Difference between revisions of "2017 AMC 12B Problems/Problem 19"
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<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 | ||
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To find the sum of digits of our number, we break it up into <math>5</math> cases, starting with <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, or <math>4</math>. | To find the sum of digits of our number, we break it up into <math>5</math> cases, starting with <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, or <math>4</math>. | ||
− | Case 1: <math>1+2+3+\cdots+9 = 45</math> | + | Case 1: <math>1+2+3+\cdots+9 = 45</math>, |
+ | |||
Case 2: <math>1+0+1+1+1+2+\cdots+1+8+1+9 = 55</math> (We add 10 to the previous cases, as we are in the next ten's place) | Case 2: <math>1+0+1+1+1+2+\cdots+1+8+1+9 = 55</math> (We add 10 to the previous cases, as we are in the next ten's place) | ||
− | Case 3: <math>2+0+2+1+\cdots+2+9 = 65</math> | + | |
− | Case 4: <math>3+0+3+1+\cdots+3+9 = 75</math> | + | Case 3: <math>2+0+2+1+\cdots+2+9 = 65</math>, |
− | Case 5: <math>4+0+4+1+\cdots+4+4 = 30</math> | + | |
+ | Case 4: <math>3+0+3+1+\cdots+3+9 = 75</math>, | ||
+ | |||
+ | Case 5: <math>4+0+4+1+\cdots+4+4 = 30</math>, | ||
Thus the sum of the digits is <math>45+55+65+75+30 = 270</math>, so the number is divisible by <math>9</math>. We notice that the number ends in "<math>4</math>", which is <math>9</math> more than a multiple of <math>5</math>. Thus if we subtracted <math>9</math> from our number it would be divisible by <math>5</math>, and <math>5\cdot 9 = 45</math>. (Multiple of n - n = Multiple of n) | Thus the sum of the digits is <math>45+55+65+75+30 = 270</math>, so the number is divisible by <math>9</math>. We notice that the number ends in "<math>4</math>", which is <math>9</math> more than a multiple of <math>5</math>. Thus if we subtracted <math>9</math> from our number it would be divisible by <math>5</math>, and <math>5\cdot 9 = 45</math>. (Multiple of n - n = Multiple of n) | ||
− | So our remainder is <math>\boxed{\textbf{(C)} 9}</math>, the value we need to add to the multiple of <math>45</math> to get to our number. | + | |
+ | So our remainder is <math>\boxed{\textbf{(C)}\,9}</math>, the value we need to add to the multiple of <math>45</math> to get to our number. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | We notice that <math>10^{k}\equiv 10 \pmod {45}</math>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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
Contents
Problem
Let be the -digit number that is formed by writing the integers from to in order, one after the other. What is the remainder when is divided by ?
Solution 1
We will consider this number and . By looking at the last digit, it is obvious that the number is . To calculate the number , note that
so it is equivalent to
Let be the remainder when this number is divided by . We know that and , so by the Chinese remainder theorem, since , , or . So the answer is .
Solution 2
We know that this number is divisible by because the sum of the digits is , which is divisible by . If we subtracted from the integer we would get , which is also divisible by and by . Thus the remainder is , or .
Solution 3 (Beginner's Method)
To find the sum of digits of our number, we break it up into cases, starting with , , , , or .
Case 1: ,
Case 2: (We add 10 to the previous cases, as we are in the next ten's place)
Case 3: ,
Case 4: ,
Case 5: ,
Thus the sum of the digits is , so the number is divisible by . We notice that the number ends in "", which is more than a multiple of . Thus if we subtracted from our number it would be divisible by , and . (Multiple of n - n = Multiple of n)
So our remainder is , the value we need to add to the multiple of to get to our number.
Solution 4
We notice that .
Hence
Choose
~ PythZhou
Solution 5 (Solution 2 but more in-depth)
The sum of all of the digits is just the sum of consecutive numbers from or . The prime factorization of is . So if a number is divisible by it has to both be divisible by and . The first number that satisfies this ends in because the ten's digit is one greater than the ten's digit in the consecutive numbers from to , and the one's digit is one less than this number, and the number ends in a 5. The difference between and is giving .
~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 |
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