Difference between revisions of "1992 AIME Problems/Problem 6"

Latest revision as of 01:03, 10 July 2021

Problem

For how many pairs of consecutive integers in $\{1000,1001,1002,\ldots,2000\}$ is no carrying required when the two integers are added?

Solution 1

For one such pair of consecutive integers, let the smaller integer be $\underline{1ABC},$ where $A,B,$ and $C$ are digits from $0$ through $9.$

We wish to count the ordered triples $(A,B,C).$ By casework, we consider all possible forms of the larger integer, as shown below. $\begin{array}{c|c|c|c|c|c|c} & & & & & & \\ [-2.5ex] \textbf{Case} & \textbf{Thousands} & \textbf{Hundreds} & \textbf{Tens} & \textbf{Ones} & \textbf{Conditions for No Carrying} & \boldsymbol{\#}\textbf{ of Ordered Triples} \\ [0.5ex] \hline & & & & & & \\ [-2ex] 0\leq C\leq 8 & 1 & A & B & C+1 & 0\leq A,B,C\leq 4 & 5^3 \\ 0\leq B\leq 8; \ C=9 & 1 & A & B+1 & 0 & 0\leq A,B\leq 4; \ C=9 & 5^2 \\ 0\leq A\leq 8; \ B=C=9 & 1 & A+1 & 0 & 0 & 0\leq A\leq 4; \ B=C=9 & 5 \\ A=B=C=9 & 2 & 0 & 0 & 0 & A=B=C=9 & 1 \end{array}$ Together, the answer is $5^3+5^2+5+1=\boxed{156}.$

~MRENTHUSIASM

Solution 2

Consider what carrying means: If carrying is needed to add two numbers with digits $abcd$ and $efgh$ , then $h+d\ge 10$ or $c+g\ge 10$ or $b+f\ge 10$ . 6. Consider $c \in \{0, 1, 2, 3, 4\}$ . $1abc + 1ab(c+1)$ has no carry if $a, b \in \{0, 1, 2, 3, 4\}$ . This gives $5^3=125$ possible solutions.

With $c \in \{5, 6, 7, 8\}$ , there obviously must be a carry. Consider $c = 9$ . $a, b \in \{0, 1, 2, 3, 4\}$ have no carry. This gives $5^2=25$ possible solutions. Considering $b = 9$ , $a \in \{0, 1, 2, 3, 4, 9\}$ have no carry. Thus, the solution is $125 + 25 + 6=\boxed{156}$ .

Solution 3

Consider the ordered pair $(1abc , 1abc - 1)$ where $a,b$ and $c$ are digits. We are trying to find all ordered pairs where $(1abc) + (1abc - 1)$ does not require carrying. For the addition to require no carrying, $2a,2b < 10$ , so $a,b < 5$ unless $1abc$ ends in $00$ , which we will address later. Clearly, if $c \in \{0, 1, 2, 3, 4 ,5\}$ , then adding $(1abc) + (1abc - 1)$ will require no carrying. We have $5$ possibilities for the value of $a$ , $5$ for $b$ , and $6$ for $c$ , giving a total of $(5)(5)(6) = 150$ , but we are not done yet.

We now have to consider the cases where $b,c = 0$ , specifically when $1abc \in \{1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000\}$ . We can see that $1100, 1200, 1300, 1400, 1500$ , and $2000$ all work, giving a grand total of $150 + 6 = \boxed{156}$ ordered pairs.

1992 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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