the sum of the digits in a three-digit number is 16.the ones digit is three times the hundreds digit.if the digits of the number are reversed ,the new number is 594 more than the original .what is the original number?
To solve this problem, let's break it down step by step.
Step 1: According to the problem, the sum of the digits in a three-digit number is 16. Let's represent the three-digit number as ABC, where A is the hundreds digit, B is the tens digit, and C is the ones digit. We are given that A + B + C = 16.
Step 2: The problem also states that the ones digit is three times the hundreds digit. So we can write C = 3A.
Step 3: The problem mentions that if we reverse the digits of the number, the new number is 594 more than the original. So if we reverse the digits ABC, it becomes CBA. We can represent this as 100C + 10B + A, which is equal to ABC + 594.
Step 4: Now, let's substitute the values from Step 2 into Step 3. We have 100(3A) + 10B + A = 100A + 10B + C + 594.
Simplifying this equation: 300A + 10B + A = 100A + 10B + 3A + 594.
Step 5: Combine like terms on both sides of the equation: 301A = 103A + 594.
Step 6: Move all the terms involving A to one side of the equation: 301A - 103A = 594.
Simplifying: 198A = 594.
Step 7: Solve for A: A = 594 / 198 = 3.
Step 8: Now that we have the value of A, we can substitute it back into Step 2 to find the value of C: C = 3A = 3 × 3 = 9.
Step 9: Finally, substitute the values of A and C into the equation from Step 1 to find B: A + B + C = 16.
3 + B + 9 = 16.
Simplifying: B = 16 - 3 - 9 = 4.
Step 10: Putting it all together, the original number is ABC = 349.