sum to 2024

The post below has been deleted. Click to close.

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

pog

4917 posts

pog

#1 h Nov 7, 2024, 5:07 PM

Y by

The number $2024$ is written as the sum of not necessarily distinct two-digit numbers. What is the least number of two-digit numbers needed to write this sum?

$\textbf{(A) }20\qquad\textbf{(B) }21\qquad\textbf{(C) }22\qquad\textbf{(D) }23\qquad\textbf{(E) }24$

This post has been edited 1 time. Last edited by jlacosta, Nov 7, 2024, 5:08 PM

Z K Y

The post below has been deleted. Click to close.

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

ConfidentKoala4

599 posts

ConfidentKoala4

#2 h Nov 7, 2024, 5:07 PM

Y by

B i think iirc

Z K Y

The post below has been deleted. Click to close.

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

pog

4917 posts

pog

#3 h Nov 7, 2024, 5:08 PM • 1 Y

Y by ranu540

We want to use as many $99$ s as possible. Then $99 \cdot 20 + 44 = 2024$ , so our answer is $20 + 1 = \boxed{\textbf{(B) }21}$ .

Z K Y

The post below has been deleted. Click to close.

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

bobjoebilly

79 posts

bobjoebilly

#4 h Nov 7, 2024, 5:09 PM • 1 Y

Y by pog

fairly certain this was also #3 on 2024 AMC 12A

Z K Y

The post below has been deleted. Click to close.

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

Jonysun

34 posts

Jonysun

#6 h Nov 7, 2024, 5:10 PM

Y by

Yeah it's $\fbox{B} \Rightarrow 21$ .

Z K Y

The post below has been deleted. Click to close.

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

HonestCat

972 posts

HonestCat

#7 h Nov 7, 2024, 5:10 PM

Y by

MAA looked at this and said “yeah, this is harder than p2”

Z K Y

The post below has been deleted. Click to close.

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

Sedro

5837 posts

Sedro

#8 h Nov 7, 2024, 5:10 PM

Y by

B confirmed.

Z K Y

The post below has been deleted. Click to close.

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

The_Reaperr

43 posts

The_Reaperr

#9 h Nov 7, 2024, 5:11 PM

Y by

Yes it is B

Z K Y

The post below has been deleted. Click to close.

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

ElaineGu

389 posts

ElaineGu

#10 h Nov 7, 2024, 5:11 PM

Y by

Yup B. 99*21>2024

Z K Y

The post below has been deleted. Click to close.

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

MathRook7817

659 posts

MathRook7817

#11 h Nov 7, 2024, 5:12 PM

Y by

ceil(2024/99)

Z K Y

The post below has been deleted. Click to close.

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

Marcus_Zhang

979 posts

Marcus_Zhang

#12 h Nov 7, 2024, 5:13 PM

Y by

legit read this one wrong as "2024 is the sum of two not necesarily distinct two-digit numbers"

rest in peace my reading skills

This post has been edited 1 time. Last edited by Marcus_Zhang, Nov 7, 2024, 5:13 PM

Z K Y

The post below has been deleted. Click to close.

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

LostDreams

145 posts

LostDreams

#13 h Nov 7, 2024, 5:14 PM

Y by

Notice how it says "not necessarily distinct"
So yea just max out with bunch of 99s and you get B

Z K Y

The post below has been deleted. Click to close.

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

golden_star_123

204 posts

golden_star_123

#15 h Nov 7, 2024, 5:19 PM

Y by

This one was extremely easy and should have been #1.

This is $\lceil \frac{2024}{99} \rceil$ , which is $\boxed{21}$ .

Z K Y

The post below has been deleted. Click to close.

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

aryanpadarthi

4995 posts

aryanpadarthi

#16 h Nov 7, 2024, 5:19 PM

Y by

this should have been #2 or #1

ez

Z K Y

The post below has been deleted. Click to close.

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

SomeonecoolLovesMaths

3197 posts

SomeonecoolLovesMaths

#17 h Nov 7, 2024, 6:07 PM

Y by

oh no I assumed them to be distinct

Z K Y

The post below has been deleted. Click to close.

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

wangzrpi

158 posts

wangzrpi

#18 h Nov 7, 2024, 6:13 PM

Y by

HonestCat wrote:

MAA looked at this and said “yeah, this is harder than p2”

didn't even read p2
B confirmed
this should be an amc8 problem

Z K Y

The post below has been deleted. Click to close.

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

OlympusHero

17020 posts

OlympusHero

#19 h Nov 7, 2024, 6:15 PM

Y by

May have misbubbled C 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.

xHypotenuse

773 posts

xHypotenuse

#20 h Nov 7, 2024, 6:23 PM

Y by

OlympusHero wrote:

May have misbubbled C for this question

Don't worry I think answer choices and questions were mixed up as an anti-cheat

Z K Y

The post below has been deleted. Click to close.

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

MC_ADe

176 posts

MC_ADe

#21 h Nov 7, 2024, 6:24 PM

Y by

sillied, said A cuz 99x20, forgot to +1 cuz of 44

Z K Y

The post below has been deleted. Click to close.

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

gicyuraok2

1059 posts

gicyuraok2

#22 h Nov 7, 2024, 11:27 PM

Y by

quite a simple problem, just $99*20+44=2024$ , but you can easily mess this one up

Z K Y

The post below has been deleted. Click to close.

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

andrewcheng

525 posts

andrewcheng

#23 h Nov 7, 2024, 11:27 PM

Y by

pog wrote:

The number $2024$ is written as the sum of not necessarily distinct two-digit numbers. What is the least number of two-digit numbers needed to write this sum?

$\textbf{(A) }20\qquad\textbf{(B) }21\qquad\textbf{(C) }22\qquad\textbf{(D) }23\qquad\textbf{(E) }24$

easier than P2 should be moved

Z K Y

The post below has been deleted. Click to close.

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