AMC 10 · 2014 · #14
Grade 6 arithmeticDanica 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≥1 and a+b+c≤7. At the end of the trip, the odometer showed cba miles. What is a2+b2+c2?
Pick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Danica drives for a whole number of hours at a steady $55$ miles per hour. Her odometer starts at the $3$-digit number $\overline{abc}=100a+10b+c$ (with $a\ge 1$ and $a+b+c\le 7$) and ends at the reversed number $\overline{cba}=100c+10b+a$. Find $a^2+b^2+c^2$.
Givens: Speed is $55$ miles per hour and the driving time is a whole number of hours $h$; Start odometer reading is $\overline{abc}=100a+10b+c$, a $3$-digit number; End odometer reading is $\overline{cba}=100c+10b+a$; Digit constraints: $a\ge 1$ and $a+b+c\le 7$; Answer choices: (A) $26$, (B) $27$, (C) $36$, (D) $37$, (E) $41$
Unknowns: The digits $a$, $b$, $c$ and hence the value of $a^2+b^2+c^2$
Understand
Restated: Danica drives for a whole number of hours at a steady $55$ miles per hour. Her odometer starts at the $3$-digit number $\overline{abc}=100a+10b+c$ (with $a\ge 1$ and $a+b+c\le 7$) and ends at the reversed number $\overline{cba}=100c+10b+a$. Find $a^2+b^2+c^2$.
Givens: Speed is $55$ miles per hour and the driving time is a whole number of hours $h$; Start odometer reading is $\overline{abc}=100a+10b+c$, a $3$-digit number; End odometer reading is $\overline{cba}=100c+10b+a$; Digit constraints: $a\ge 1$ and $a+b+c\le 7$; Answer choices: (A) $26$, (B) $27$, (C) $36$, (D) $37$, (E) $41$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #8 Analyze the Units, #3 Eliminate Possibilities, #14 Extreme Principle
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 $a+b+c\le 7$ is a boundary condition, so Tool #14 (Extreme Principle) squeezes the digits to a single choice.
Execute — Answer: D
6.EE.B.6 Step 1 Name the digits and the driving time
- Let the starting reading be $\overline{abc}=100a+10b+c$ and the ending reading be the reversed number $\overline{cba}=100c+10b+a$.
- Let $h$ be the whole number of hours Danica drove.
- Writing each reading by place value is what lets the trip become an equation instead of a story.
💡 Giving the unknown digits letters lets place value carry the meaning for you.
6.RP.A.3 Step 2 Write the distance two ways
- Distance is speed times time, so the trip covers $55\times h=55h$ miles.
- The same distance is how much the odometer climbed, namely end minus start.
- Subtracting, the $10b$ terms cancel and the hundreds and units swap, leaving $99(c-a)$.
- Setting the two expressions for distance equal gives the master equation.
💡 Distance is one number, so its rate form $55h$ and its odometer form $99(c-a)$ must match.
4.OA.B.4 Step 3 Simplify and pin down the digit gap
- Both $99$ and $55$ share the factor $11$, so divide the equation by $11$ to get $9(c-a)=5h$.
- The right side is a multiple of $5$, so the left side must be too; since $9$ is not a multiple of $5$, the gap $c-a$ must be.
- The odometer grew, so $c>a$, making $c-a$ a positive one-digit number.
- The only multiple of $5$ that fits is $c-a=5$, and then $9\cdot 5=5h$ gives $h=9$.
💡 A multiple of $5$ built from $9\times(\text{gap})$ can only work if the gap itself is a multiple of $5$.
6.EE.B.8 Step 4 Use the digit budget to fix the digits
- From $c-a=5$ we have $c=a+5$.
- Plug this into the budget $a+b+c\le 7$: it becomes $a+b+(a+5)\le 7$, i.e.
- $2a+b\le 2$.
- Since $a\ge 1$, already $2a\ge 2$, leaving no room, so $b=0$ and $a=1$.
- Then $c=a+5=6$.
- The digits are forced to the single triple $(a,b,c)=(1,0,6)$.
💡 The budget $a+b+c\le 7$ is so tight that the boundary leaves exactly one legal set of digits.
6.EE.A.1 Step 5 Evaluate the requested sum of squares
- With $a=1$, $b=0$, $c=6$, compute $a^2+b^2+c^2=1^2+0^2+6^2=1+0+36=37$.
- This matches choice (D).
💡 Once the digits are known, the answer is just three squares added up.
6.EE.B.6 Let the starting reading be $\overline{abc}=100a+10b+c$ and the ending reading b 6.RP.A.3 Distance is speed times time, so the trip covers $55\times h=55h$ miles. The sam 4.OA.B.4 Both $99$ and $55$ share the factor $11$, so divide the equation by $11$ to get 6.EE.B.8 From $c-a=5$ we have $c=a+5$. Plug this into the budget $a+b+c\le 7$: it becomes 6.EE.A.1 With $a=1$, $b=0$, $c=6$, compute $a^2+b^2+c^2=1^2+0^2+6^2=1+0+36=37$. This matc Review
Reasonableness: 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, $495\div 55=9$, a whole number of hours, so the $55$ mph condition holds exactly. The digit budget gives $a+b+c=1+0+6=7\le 7$ and $a=1\ge 1$, both satisfied. Everything is consistent, and $a^2+b^2+c^2=37$ agrees with answer (D).
Alternative: Instead of the divisibility argument, note the distance $99(c-a)$ must be a multiple of $55$. Testing $c-a=1,2,\dots$ gives $99,198,297,396,495,\dots$; the first that is a multiple of $55$ is $495=55\cdot 9$ at $c-a=5$. From there the budget $a+b+c\le 7$ with $c=a+5$ again forces $(1,0,6)$, so $a^2+b^2+c^2=37$, choice (D).
CCSS standards used (min grade 6)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the digits $a,b,c$ and the hours $h$, and writing both odometer readings by place value.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Turning $55$ miles per hour over $h$ whole hours into the distance $55h$ and equating it to the odometer difference.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Reasoning that $9(c-a)$ is a multiple of $5$ forces the gap $c-a$ to be a multiple of $5$, so $c-a=5$.)6.EE.B.8Write an inequality of the form x > c or x < c and graph on a number line (Using the constraint $a+b+c\le 7$ with $c=a+5$ to bound the digits and force $a=1,\,b=0,\,c=6$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating $a^2+b^2+c^2=1^2+0^2+6^2=37$.)
⭐ 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 $55\times\text{hours}$ pins the digits down.
⭐ 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 $55\times\text{hours}$ pins the digits down.
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