AMC 10 · 2014 · #10
Grade 6 arithmeticIn the addition shown below A, B, C, and D are distinct digits. How many different values are possible for D?
\begin{array}[t]{r} ABBCB \\ + \ BCADA \\ \hline DBDDD \end{array}
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: In the column addition $\begin{array}{r} ABBCB \\ +\,BCADA \\ \hline DBDDD \end{array}$ the letters $A$, $B$, $C$, $D$ stand for four distinct digits. Count how many different digits $D$ could possibly be.
Givens: The two five-digit addends are $ABBCB$ and $BCADA$; their sum is the five-digit number $DBDDD$; $A$, $B$, $C$, $D$ are digits ($0$ through $9$) and are all distinct from one another; $A$, $B$, $D$ are leading digits, so none of them is $0$; Answer choices: (A) $2$, (B) $4$, (C) $7$, (D) $8$, (E) $9$
Unknowns: The number of different values the digit $D$ can take
Understand
Restated: In the column addition $\begin{array}{r} ABBCB \\ +\,BCADA \\ \hline DBDDD \end{array}$ the letters $A$, $B$, $C$, $D$ stand for four distinct digits. Count how many different digits $D$ could possibly be.
Givens: The two five-digit addends are $ABBCB$ and $BCADA$; their sum is the five-digit number $DBDDD$; $A$, $B$, $C$, $D$ are digits ($0$ through $9$) and are all distinct from one another; $A$, $B$, $D$ are leading digits, so none of them is $0$; Answer choices: (A) $2$, (B) $4$, (C) $7$, (D) $8$, (E) $9$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #14 Extreme Principle
A cryptarithm is solved by naming the hidden quantities and letting the columns speak. Tool #4 (Introduce a Variable) adds one carry digit $c_1,\dots,c_4$ between the columns, turning the picture into five exact equations. Tool #7 (Identify Subproblems) treats each column as its own tiny equation, and comparing two of them collapses the carries to zero and pins $C=0$. Once the puzzle reduces to $A+B=D$, Tool #14 (Extreme Principle) finds the smallest and largest $D$ can be, and everything between is reachable — so counting the values is just measuring that range.
Execute — Answer: C
6.EE.B.6 Step 1 Name a carry for each column
- Line the addition up by place value and let $c_1,c_2,c_3,c_4$ be the carries out of the units, tens, hundreds, and thousands columns.
- Each carry is $0$ or $1$.
- Reading each column (right to left) as digit-plus-carry equals result-digit-plus-ten-times-next-carry gives five equations.
- The tens and thousands columns simplify right away because $D$ (tens) and $B$ (thousands) appear on both sides.
💡 A single unknown carry per column converts the picture into equations you can manipulate instead of guess.
6.EE.B.5 Step 2 Compare units and hundreds
- The units column and the hundreds column both add $A+B$.
- Subtract the units equation from the hundreds equation: the $A+B$ and the $D$ cancel, leaving $c_2 = 10c_3 - 10c_1 = 10(c_3-c_1)$.
- But $c_2$ is only $0$ or $1$, and $10(c_3-c_1)$ is a multiple of $10$.
- The only way is $c_2 = 0$, which forces $c_3 = c_1$.
💡 A number that is both a single carry and a multiple of ten has no choice but to be zero.
6.EE.B.5 Step 3 Force every carry to zero and C to zero
- Put $c_2=0$ into the tens equation $C+c_1=10c_2$: it becomes $C+c_1=0$.
- Digits are non-negative, so $C=0$ and $c_1=0$.
- Then $c_3=c_1=0$.
- Finally the thousands equation $C+c_3=10c_4$ reads $0=10c_4$, so $c_4=0$.
- Every carry is $0$ and $C=0$.
💡 With no carry allowed to leak, the middle letter $C$ is squeezed all the way down to $0$.
4.NBT.B.4 Step 4 Reduce the whole puzzle to A + B = D
- With all carries $0$, the units, hundreds, and ten-thousands equations all say the same thing: $A+B=D$.
- There is no carry, so $A+B\le 9$.
- Because $A$ and $B$ are non-zero leading digits, $D=A+B$ is automatically bigger than each of them, so $D$ differs from $A$ and $B$; and $D\ge 3>0=C$, so $D$ differs from $C$ too.
- Distinctness takes care of itself as long as $A\ne B$.
💡 Once the carries vanish, the five-letter tower is really just one small addition fact.
4.NBT.B.4 Step 5 Sweep D from smallest to largest
- Now use the boundary cases.
- The smallest $D=A+B$ with two different non-zero digits is $1+2=3$.
- The largest, staying at or below $9$, is a pair summing to $9$, such as $1+8=9$.
- Every whole number from $3$ up to $9$ is a reachable sum of two distinct non-zero digits (e.g.
- $3=1+2$, $4=1+3$, $5=1+4$, $6=2+4$, $7=3+4$, $8=1+7$, $9=4+5$).
- So $D\in\{3,4,5,6,7,8,9\}$, which is $7$ values.
💡 Find the two extremes and check nothing in between is skipped, and the count is just the length of the range.
6.EE.B.6 Line the addition up by place value and let $c_1,c_2,c_3,c_4$ be the carries out 6.EE.B.5 The units column and the hundreds column both add $A+B$. Subtract the units equa 6.EE.B.5 Put $c_2=0$ into the tens equation $C+c_1=10c_2$: it becomes $C+c_1=0$. Digits a 4.NBT.B.4 With all carries $0$, the units, hundreds, and ten-thousands equations all say t 4.NBT.B.4 Now use the boundary cases. The smallest $D=A+B$ with two different non-zero dig Review
Reasonableness: Test one case fully. Take $A=1$, $B=2$, so $C=0$ and $D=A+B=3$. Then $ABBCB=12202$ and $BCADA=20131$, and $12202+20131=32333$, which is exactly $DBDDD=32333$ with $D=3$, $B=2$. The pattern holds, and the four digits $1,2,0,3$ are distinct. Since the same $A+B=D$ rule works for every valid pair, and $D$ runs over the seven integers $3$ through $9$, the count of $7$ is consistent with answer (C).
Alternative: Skip the algebra of carries and reason from the addend $BCADA$ having leading digit $B$ while the sum $DBDDD$ has leading digit $D$: since adding a five-digit number to $ABBCB$ keeps the result five digits, no carry can escape the top column, so the top column gives $A+B=D$ directly. The tens column $C+D=D$ then forces $C=0$ with no carry, and the rest follows. Counting $D=A+B$ over distinct non-zero digits with $A+B\le 9$ again gives $\{3,\dots,9\}$, seven values, answer (C).
CCSS standards used (min grade 6)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Introducing carry variables $c_1,\dots,c_4$ and writing one equation per column of the addition.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Testing which carry values can satisfy $c_2=10(c_3-c_1)$ and $C+c_1=0$, forcing every carry and $C$ to zero.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Reducing the columns to $A+B=D$ with no carry and checking the concrete sum $12202+20131=32333$.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Reading each labelled column by place value and enforcing that leading digits $A$, $B$, $D$ are non-zero.)
⭐ Give each hidden carry a name; the columns then force $C=0$ and $A+B=D$, so $D$ can be any digit from $3$ to $9$ — seven values in all.
⭐ Give each hidden carry a name; the columns then force $C=0$ and $A+B=D$, so $D$ can be any digit from $3$ to $9$ — seven values in all.
More like this
Same archetype — closest grade level first.