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

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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.
 
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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=== Solution 3 ===
 
=== Solution 3 ===
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Thus, since there should be no carrying, in <math>O</math> only integers <math>0</math> to <math>4</math> is possible
 
Thus, since there should be no carrying, in <math>O</math> only integers <math>0</math> to <math>4</math> is possible
 
Therefore, the answer is <math>{156}</math>
 
Therefore, the answer is <math>{156}</math>
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{{AIME box|year=1992|num-b=5|num-a=7}}
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[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Revision as of 19:00, 14 August 2016

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.


Solution 3

There are 3 forms possible. $1OOO$ : $5^3$ , $1OO9$ : $5^2$ , $1O99$ : $5^1$ , $1999$ : $5^0 or 1$ --- Thus, since there should be no carrying, in $O$ only integers $0$ to $4$ is possible Therefore, the answer is ${156}$


1992 AIME (Problems • Answer Key • Resources)
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Problem 7
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