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Difference between revisions of "2025 AIME I Problems/Problem 5"

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m (→‎Video Solution by SpreadTheMathLove)
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==Video Solution by SpreadTheMathLove==
 
==Video Solution by SpreadTheMathLove==
 
https://www.youtube.com/watch?v=P6siafb6rsI
 
https://www.youtube.com/watch?v=P6siafb6rsI
 +
(also the person in the Youtube video wrote the final answer wrong, it was supposed to be 279 and he accidentally wrote it as 729)
  
 
==See also==
 
==See also==

Revision as of 12:26, 18 February 2025

Problem

There are $8!= 40320$ eight-digit positive integers that use each of the digits $1, 2, 3, 4, 5, 6, 7, 8$ exactly once. Let $N$ be the number of these integers that are divisible by $22$. Find the difference between $N$ and $2025$.

Solution 1

Notice that if the 8-digit number is divisible by $22$, it must have an even units digit. Therefore, we can break it up into cases and let the last digit be either $2, 4, 6,$ or $8$. Due to symmetry, upon finding the total count of one of these last digit cases (we look at last digit $2$ here), we may multiply the resulting value by $4$.


Now, we just need to find the number of positions of the remaining numbers such that the units digit is $2$ and the number is divisible by $11$. Denote the odd numbered positions to be $a_1, a_3, a_5, a_7$ and the even numbered positions to be $a_2, a_4, a_6$ (recall $a_8=2$). By the divisibility rule of $11$, we must have: \[(a_1 + a_3 + a_5 + a_7) - (a_2 + a_4 + a_6 + 2)\] which is congruent to $0\hspace{2mm}(\text{mod}\hspace{1mm}11)$. Therefore, after simplifying, we must have: \[a_1 - a_2 + a_3 - a_4 + a_5 - a_6 + a_7\equiv2\hspace{2mm}(\text{mod}\hspace{1mm}11)\] Now consider $a_1+ a_2 +\ldots + a_7=1+2+\ldots+8-2=34\equiv1\hspace{2mm}(\text{mod}\hspace{1mm}11)$. Therefore, \[(a_1 + a_2 + \ldots+ a_7) - 2(a_2 + a_4 + a_6)\equiv2\hspace{2mm}(\text{mod}\hspace{1mm}11)\] which means that \[a_2 + a_4 + a_6\equiv5\hspace{2mm}(\text{mod}\hspace{1mm}11)\] Notice that the minimum of $a_2+a_4+a_6$ is $1 + 3 + 4 = 8$ and the maximum is $6 + 7 + 8 = 21$. The only possible number congruent to $5\hspace{2mm}(\text{mod}\hspace{1mm}11)$ in this range is $16$. All that remains is to count all the possible sums of $16$ using the values $1, 3, 4, 5, 6, 7, 8$. There are a total of four possibilities: \[(1, 7, 8), (3, 5, 8), (3, 6, 7), (4, 5, 7)\] The arrangement of the odd-positioned numbers ($a_1,a_3,a_5,a_7$) does not matter, so there are $4!=24$ arrangements of these numbers. Recall that the $4$ triplets above occupy $a_2,a_4,a_6$; the number of arrangements is $3!=6$. Thus, we have $24\cdot6\cdot4=576$ possible numbers such that the units digit is $2$. Since we claimed symmetry over the rest of the units digits, we must multiply by $4$, resulting in $576\cdot4=2304$ eight-digit positive integers. Thus, the positive difference between $N$ and $2025$ is $2304 - 2025 = \boxed{279}$.

~ilikemath247365

~LaTeX by eevee9406

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=P6siafb6rsI (also the person in the Youtube video wrote the final answer wrong, it was supposed to be 279 and he accidentally wrote it as 729)

See also

2025 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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

1. To be multiple of 11: Total of 1,2,3,4,5,6,7,8 is 36, dividing into two groups of 4 numbers, the difference of sum of two group x and y need to be 0 or multiple of 11, i.e. x+y=36, x-y=0,11,22… only x=y=18 is possible. Number 8 can only be with (8,1,4,5),(8,1,2,7),(8,1,3,6),(8,2,3,5). One group of 4 numbers make 4! different arrangement, two groups make 4!*4!, the 2 group makes 2! arrangement. The two group of numbers are alternating by digits. Total number of multiple of 11 is 4*2!*4!*4! 2. To be multiple of 2: We noticed in each number group, there are two odd two even. So the final answer is above divided by 2, 4*2!*4!*4!/2=2304. 2304-2025=279.

~Mathzu.club

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