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Difference between revisions of "2024 AIME II Problems/Problem 3"

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==Problem==
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Find the number of ways to place a digit in each cell of a 2x3 grid so that the sum of the two numbers formed by reading left to right is <math>999</math>, and the sum of the three numbers formed by reading top to bottom is <math>99</math>. The grid below is an example of such an arrangement because <math>8+991=999</math> and <math>9+9+81=99</math>.
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<cmath>
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\begin{array}{|c|c|c|} \hline
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0 & 0 & 8 \\ \hline
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9 & 9 & 1 \\ \hline
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\end{array}
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</cmath>
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==Solution 1==
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Consider this table:
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<math>
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\begin{array}{|c|c|c|} \hline
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a & b & c \\ \hline
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d & e & f\\ \hline
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\end{array}
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</math>
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We note that <math>c+f = 9</math>, because <math>c+f \leq 18</math>, meaning it never achieves a unit's digit sum of <math>9</math> otherwise. Since no values are carried onto the next digit, this implies <math>b+e=9</math> and <math>a+d=9</math>. We can then simplify our table into this:
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<math>
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\begin{array}{|c|c|c|} \hline
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a & b & c \\ \hline
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9-a & 9-b & 9-c \\ \hline
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\end{array}
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</math>
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We want <math>10(a+b+c) + (9-a+9-b+9-c) = 99</math>, or <math>9(a+b+c+3) = 99</math>, or <math>a+b+c=8</math>. Since zeroes are allowed, we just need to apply stars and bars on <math>a, b, c</math>, to get <math>\tbinom{8+3-1}{3-1} = \boxed{045}</math>. ~akliu
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==Video Solution==
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https://youtu.be/nKRfXAHaQvA
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==See also==
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{{AIME box|year=2024|num-b=2|num-a=4|n=II}}
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[[Category:Intermediate Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 17:08, 5 March 2024

Problem

Find the number of ways to place a digit in each cell of a 2x3 grid so that the sum of the two numbers formed by reading left to right is $999$, and the sum of the three numbers formed by reading top to bottom is $99$. The grid below is an example of such an arrangement because $8+991=999$ and $9+9+81=99$.

\[\begin{array}{|c|c|c|} \hline 0 & 0 & 8 \\ \hline 9 & 9 & 1 \\ \hline \end{array}\]

Solution 1

Consider this table:

$\begin{array}{|c|c|c|} \hline a & b & c \\ \hline d & e & f\\ \hline \end{array}$

We note that $c+f = 9$, because $c+f \leq 18$, meaning it never achieves a unit's digit sum of $9$ otherwise. Since no values are carried onto the next digit, this implies $b+e=9$ and $a+d=9$. We can then simplify our table into this:

$\begin{array}{|c|c|c|} \hline a & b & c \\ \hline 9-a & 9-b & 9-c \\ \hline \end{array}$

We want $10(a+b+c) + (9-a+9-b+9-c) = 99$, or $9(a+b+c+3) = 99$, or $a+b+c=8$. Since zeroes are allowed, we just need to apply stars and bars on $a, b, c$, to get $\tbinom{8+3-1}{3-1} = \boxed{045}$. ~akliu

Video Solution

https://youtu.be/nKRfXAHaQvA

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2024 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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