem and prove this step by step:
Step 1: Express the number in terms of its digits.
- The given number is in the form of 'a0b.'
- We can represent this number using the algebraic expression: 100a + 10(0) + b = 100a + b.
Step 2: Calculate the sum of the digits.
- The digit representation of the given number is a + 0 + b = a + b.
- So, the sum of the digits is (a + b).
Step 3: Divide the sum of digits by 99 and check the result.
- Divide (a + b) by 99: (a + b) / 99.
- This division will result in a quotient, which we need to prove is always equal to 'a.'
Step 4: Determine the quotient (a + b) / 99.
- Let's express (a + b) / 99 as a mixed fraction: q + r/99, where q is the quotient and r is the remainder.
- We need to show that the remainder, r, is always 0, and q is always equal to 'a.'
Step 5: Express (a + b) / 99 in terms of q and r.
- (a + b) can be written as a multiple of 99 plus the remainder: (a + b) = 99q + r.
Step 6: Analyze the properties of (a + b) = 99q + r.
- Since a is a digit, the maximum value of a can be 9.
- Therefore, a + b can have a maximum value of (9 + 9) = 18.
Step 7: Check the values of q and r.
- If r = 0, then (a + b) / 99 = q.
- We need to prove that r is always equal to 0 and q is always equal to 'a.'
Step 8: Prove that r is always 0.
- From step 6, we know that (a + b) can have a maximum value of 18.
- Since the maximum value of (a + b) is less than 99, the maximum remainder, r, can be obtained by subtracting 99q from (a + b).
- Therefore, r = (a + b) - 99q.
- Since (a + b) is less than 99, subtracting 99q from (a + b) will always result in a non-negative number with a maximum value of 18.
- As a result, r can only be 0.
Step 9: Prove that q is always equal to 'a.'
- From Step 5, we have (a + b) = 99q.
- Since r is always 0 (from Step 8), we can rewrite (a + b) = 99q + 0, which simplifies to (a + b) = 99q.
- This equation implies that (a + b) is divisible evenly by 99, which makes q the quotient.
- Since (a + b) = 99q, and we know that r is always 0, q can be derived by dividing (a + b) by 99.
- Hence, q is always equal to 'a.'
Step 10: Conclude the proof.
- Since q is always equal to 'a' (from Step 9), we have proved that if the sum of the digits of a three-digit number in the form of 'a0b' is divided by 99, the quotient will always be equal to 'a.'
Thus, it is proven that, for any three-digit number in the form of 'a0b,' where the digit at the tens place is 0, if the sum of its digits is divided by 99, the result will always be equal to 'a.'
a0b has the value 100a+0+b
100a+b - (a+b) = 99a
divide by 99, and you always get a