2014 AMC 10B Problem 14

P11-15~4 minPrint

Bases & Digits Divisibility & Factors Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Bound the quantity above/below or estimate to pin it down

Danica drove her new car on a trip for a whole number of hours, averaging $55$ miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a 3-digit number with $a \ge 1$ and $a + b + c \le 7$ . At the end of the trip, the odometer showed $cba$ miles. What is $a^2 + b^2 + c^2$ ?

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Key insight

Reversing a 3-digit number changes it by 99 times (c - a). That must equal 55 times the hours, forcing c - a = 5.

Solution

1. How far did she drive? The odometer went from $abc$ to $cba$ . Write each as a number: $abc = 100a + 10b + c$ and $cba = 100c + 10b + a$ .
2. Subtract. The middle digit cancels:
$(100c + 10b + a) - (100a + 10b + c) = 99c - 99a = 99(c - a).$
(Check with an example: $601 - 106 = 495 = 99 \times 5$ .)
3. Use the speed. She drove $h$ whole hours at $55$ miles per hour, so $99(c - a) = 55h$ .
4. Both $99$ and $55$ are multiples of $11$ . Divide by $11$ : $9(c - a) = 5h$ .
5. The left side is a multiple of $9$ and the right side is a multiple of $5$ . Since $9$ has no factor $5$ , the number $c - a$ must be a multiple of $5$ .
6. She actually drove, so $c - a$ is positive. It is a difference of two digits, so it is at most $9$ . The only positive multiple of $5$ up to $9$ is $5$ . So $c = a + 5$ (and $h = 9$ hours).
7. Use the digit limit. $a + b + c \le 7$ becomes $a + b + (a + 5) \le 7$ , so $2a + b \le 2$ . Since $a \ge 1$ , the only choice is $a = 1$ , $b = 0$ . Then $c = 6$ .
8. Check: $601 - 106 = 495 = 55 \times 9$ . Then $a^2 + b^2 + c^2 = 1 + 0 + 36 = 37$ .

The answer is $\boxed{\textbf{(D)}\ 37}$ .

Why this works

Reversing the digits of a 3-digit number always changes it by a multiple of $99$ , because the middle digit stays put. Matching that with a multiple of $55$ leaves only one possible gap between the outer digits. A strange-looking condition like $a + b + c \le 7$ is usually there to pick out exactly one answer at the end.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

If the trip lasted $h$ hours, the distance is $55h$ , so
$\overline{cba} - \overline{abc} = (100c + 10b + a) - (100a + 10b + c) = 99(c - a) = 55h.$

Divide by $11$ : $9(c - a) = 5h$ . Since $\gcd(9, 5) = 1$ , the number $5$ must divide $c - a$ . The trip was real ( $h \ge 1$ ), so $c - a > 0$ ; and as a difference of digits $c - a \le 9$ , so $c - a = 5$ (and $h = 9$ ).

Now use $a \ge 1$ and $a + b + c \le 7$ . With $c = a + 5$ , the sum is $2a + b + 5 \le 7$ , so $2a + b \le 2$ . Since $a \ge 1$ , this forces $a = 1$ and $b = 0$ , hence $c = 6$ .

Check: $601 - 106 = 495 = 55 \cdot 9$ . Then $a^2 + b^2 + c^2 = 1 + 0 + 36 = 37$ .

The answer is $\boxed{\textbf{(D)}\ 37}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 8: Base Numbers
Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Algebra, Chapter 11: Special Factorizations
Introduction to Number Theory, Chapter 9: Base Number Arithmetic