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? | ||
− | == Solution == | + | == 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> | + | 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> |
+ | |||
+ | 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. | ||
+ | <cmath>\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}</cmath> | ||
+ | Together, the answer is <math>5^3+5^2+5+1=\boxed{156}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | == Solution 2 == | ||
+ | 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>. | ||
+ | == Solution 3 == | ||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | ==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 is no carrying required when the two integers are added?
Solution 1
For one such pair of consecutive integers, let the smaller integer be where and are digits from through
We wish to count the ordered triples By casework, we consider all possible forms of the larger integer, as shown below. Together, the answer is
~MRENTHUSIASM
Solution 2
Consider what carrying means: If carrying is needed to add two numbers with digits and , then or or . 6. Consider . has no carry if . This gives possible solutions.
With , there obviously must be a carry. Consider . have no carry. This gives possible solutions. Considering , have no carry. Thus, the solution is .
Solution 3
Consider the ordered pair where and are digits. We are trying to find all ordered pairs where does not require carrying. For the addition to require no carrying, , so unless ends in , which we will address later. Clearly, if , then adding will require no carrying. We have possibilities for the value of , for , and for , giving a total of , but we are not done yet.
We now have to consider the cases where , specifically when . We can see that , and all work, giving a grand total of 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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