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 "1999 AMC 8 Problems/Problem 24"
m (→‎Solution 2)
 
(14 intermediate revisions by 7 users not shown)
Line 7: Line 7:
  
 
==Solution 1==
 
==Solution 1==
−
Note that the units digits of the powers of 9 have a pattern: <math>9^1 = {\bf 9}</math>,<math>9^2 = 8{\bf 1}</math>,<math>9^3 = 72{\bf 9}</math>,<math>9^4 = 656{\bf 1}</math>, and so on. Since all natural numbers with the same last digit have the same remainder when divided by 5, the entire number doesn't matter, just the last digit. For even powers of <math>9</math>, the number ends in a <math>1</math>. Since the exponent is even, the final digit is <math>1</math>. Note that all natural numbers that end in <math>1</math> have a remainder of <math>1</math> when divided by <math>5</math>. So, our answer is <math>\boxed{\text{(B)}\ 1}</math>.
+
Note that the units digits of the powers of 9 have a pattern: <math>9^1 = {\bf 9}</math>, <math>9^2 = 8{\bf 1}</math>, <math>9^3 = 72{\bf 9}</math>, <math>9^4 = 656{\bf 1}</math>, and so on. Since all natural numbers with the same last digit have the same remainder when divided by 5, the entire number doesn't matter, just the last digit. For even powers of <math>9</math>, the number ends in a <math>1</math>. Since the exponent is even, the final digit is <math>1</math>. Note that all natural numbers that end in <math>1</math> have a remainder of <math>1</math> when divided by <math>5</math>. So, our answer is <math>\boxed{\text{(B)}\ 1}</math>.
  
 
==Solution 2==
 
==Solution 2==
  
−
Write <math>1999</math> as <math>2000-1</math>. We are taking <math>(2000-1)^{2000} \mod{10}</math>. Using the binomial theorem, we see that ALL terms in this expansion are divisible by <math>2000</math> except for the very last term, which is just <math>(-1)^{2000}</math>. This is clear because the binomial expansion is just choosing how many <math>2000</math>s and how many <math>-1</math>s there are for each term. Using this, we can take the entire polynomial <math>\mod{10}</math>, which leaves just <math>(-1)^{2000}=\boxed{\text{(B)}\ 1}</math>.
+
Write <math>1999</math> as <math>2000-1</math>. We are taking <math>(2000-1)^{2000} \mod{10}</math>. Using the binomial theorem, we see that ALL terms in this expansion are divisible by <math>2000</math>, which is in turn divisible by <math>5</math>, except for the very last term, which is just <math>(-1)^{2000}</math>. This is clear because the binomial expansion is just choosing how many <math>2000</math>s and how many <math>-1</math>s there are for each term. Using this, we can take the entire polynomial <math>\mod{10}</math>, which leaves just <math>(-1)^{2000}=\boxed{\text{(B)}\ 1}</math>.
  
 
==Solution 3==
 
==Solution 3==
Line 17: Line 17:
  
 
==Solution 4==
 
==Solution 4==
−
A sum/product of any two natural numbers has the same remainder, when divided by <math>n \in \mathbb{N}</math>, as the sum/product of their remainders. Thus, we can use the basic definition of an exponent and view the problem as <math>1999 \cdot 1999 \cdot \cdot \cdot 1999</math>.  
+
A sum/product of any two natural numbers has the same remainder, when divided by <math>n \in \{\mathbb N}</math>, as the sum/product of their remainders. Thus, we can use the basic definition of an exponent and view the problem as <math>1999 \cdot 1999 \cdot \cdot \cdot 1999</math>.  
  
 
Using the fact stated in the first sentence, we see that the remainder of <math>1999</math>, when divided by <math>5</math>, is <math>4</math>. The problem can now be written viewed as finding the remainder of <math>4^{2000}</math> when it is divided by <math>5</math>, which is already much simpler.  
 
