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 "2017 AMC 12B Problems/Problem 19"
(→‎Solution 5)
 
(10 intermediate revisions by 5 users not shown)
Line 4: Line 4:
 
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math>
 
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math>
  
−
==Solution==
+
==Solution 1==
  
 
We will consider this number <math>\bmod\ 5</math> and <math>\bmod\ 9</math>. By looking at the last digit, it is obvious that the number is <math>\equiv 4\bmod\ 5</math>. To calculate the number <math>\bmod\ 9</math>, note that
 
We will consider this number <math>\bmod\ 5</math> and <math>\bmod\ 9</math>. By looking at the last digit, it is obvious that the number is <math>\equiv 4\bmod\ 5</math>. To calculate the number <math>\bmod\ 9</math>, note that
Line 40: Line 40:
  
 
==Solution 4==
 
==Solution 4==
−
This Python code works:
 
  
−
def createNumberString(n):
+
We notice that <math>10^{k}\equiv 10 \pmod {45}</math>.
−
    final = ""
 
−
    for i in range (1, n+1)
 
−
        final += i
 
−
    return final
 
  
−
N = int(createNumberString(44))
+
Hence <math>N = 44+43\cdot10^{2}+42\cdot10^{4}+\cdots+10^{78} \equiv 44+10\cdot(1+2+3+\cdots+43)\equiv 9 \pmod {45}.</math>
−
remainder = N % 45
+
 
−
print(remainder)
+
Choose <math>\boxed{\textbf{(C)}\,9}</math>
 +
 
 +
~ PythZhou
 +
 
 +
==Solution 5 (Solution 2 but more in-depth)==
 +
 
 +
The sum of all of the digits is just the sum of consecutive numbers from <math>1 -> 44</math> or <math>\frac{44}{2}(45) = 990</math>. The prime factorization of <math>45</math> is <math>3^2 * 5</math>. So if a number is divisible by <math>45</math> it has to both be divisible by <math>9</math> and <math>5</math>. The first number that satisfies this ends in <math>35</math> because the ten's digit is one greater than the ten's digit in the consecutive numbers from <math>1</math> to <math>44</math>, and the one's digit is one less than this number, and the number ends in a 5. The difference between <math>35</math> and <math>44</math> is <math>9</math> giving <math>\boxed{\textbf{(C)}\,9}</math>.
 +
 
 +
~PeterDoesPhysics
 +
 
 +
== Video Solution by OmegaLearn==
 +
https://youtu.be/zfChnbMGLVQ?t=3342
 +
 
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 16:06, 2 September 2024

Problem

Let $N=123456789101112\dots4344$ be the $79$-digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$

Solution 1

We will consider this number $\bmod\ 5$ and $\bmod\ 9$. By looking at the last digit, it is obvious that the number is $\equiv 4\bmod\ 5$. To calculate the number $\bmod\ 9$, note that

\[123456\cdots 4344 \equiv 1+2+3+4+5+6+7+8+9+(1+0)+(1+1)+\cdots+(4+3)+(4+4) \equiv 1+2+\cdots+44 \bmod\ 9,\]

so it is equivalent to

\[\frac{44\cdot 45}{2} = 22\cdot 45 \equiv 0\bmod\ 9.\]

Let $x$ be the remainder when this number is divided by $45$. We know that $x\equiv 0 \pmod {9}$ and $x\equiv 4 \pmod {5}$, so by the Chinese remainder theorem, since $9(-1)\equiv 1 \pmod{5}$, $x\equiv 5(0)+9(-1)(4) \pmod {5\cdot 9}$, or $x\equiv -36 \equiv 9 \pmod {45}$. So the answer is $\boxed {\textbf {(C)}}$.

Solution 2

We know that this number is divisible by $9$ because the sum of the digits is $270$, which is divisible by $9$. If we subtracted $9$ from the integer we would get $1234 \cdots 4335$, which is also divisible by $5$ and by $45$. Thus the remainder is $9$, or $\boxed{\textbf{C}}$.

Solution 3 (Beginner's Method)

To find the sum of digits of our number, we break it up into $5$ cases, starting with $0$, $1$, $2$, $3$, or $4$.

Case 1: $1+2+3+\cdots+9 = 45$,

Case 2: $1+0+1+1+1+2+\cdots+1+8+1+9 = 55$ (We add 10 to the previous cases, as we are in the next ten's place)

Case 3: $2+0+2+1+\cdots+2+9 = 65$,

Case 4: $3+0+3+1+\cdots+3+9 = 75$,

Case 5: $4+0+4+1+\cdots+4+4 = 30$,

Thus the sum of the digits is $45+55+65+75+30 = 270$, so the number is divisible by $9$. We notice that the number ends in "$4$", which is $9$ more than a multiple of $5$. Thus if we subtracted $9$ from our number it would be divisible by $5$, and $5\cdot 9 = 45$. (Multiple of n - n = Multiple of n)


So our remainder is $\boxed{\textbf{(C)}\,9}$, the value we need to add to the multiple of $45$ to get to our number.

Solution 4

We notice that $10^{k}\equiv 10 \pmod {45}$.

Hence $N = 44+43\cdot10^{2}+42\cdot10^{4}+\cdots+10^{78} \equiv 44+10\cdot(1+2+3+\cdots+43)\equiv 9 \pmod {45}.$

Choose $\boxed{\textbf{(C)}\,9}$

~ PythZhou

Solution 5 (Solution 2 but more in-depth)

The sum of all of the digits is just the sum of consecutive numbers from $1 -> 44$ or $\frac{44}{2}(45) = 990$. The prime factorization of $45$ is $3^2 * 5$. So if a number is divisible by $45$ it has to both be divisible by $9$ and $5$. The first number that satisfies this ends in $35$ because the ten's digit is one greater than the ten's digit in the consecutive numbers from $1$ to $44$, and the one's digit is one less than this number, and the number ends in a 5. The difference between $35$ and $44$ is $9$ giving $\boxed{\textbf{(C)}\,9}$.

~PeterDoesPhysics

Video Solution by OmegaLearn

https://youtu.be/zfChnbMGLVQ?t=3342

~ pi_is_3.14

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 AMC 12 Problems and Solutions
2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 AMC 10 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=2017_AMC_12B_Problems/Problem_19&oldid=226237"