AMC 8 Daily Practice Round 11

Complete problem set with solutions and individual problem pages

Problem 18 Easy

There is a three-digit number \overline{abc}, the sum of a and c is 10, and \overline{abc} - \overline{bca} = 486. What is the value of a+b-c?

  • A.

    4

  • B.

    5

  • C.

    6

  • D.

    7

  • E.

    None of the above

Answer:C

We can solve this problem using place value .

Step 1: Considering the Units Digit

If no borrowing occurs, we get:

c - a = 6

c + a = 10

Solving these equations:

c = \frac{10 + 6}{2} = 8, \quad a = 8 - 6 = 2.

However, if the minuend’s hundreds digit is 2, the difference cannot be 486, so borrowing must have occurred.

Step 2: Considering Borrowing

With borrowing, we modify the equation:

10 + c - a = 6.

Rearranging:

a - c = 4.

Since a + c = 10, solving for a and c:

a = \frac{10 + 4}{2} = 7, \quad c = 7 - 4 = 3.

Step 3: Finding b

By analyzing the tens and hundreds digits, we determine:

b = 2.

Thus, the three-digit number is 723.

Final Answer:

Thus, the correct answer is 7+2-3=6, which is \boxed{C}.

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