,
?
Y by byron-aj-tom
Prove that for all real numbers 
and 
verifying 
. 
and 
verifying 
. 
G
Topic
First Poster
Last Poster
9 MATHCOUNTS STATE difficulty
Eddie_tiger 61
N
by DhruvJha
I personally thought the problems were much easier than last year, but I didn't really improve as much as I would of liked to improve.
61 replies
Easy Counting?
orangefronted 2
N
by Mathematicalprodigy37
Lucien randomly picks two distinct numbers at the same time from the following group of numbers: 
. If the product of the 2 numbers is also a member of this group, how many selections can be done?
. If the product of the 2 numbers is also a member of this group, how many selections can be done?
2 replies
Factoring Marathon
pican 1454
N
by AVY2024
Hello guys,
I think we should start a factoring marathon. Post your solutions like this , and your problems like this . Please make your own problems, and I'll start off simple:
I think we should start a factoring marathon. Post your solutions like this , and your problems like this . Please make your own problems, and I'll start off simple:
1454 replies
Square Track
orangefronted 1
N
by Lankou
Ziron and Zane are running around a square track at constant speeds. Ziron runs at a speed of 2 meters per second, while Zane runs twice as fast as Ziron. They begin at the same point and start running in opposite directions. If they meet after exactly 8 seconds, what is the area enclosed by the square track?
1 reply
Probability NOT a perfect square
orangefronted 4
N
by ilikemath247365
Mike decides to play a game with himself. He begins with a score of 0 and proceeds to flip a fair coin. If he lands on heads, he adds 2 to his score. If he lands on tails, he subtracts 1 from his score. After 5 flips, what is the probability that Mike’s score is not a perfect square?
4 replies
prime factorization formula questions
Soupboy0 8
N
by martianrunner
If i have a number, say 
with prime factorization 
, and I want to find 
numbers that multiply to , does anybody know a formula for this?
with prime factorization 
, and I want to find 
numbers that multiply to , does anybody know a formula for this?
8 replies
2025 MATHCOUNTS State Hub
SirAppel 217
N
by morestuf
Previous Years' "Hubs": (2022) (2023) (2024)
Now that it's April and we're allowed to discuss, and no one else has made this yet ...
[list=disc]
[*] CA: 43 (45 44 43 43 43 42 42 41 41 41)
[*] NJ: 43 (45 44 44 43 39 42 40 40 39 38) *
[*] NY: 42 (43 42 42 42 41 40)
[*] TX: 42 (43 43 43 42 42 40 40 38 38 38)
[*] MA: 41 (45 43 42 41)
[*] WA: 41 (41 45 42 41 41 41 41 41 41 40) *
[*] FL: 39 (42 41 40 39 38 37 37)
[*] IN: 39 (41 40 40 39 ?? 35)
[*] NC: 39 (42 42 41 39)
[*] IL: 38 (41 40 39 38 38 38)
[*] OR: 38 (44 41? 38 38)
[*] PA: 38 (41 40 40 38 38 37 36 36 34 34) *
[*] MD: 37 (43 39 39 37 37 37)
[*] CT: 36 (44 38 36 34 34 34 34)
[*] MI: 36 (39 41 41 36 37 37 36 36 36 36) *
[*] MN: 36 (40 36 36 36 35 35 35 34)
[*] CO: 35 (41 37 37 35 35 35 ?? 31 31 30) *
[*] GA: 35 (38 37 36 35 34 34 34 34 34 33)
[*] OH: 35 (41 37 36 35)
[*] AR: 34 (46 45 35 34 33 31 31 31 29 29)
[*] WI: 34 (40 37 37 34 35 30 28 29 29 29) *
[*] NH: 31 (42 35 33 31 30)
[*] DE: 30 (34 33 32 30 30 29 28 27 26? 24)
[*] SC: 30 (33 33 31 30)
[*] IA: 29 (33 30 31 29 29 29 29 29 29 29 29 29) *
[*] NE: 28 (34 30 28 28 27 27 26 26 25 25)
[*] SD: 22 (30 29 24 22 22 22 21 21 20 20)
[/list]
* means that CDR is official in that state.
For those asking about the removal of the tiers, I'd like to quote Jason himself:
[quote=peace09]
learn from my mistakes
[/quote]
Help contribute by sharing your state's cutoffs!
As per last year's guidelines, refrain from problem discussion until their official release on the MATHCOUNTS website.
Now that it's April and we're allowed to discuss, and no one else has made this yet ...
[list=disc]
[*] CA: 43 (45 44 43 43 43 42 42 41 41 41)
[*] NJ: 43 (45 44 44 43 39 42 40 40 39 38) *
[*] NY: 42 (43 42 42 42 41 40)
[*] TX: 42 (43 43 43 42 42 40 40 38 38 38)
[*] MA: 41 (45 43 42 41)
[*] WA: 41 (41 45 42 41 41 41 41 41 41 40) *
[*] FL: 39 (42 41 40 39 38 37 37)
[*] IN: 39 (41 40 40 39 ?? 35)
[*] NC: 39 (42 42 41 39)
[*] IL: 38 (41 40 39 38 38 38)
[*] OR: 38 (44 41? 38 38)
[*] PA: 38 (41 40 40 38 38 37 36 36 34 34) *
[*] MD: 37 (43 39 39 37 37 37)
[*] CT: 36 (44 38 36 34 34 34 34)
[*] MI: 36 (39 41 41 36 37 37 36 36 36 36) *
[*] MN: 36 (40 36 36 36 35 35 35 34)
[*] CO: 35 (41 37 37 35 35 35 ?? 31 31 30) *
[*] GA: 35 (38 37 36 35 34 34 34 34 34 33)
[*] OH: 35 (41 37 36 35)
[*] AR: 34 (46 45 35 34 33 31 31 31 29 29)
[*] WI: 34 (40 37 37 34 35 30 28 29 29 29) *
[*] NH: 31 (42 35 33 31 30)
[*] DE: 30 (34 33 32 30 30 29 28 27 26? 24)
[*] SC: 30 (33 33 31 30)
[*] IA: 29 (33 30 31 29 29 29 29 29 29 29 29 29) *
[*] NE: 28 (34 30 28 28 27 27 26 26 25 25)
[*] SD: 22 (30 29 24 22 22 22 21 21 20 20)
[/list]
* means that CDR is official in that state.
