2025 AMC 8 Problem

Art of Problem Solving

AoPS Online

Contests & Programs AMC and other contests, summer programs, etc.

AMC and other contests, summer programs, etc.

3 M G

BBookmark VNew Topic kLocked

Contests & Programs AMC and other contests, summer programs, etc.

AMC and other contests, summer programs, etc.

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

2025 AMC 8 Problem

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: 2025 AMC 8 Problem #15

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

172 posts

#1 h Jan 31, 2025, 3:29 AM

Y by

Kei draws a $6$ -by- $6$ grid. He colors $13$ of the unit squares silver and the remaining squares gold. Kei then folds the grid in half vertically, forming pairs of overlapping unit squares. Let $m$ and $M$ equal equal the least and greatest possible number of gold-on-gold pairs, respectively. What is the value of $m+M$ ?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 16\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 20$

Had fun doing this one!

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

958 posts

#2 h Jan 31, 2025, 3:36 AM

Y by

I got this wrong

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4216 posts

#3 h Jan 31, 2025, 3:42 AM

Y by

aight so we have $18$ pairs

these pairs are one of gold-gold, gold-silver, silver-silver

let $a$ be gold-gold, $b$ be gold-silver, $c$ be silver-silver

$a+b+c=18$ and $b+2c=13$

subtracting both equations gives $a-c=5$ , so $a\ge 5$

we substitute to get $2a+b=23$ so $a\le 11$

our answer is $5+11=\boxed{16}$

the specific constructions are $(a,b,c)=(5,13,0)$ and $(a,b,c)=(11,1,6)$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

749 posts

#4 h Jan 31, 2025, 4:14 AM

Y by

Wait I might have put 11+3. Shoot….

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

143 posts

KMSONI

#5 h Feb 1, 2025, 12:43 AM

Y by

Thank god, i put 16. WHat question number was this again?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1555 posts

#6 h Feb 1, 2025, 1:10 AM

Y by

15; it says at the top where it says the source

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

172 posts

#7 h Feb 1, 2025, 2:03 AM

Y by

@RedFireTruck
No offence, but you should probably hide your solutions so people don't get spoiled.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1057 posts

#8 h Feb 1, 2025, 2:16 AM

Y by

I got Answer for this question

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3 posts

CJB19

#9 h Mar 15, 2025, 3:00 AM

Y by

I remember this one! Don't read this if you want to solve it yourself (Idk how to hide stuff)! So basically, the most overlapping gold occurs when we maximize silver overlap. This can be don't by having 6 pairs of silver-on-silver and 1 pair of silver-on-gold. this leaves 18-(6+1)=11 pairs as our maximum. To minimize gold-on-gold, we want to maximize silver-on-gold. This gives 13-silver-on-gold, and 5 gold-on-gold. Therefore, our answer is 5+11=16

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a