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

m (→‎Video Solution by SpreadTheMathLove)
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)
 
(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)
 +
 +
~Mathycoder
  
 
==See also==
 
==See also==

Latest revision as of 12:28, 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)

~Mathycoder

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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