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Difference between revisions of "1992 AIME Problems/Problem 6"
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== Solution ==
 
== Solution ==
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=== Solution 1 ===
 
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.  
 
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 2 ===
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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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{{AIME box|year=1992|num-b=5|num-a=7}}
 
{{AIME box|year=1992|num-b=5|num-a=7}}

Revision as of 19:11, 5 August 2014

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

Solution 1

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 2

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)
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Problem 7
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