Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 8 Problems/Problem 7"

(→‎Solution 1)
(→‎Video Solution (💩Fast💩))
 
(35 intermediate revisions by 23 users not shown)
Line 1: Line 1:
−
==Problem 7==
+
==Problem==
−
How many integers between <math>2020</math> and <math>2400</math> have four distinct digits arranged in increasing order? (For example, <math>2347</math> is one integer.)
+
How many integers between <math>2020</math> and <math>2400</math> have four distinct digits arranged in increasing order? (For example, <math>2347</math> is one integer.).
  
 
<math>\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}</math>
 
<math>\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}</math>
  
 
==Solution 1==
 
==Solution 1==
−
First, observe that the second digit of the four digit number cannot be a <math>1</math> or a <math>2</math> because the digits must be distinct and increasing. The second digit also cannot be a <math>4</math> because the number must be less than <math>2400</math>. Thus, the second digit must be <math>3</math>. If we place a <math>4</math> in the third digit then there are 5 ways to select the last digit, namely the last digit could then be <math>5,6,7,8,</math> or <math>9</math>. If we place a <math>5</math> in the third digit, there are 4 ways to select the last digit, namely the last digit could then be <math>6,7,8,</math> or <math>9</math>. Similarly, if the third digit is <math>6</math>, there are 3 ways to select the last digit, etc. Thus, it follows that the total number of valid numbers is <math>5+4+3+2+1=15\implies\boxed{\textbf{(C) }15}</math>.<br>
+
Firstly, observe that the second digit of such a number cannot be <math>1</math> or <math>2</math>, because the digits must be distinct and increasing. The second digit also cannot be <math>4</math> as the number must be less than <math>2400</math>, so it must be <math>3</math>. It remains to choose the latter two digits, which must be <math>2</math> distinct digits from <math>\left\{4,5,6,7,8,9\right\}</math>. That can be done in <math>\binom{6}{2} = \frac{6 \cdot 5}{2 \cdot 1} = 15</math> ways; there is then only <math>1</math> way to order the digits, namely in increasing order. This means the answer is <math>\boxed{\textbf{(C) }15}</math>.
−
~jmansuri
 
  
−
==See also==  
+
==Solution 2 (without using the "choose" function)==
−
{{AMC8 box|year=2020|num-b=6|num-a=8}}
+
As in Solution 1, we find that the first two digits must be <math>23</math>, and the third digit must be at least <math>4</math>. If it is <math>4</math>, then there are <math>5</math> choices for the last digit, namely <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, or <math>9</math>. Similarly, if the third digit is <math>5</math>, there are <math>4</math> choices for the last digit, namely <math>6</math>, <math>7</math>, <math>8</math>, and <math>9</math>; if <math>6</math>, there are <math>3</math> choices; if <math>7</math>, there are <math>2</math> choices; and if <math>8</math>, there is <math>1</math> choice. It follows that the total number of such integers is <math>5+4+3+2+1=\boxed{\textbf{(C) }15}</math>.
−
{{MAA Notice}}
+
 
 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=8hgK6rESdek&t=9s
 +
 
 +
~NiuniuMaths
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=xBEcgPkL367f3Zp8&t=713
 +
 
 +
~Math-X
 +
 
 +
==Video Solution (🚀Fast🚀)==
 +
https://youtu.be/QqBpLTQojHg
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/FjmBtgrGfCs
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution==
 +
https://youtu.be/61c1MR9tne8 ~ The Learning Royal
  
 +
==Video Solution by Interstigation==
 +
https://youtu.be/YnwkBZTv5Fw?t=251
  
−
The thousands place (first digit) has to be a 2 (2020-2400).
+
~Interstigation
−
Since the thousands digit is 2, the next digit must be a 3 (not 4 or onwards because that will go over the range given).
 
  
−
The next digit has to be from 4, 5, 6, 7, or 8. For each of the cases, you get a total of 15 possibilities, which gives you the answer C.
+
==Video Solution by STEMbreezy==
 +
https://youtu.be/U27z1hwMXKY?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=85
  
 +
~STEMbreezy
  
−
~itsmemasterS
+
==See also==
 +
{{AMC8 box|year=2020|num-b=6|num-a=8}}
 +
{{MAA Notice}}

Latest revision as of 19:02, 12 June 2024

Problem

How many integers between $2020$ and $2400$ have four distinct digits arranged in increasing order? (For example, $2347$ is one integer.).

$\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}$

Solution 1

Firstly, observe that the second digit of such a number cannot be $1$ or $2$, because the digits must be distinct and increasing. The second digit also cannot be $4$ as the number must be less than $2400$, so it must be $3$. It remains to choose the latter two digits, which must be $2$ distinct digits from $\left\{4,5,6,7,8,9\right\}$. That can be done in $\binom{6}{2} = \frac{6 \cdot 5}{2 \cdot 1} = 15$ ways; there is then only $1$ way to order the digits, namely in increasing order. This means the answer is $\boxed{\textbf{(C) }15}$.

Solution 2 (without using the "choose" function)

As in Solution 1, we find that the first two digits must be $23$, and the third digit must be at least $4$. If it is $4$, then there are $5$ choices for the last digit, namely $5$, $6$, $7$, $8$, or $9$. Similarly, if the third digit is $5$, there are $4$ choices for the last digit, namely $6$, $7$, $8$, and $9$; if $6$, there are $3$ choices; if $7$, there are $2$ choices; and if $8$, there is $1$ choice. It follows that the total number of such integers is $5+4+3+2+1=\boxed{\textbf{(C) }15}$.

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=8hgK6rESdek&t=9s

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=xBEcgPkL367f3Zp8&t=713

~Math-X

Video Solution (🚀Fast🚀)

https://youtu.be/QqBpLTQojHg

~Education, the Study of Everything

Video Solution by WhyMath

https://youtu.be/FjmBtgrGfCs

~savannahsolver

Video Solution

https://youtu.be/61c1MR9tne8 ~ The Learning Royal

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=251

~Interstigation

Video Solution by STEMbreezy

https://youtu.be/U27z1hwMXKY?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=85

~STEMbreezy

See also

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_7&oldid=220419"