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1992 AIME Problems/Problem 6

Revision as of 21:59, 9 July 2021 by MRENTHUSIASM (talk | contribs) (→‎Solution 1)

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.

See also

1992 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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