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

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For how many pairs of consecutive integers in <math>\{1000,1001,1002,\ldots,2000\}</math> is no carrying required when the two integers are added?
 
For how many pairs of consecutive integers in <math>\{1000,1001,1002,\ldots,2000\}</math> is no carrying required when the two integers are added?
  
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== Solution ==
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== Solution 1 ==
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Consider what carrying means: If carrying is needed to add two numbers with digits <math>abcd</math> and <math>efgh</math>, then <math>h+d\ge 10</math> or <math>c+g\ge 10</math> or <math>b+f\ge 10</math>. 6.  Consider <math>c \in \{0, 1, 2, 3, 4\}</math>. <math>1abc + 1abc+1</math> has no carry if <math>a, b \in \{0, 1, 2, 3, 4\}</math>. This gives <math>5^3=125</math> possible solutions.  
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For one such pair of consecutive integers, let the smaller integer be <math>\underline{1ABC},</math> where <math>A,B,</math> and <math>C</math> are digits from <math>0</math> through <math>9.</math>
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We wish to count the ordered triples <math>(A,B,C).</math> By casework, we consider all possible forms of the larger integer, as shown below.
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<cmath>\begin{array}{c|c|c|c|c|c|c}
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& & & & & & \\ [-2.5ex]
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\textbf{Case} & \textbf{Thousands} & \textbf{Hundreds} & \textbf{Tens} & \textbf{Ones} & \textbf{Conditions for No Carrying} & \boldsymbol{\#}\textbf{ of Ordered Triples} \\ [0.5ex]
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\hline
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& & & & & & \\ [-2ex]
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0\leq C\leq 8 & 1 & A & B & C+1 & 0\leq A,B,C\leq 4 & 5^3 \\
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0\leq B\leq 8; \ C=9 & 1 & A & B+1 & 0 & 0\leq A,B\leq 4; \ C=9 & 5^2 \\
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0\leq A\leq 8; \ B=C=9 & 1 & A+1 & 0 & 0 & 0\leq A\leq 4; \ B=C=9 & 5 \\
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A=B=C=9 & 2 & 0 & 0 & 0 & A=B=C=9 & 1
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\end{array}</cmath>
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Together, the answer is <math>5^3+5^2+5+1=\boxed{156}.</math>
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~MRENTHUSIASM
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== Solution 2 ==
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Consider what carrying means: If carrying is needed to add two numbers with digits <math>abcd</math> and <math>efgh</math>, then <math>h+d\ge 10</math> or <math>c+g\ge 10</math> or <math>b+f\ge 10</math>. 6.  Consider <math>c \in \{0, 1, 2, 3, 4\}</math>. <math>1abc + 1ab(c+1)</math> has no carry if <math>a, b \in \{0, 1, 2, 3, 4\}</math>. This gives <math>5^3=125</math> possible solutions.  
  
 
With <math>c \in \{5, 6, 7, 8\}</math>, there obviously must be a carry. Consider <math>c = 9</math>. <math>a, b \in \{0, 1, 2, 3, 4\}</math> have no carry. This gives <math>5^2=25</math> possible solutions. Considering <math>b = 9</math>,  <math>a \in \{0, 1, 2, 3, 4, 9\}</math> have no carry. Thus, the solution is <math>125 + 25 + 6=\boxed{156}</math>.  
 
With <math>c \in \{5, 6, 7, 8\}</math>, there obviously must be a carry. Consider <math>c = 9</math>. <math>a, b \in \{0, 1, 2, 3, 4\}</math> have no carry. This gives <math>5^2=25</math> possible solutions. Considering <math>b = 9</math>,  <math>a \in \{0, 1, 2, 3, 4, 9\}</math> have no carry. Thus, the solution is <math>125 + 25 + 6=\boxed{156}</math>.  
  
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== Solution 3 ==
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Consider the ordered pair <math>(1abc , 1abc - 1)</math> where <math>a,b</math> and <math>c</math> are digits. We are trying to find all ordered pairs where <math>(1abc) + (1abc - 1)</math> does not require carrying. For the addition to require no carrying, <math>2a,2b < 10</math>, so <math>a,b < 5</math> unless <math>1abc</math> ends in <math>00</math>, which we will address later.  Clearly, if <math>c \in \{0, 1, 2, 3, 4 ,5\}</math>, then adding <math>(1abc) + (1abc - 1)</math> will require no carrying. We have <math>5</math> possibilities for the value of <math>a</math>, <math>5</math> for <math>b</math>, and <math>6</math> for <math>c</math>, giving a total of <math>(5)(5)(6) = 150</math>, but we are not done yet.
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We now have to consider the cases where <math>b,c = 0</math>, specifically when <math>1abc \in \{1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000\}</math>. We can see that <math>1100, 1200, 1300, 1400, 1500</math>, and <math>2000</math> all work, giving a grand total of <math>150 + 6 = \boxed{156}</math> ordered pairs.
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==See also==
 
{{AIME box|year=1992|num-b=5|num-a=7}}
 
{{AIME box|year=1992|num-b=5|num-a=7}}
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

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.

See also

1992 AIME (Problems • Answer Key • Resources)
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Problem 5 		Followed by
Problem 7
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All AIME Problems and Solutions

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