For those asking about the removal of the tiers, I'd like to quote Jason himself:
[quote=peace09]
learn from my mistakes
[/quote]
Help contribute by sharing your state's cutoffs!
As per last year's guidelines, refrain from problem discussion until their official release on the MATHCOUNTS website.
217 replies
I cant find one problem
tanujkundu 9
N
by vsarg
Does anybody know which problem is about when a number is a meteor and when a number is a shooting star?
9 replies
Cool one
G
H
G
H
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Y by
Very easy , Just use the geometric Serie Expansion
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
This post has been edited 1 time. Last edited by vsarg, Mar 15, 2025, 10:12 PM
Reason: Many mistake for me. I edit to ffix it
Reason: Many mistake for me. I edit to ffix it
Z
K
Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
#3
Mar 15, 2025, 10:32 PM
Y by
vsarg wrote:
Very easy , Just use the geometric Serie Expansion
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
Not valid you didnt prove.
Z
K
Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Y by
vsarg wrote:
Very easy , Just use the geometric Serie Expansion
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
EXACTLY WHY IS IT GREATER THAT
?This post has been edited 1 time. Last edited by MTA_2024, Mar 15, 2025, 10:42 PM
Reason: Miswriting
Reason: Miswriting
Z
K
Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
#5
Mar 15, 2025, 11:55 PM
Y by
EaZ_Shadow wrote:
vsarg wrote:
Very easy , Just use the geometric Serie Expansion
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
Not valid you didnt prove.
His proof is valid (and imo explained thoroughly enough).
Z
K
Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
#6
Mar 15, 2025, 11:59 PM
Y by
MTA_2024 wrote:
Prove that for all real numbers and verifying .
and verifying . bro! the middle term is a^n + a^{n-1}b + ... + b^n, so it has 11 terms made up of mixed powers of a and b. Because a>b, then the middle term is larger than (n+1)b^n. Because b<a, the middle term is less than (n+1)a^n. Easy Peasy Lemon Squeezy! Now that's rizz!
I agree! vsarg's proof is thorough and complete! Anyone who says the opposite is just a hater unwilling to embrace the fundamental property of multiplication and factorisations!
This post has been edited 1 time. Last edited by ohiorizzler1434, Mar 15, 2025, 11:59 PM
Z
K
Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
#8
Mar 20, 2025, 8:52 AM
Y by
This is the natural solution above! I have an idea that is not only valid for natural numbers 
! Let's see! Consider the function 
on the interval ![$[a;b]$](//latex.artofproblemsolving.com/0/0/6/0066f4b2714f88cdaf883b00f877aa1355b7f9ef.png)
. If we apply Lagrange's finite growth theorem, then there is a 
in the interval 
for which 
But since 
we already got the problem. And we didn't have to use the abbreviated calculation formula anywhere, which is only valid for natural numbers !
! Let's see! Consider the function 
on the interval ![$[a;b]$](http://latex.artofproblemsolving.com/0/0/6/0066f4b2714f88cdaf883b00f877aa1355b7f9ef.png)
. If we apply Lagrange's finite growth theorem, then there is a 
in the interval 
for which 
But since 
we already got the problem. And we didn't have to use the abbreviated calculation formula anywhere, which is only valid for natural numbers !This post has been edited 1 time. Last edited by invisibleman, Mar 20, 2025, 8:53 AM
Z
K
Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
#9
Mar 20, 2025, 9:58 AM
Y by
what the sigam? what is lagrange finite growth theorem? it does not appear in google saerch.
Z
K
Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
#10
Mar 20, 2025, 10:56 AM
Y by
ohiorizzler1434 wrote:
what the sigam? what is lagrange finite growth theorem? it does not appear in google saerch.
Dear friend! Maybe I didn't express myself clearly. I meant this:
https://en.wikipedia.org/wiki/Mean_value_theorem
Z
K
Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Y by
ohiorizzler1434 wrote:
MTA_2024 wrote:
Prove that for all real numbers and verifying .
and verifying . bro! the middle term is a^n + a^{n-1}b + ... + b^n, so it has 11 terms made up of mixed powers of a and b. Because a>b, then the middle term is larger than (n+1)b^n. Because b<a, the middle term is less than (n+1)a^n. Easy Peasy Lemon Squeezy! Now that's rizz!
I agree! vsarg's proof is thorough and complete! Anyone who says the opposite is just a hater unwilling to embrace the fundamental property of multiplication and factorisations!
Very correct Ohifrizzler thank yod for baking me up from meanies like Eazy. In Kentucky we would
horse race them to seeee whos the better one.
Z
K
Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Y by
MTA_2024 wrote:
Prove that for all real numbers and verifying .
and verifying . https://math.stackexchange.com/questions/3890494/how-could-someone-conceive-of-using-this-inequality-for-this-proof?noredirect=1
Z
K
Y
N Quick Reply
G
H
