2014 AMC 10B Problem 10

P1-10~5 minPrint

Bases & Digits Basic Counting Split into exhaustive cases and handle each Set up the equation/formula and compute; no special trick needed

In the addition shown below $A, B, C,$ and $D$ are distinct digits. How many different values are possible for $D$ ?

$\begin{array}{lr}&ABBCB\\ +& BCADA\\ \hline & DBDDD\end{array}$

Show solution without answering

Key insight

Columns where the answer digit repeats an added digit force C = 0 and no carrying, so D = A + B for different nonzero digits A and B.

Solution

Label the columns from the right: units, tens, hundreds, thousands, ten-thousands. Reading down each column:

column	top	bottom	answer
units	$B$	$A$	$D$
tens	$C$	$D$	$D$
hundreds	$B$	$A$	$D$
thousands	$B$	$C$	$B$
ten-thousands	$A$	$B$	$D$

1. Start with the tens column, because the answer digit $D$ is also one of the digits being added. $C + D$ (plus any carry from the units) must end in $D$ . So $C$ plus the carry adds nothing to the last digit: either $C = 0$ with no carry, or $C = 9$ with a carry of $1$ .
2. Rule out $C = 9$ . A carry into the tens column means $B + A$ is $10$ or more, so $B + A = D + 10$ . The tens column $9 + D + 1$ also carries $1$ . Then the hundreds column is $B + A + 1 = D + 11$ , which ends in $D + 1$ , not $D$ . That fails.
3. So $C = 0$ and nothing carries out of the units column. Now every column is easy:
- units: $B + A = D$ ;
- tens: $0 + D = D$ ;
- hundreds: $B + A = D$ again;
- thousands: $B + 0 = B$ ;
- ten-thousands: $A + B = D$ .
Everything fits. The only rule is $D = A + B$ , with no carrying anywhere.
4. Which values can $D$ take? $A$ and $B$ are different digits, and neither is $0$ (they must differ from $C = 0$ ; also they start the two numbers). The smallest sum is $1 + 2 = 3$ and the largest allowed is $9$ (no carry).
5. Every value from $3$ to $9$ works, for example $1 + 2, 1 + 3, 1 + 4, 1 + 5, 1 + 6, 1 + 7, 1 + 8$ . In each case $D$ is bigger than both $A$ and $B$ and is not $0$ , so all four letters are different.

That is $3, 4, 5, 6, 7, 8, 9$ : seven values.

The answer is $\boxed{\textbf{(C)}\ 7}$ .

Why this works

In a letter-addition puzzle, the most useful columns are the ones where the answer digit also appears among the digits being added. There the other digit plus the carry must be $0$ or $10$ , which leaves only two cases. Settling those columns first shrinks the whole puzzle to one simple rule.

Concise solution

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

Work column by column from the right, tracking carries.

Units: $B + A$ ends in $D$ ; call the carry $c_1$ .

Tens: $C + D + c_1$ ends in $D$ , so $C + c_1$ is $0$ or $10$ . Either $C = 0$ and $c_1 = 0$ , or $C = 9$ and $c_1 = 1$ .

If $C = 9$ , $c_1 = 1$ : then $A + B = D + 10$ and the tens column produces a carry of $1$ . The hundreds column reads $B + A + 1 = D + 11$ , which ends in $D + 1$ , not $D$ . Impossible.

If $C = 0$ , $c_1 = 0$ : then $A + B = D$ with no carry, the tens column is $0 + D = D$ with no carry, the hundreds column is $B + A = D$ again with no carry, the thousands column is $B + 0 = B$ , and the ten-thousands column is $A + B = D$ . Everything is consistent.

So the only requirement is $D = A + B$ where $A$ , $B$ are distinct digits, both nonzero (they differ from $C = 0$ ), and $D \le 9$ . Since $A \ne B$ , the sum $A + B$ automatically differs from both. The possible sums range from $1 + 2 = 3$ up to $9$ , and every value $3, 4, \dots, 9$ is achievable (for example $1 + 2, 1 + 3, 1 + 4, 1 + 5, 1 + 6, 1 + 7, 1 + 8$ ).

That is $7$ values, so the answer is $\boxed{\textbf{(C)}\ 7}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 8: Base Numbers
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Number Theory, Chapter 9: Base Number Arithmetic