AMC 10 · 2014 · #14

Grade 6 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
digit-decompositiondivisibility-rulesdigit-constraints convert-to-algebra ↑ Prerequisites: digit-decomposition
📏 Medium solution 💡 3 insights
Problem
Danica drives for a whole number of hours at a steady 55 miles per hour. Her odometer starts at the 3-digit number abc=100a+10b+c (with a ≥ 1 and a+b+c ≤ 7) and ends at the reversed number cba=100c+10b+a. Find a²+b²+c².

Pick an answer.

(A)
26
(B)
27
(C)
36
(D)
37
(E)
41

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The trip hides three digits and a driving time, so Tool #4 (Introduce a Variable) names them a,b,c,h and rewrites both odometer readings by place value. Tool #8 (Analyze the Units) turns 'miles per hour times whole hours' into a clean distance 55h that must equal the odometer difference. That difference is 99(c-a), and matching it to 55h forces a divisibility fact, where Tool #3 (Eliminate Possibilities) kills every digit gap except c-a=5. Finally the tight budget a+b+c ≤ 7 is a boundary condition, so Tool #14 (Extreme Principle) squeezes the digits to a single choice.

1STEP 1

Name the digits and the driving time

Let the start be abc=100a+10b+c, the end be cba=100c+10b+a, and h be the whole number of hours driven.

abc=100a+10b+c, cba=100c+10b+a, h∈{1,2,3,…}
2STEP 2

Write the distance two ways

Distance is 55h miles, and it also equals the odometer rise cba-abc=99(c-a), so 99(c-a)=55h.

cba-abc=(100c+10b+a)-(100a+10b+c)=99(c-a)=55h
3STEP 3

Simplify and pin down the digit gap

Divide by 11 to get 9(c-a)=5h, so 5 divides c-a; with 0 ＜ c-a ≤ 9 this forces c-a=5 and h=9.

9(c-a)=5h → 5 ∣ (c-a), 1 ≤ c-a ≤ 9 → c-a=5, h=9
4STEP 4

Use the digit budget to fix the digits

c=a+5 turns a+b+c ≤ 7 into 2a+b ≤ 2, and a ≥ 1 leaves no room, so a=1, b=0, c=6.

c=a+5, 2a+b ≤ 2, a ≥ 1 → a=1, b=0, c=6
5STEP 5

Evaluate the requested sum of squares

With a=1, b=0, c=6, a²+b²+c²=1²+0²+6²=1+0+36=37, which is choice (D).

a²+b²+c²=1²+0²+6²=1+0+36=37 → (D)
Answer
37
Check the whole trip with (a,b,c)=(1,0,6): the odometer starts at 106 and ends at 601, a rise of 495 miles. Dividing, 495 ÷ 55=9, a whole number of hours, so the 55 mph condition holds exactly. The digit budget gives a+b+c=1+0+6=7 ≤ 7 and a=1 ≥ 1, both satisfied. Everything is consistent, and a²+b²+c²=37 agrees with answer (D).
💡Key takeaway

Write the distance two ways: a reversed 3-digit number minus itself is always 99 times the end-minus-start digit gap, and matching that to 55×hours pins the digits down.

  • Name the digits and the driving time
  • Write the distance two ways
  • Simplify and pin down the digit gap
  • Use the digit budget to fix the digits
  • Evaluate the requested sum of squares

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