Using the fact stated in the first sentence, we see that the remainder of <math>1999</math>, when divided by <math>5</math>, is <math>4</math>. The problem can now be written viewed as finding the remainder of <math>4^{2000}</math> when it is divided by <math>5</math>, which is already much simpler.  
  
 
Now, toying with the simplified problem a little, see notice that powers of <math>4</math> alternate from ending in <math>4</math> and <math>6</math>. We notice that even powers of <math>4</math> always end in <math>6</math>, and also that <math>2000</math> is even. Thus <math>4^{2000}</math> must end in <math>6</math>, which, when divided by <math>5</math> gives a remainder of <math>\boxed{\text{(B)}\ 1}</math>.
 
Now, toying with the simplified problem a little, see notice that powers of <math>4</math> alternate from ending in <math>4</math> and <math>6</math>. We notice that even powers of <math>4</math> always end in <math>6</math>, and also that <math>2000</math> is even. Thus <math>4^{2000}</math> must end in <math>6</math>, which, when divided by <math>5</math> gives a remainder of <math>\boxed{\text{(B)}\ 1}</math>.
 +
 +
==Video Solution by Soo==
 +
https://youtu.be/jCHiQMK7k3w - Soo, DRMS, NM
 +
 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 +
 +
https://www.youtube.com/watch?v=U_WOZpd7eYQ
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=1999|num-b=23|num-a=25}}
 
{{AMC8 box|year=1999|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:56, 23 August 2024

Problem

When $1999^{2000}$ is divided by $5$, the remainder is

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$


Solution 1

Note that the units digits of the powers of 9 have a pattern: $9^1 = {\bf 9}$, $9^2 = 8{\bf 1}$, $9^3 = 72{\bf 9}$, $9^4 = 656{\bf 1}$, and so on. Since all natural numbers with the same last digit have the same remainder when divided by 5, the entire number doesn't matter, just the last digit. For even powers of $9$, the number ends in a $1$. Since the exponent is even, the final digit is $1$. Note that all natural numbers that end in $1$ have a remainder of $1$ when divided by $5$. So, our answer is $\boxed{\text{(B)}\ 1}$.

Solution 2

Write $1999$ as $2000-1$. We are taking $(2000-1)^{2000} \mod{10}$. Using the binomial theorem, we see that ALL terms in this expansion are divisible by $2000$, which is in turn divisible by $5$, except for the very last term, which is just $(-1)^{2000}$. This is clear because the binomial expansion is just choosing how many $2000$s and how many $-1$s there are for each term. Using this, we can take the entire polynomial $\mod{10}$, which leaves just $(-1)^{2000}=\boxed{\text{(B)}\ 1}$.

Solution 3

As $1999 \equiv -1 \pmod{5}$, we have $1999^{2000} \equiv (-1)^{2000} \equiv 1 \pmod{5}$. Thus, the answer is $\boxed{\text{(B)}\ 1}$.

Solution 4

A sum/product of any two natural numbers has the same remainder, when divided by $n \in \{\mathbb N}$ (Error compiling LaTeX. Unknown error_msg), as the sum/product of their remainders. Thus, we can use the basic definition of an exponent and view the problem as $1999 \cdot 1999 \cdot \cdot \cdot 1999$.

Using the fact stated in the first sentence, we see that the remainder of $1999$, when divided by $5$, is $4$. The problem can now be written viewed as finding the remainder of $4^{2000}$ when it is divided by $5$, which is already much simpler.

Now, toying with the simplified problem a little, see notice that powers of $4$ alternate from ending in $4$ and $6$. We notice that even powers of $4$ always end in $6$, and also that $2000$ is even. Thus $4^{2000}$ must end in $6$, which, when divided by $5$ gives a remainder of $\boxed{\text{(B)}\ 1}$.

Video Solution by Soo

https://youtu.be/jCHiQMK7k3w - Soo, DRMS, NM

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=U_WOZpd7eYQ

See Also

1999 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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=1999_AMC_8_Problems/Problem_24&oldid=225625"