Difference between revisions of "2018 AMC 10B Problems/Problem 13"
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== Solution 1 == | == Solution 1 == | ||
− | The number <math>10^n+1</math> is divisible by 101 if and only if <math>10^n\equiv -1\pmod{101}</math>. We note that <math>(10,10^2,10^3,10^4)\equiv (10,-1,- | + | The number <math>10^n+1</math> is divisible by 101 if and only if <math>10^n\equiv -1\pmod{101}</math>. We note that <math>(10,10^2,10^3,10^4)\equiv (10,-1,-9,1)\pmod{101}</math>, so the powers of 10 are 4-periodic mod 101. It follows that <math>10^n\equiv -1\pmod{101}</math> if and only if <math>n\equiv 2\pmod 4</math>. |
In the given list, <math>10^2+1,10^3+1,10^4+1,\dots,10^{2019}+1</math>, the desired exponents are <math>2,6,10,\dots,2018</math>, and there are <math>2020/4=\boxed{\textbf{(C) } 505}</math> numbers in that list. | In the given list, <math>10^2+1,10^3+1,10^4+1,\dots,10^{2019}+1</math>, the desired exponents are <math>2,6,10,\dots,2018</math>, and there are <math>2020/4=\boxed{\textbf{(C) } 505}</math> numbers in that list. | ||
+ | <br> | ||
+ | Minor fixes by fasterthanlight | ||
==Solution 2== | ==Solution 2== |
Revision as of 22:55, 23 May 2020
Problem
How many of the first numbers in the sequence are divisible by ?
Solution 1
The number is divisible by 101 if and only if . We note that , so the powers of 10 are 4-periodic mod 101. It follows that if and only if .
In the given list, , the desired exponents are , and there are numbers in that list.
Minor fixes by fasterthanlight
Solution 2
Note that for some odd will suffice . Each , so the answer is (AOPS12142015)
Solution 3
If we divide each number by , we see a pattern occuring in every 4 numbers. . We divide by to get with left over. Looking at our pattern of four numbers from above, the first number is divisible by . This means that the first of the left over will be divisible by , so our answer is .
Solution 4
Note that is divisible by , and thus is too. We know that is divisible and isn't so let us start from . We subtract to get 2. Likewise from we subtract, but we instead subtract times or to get . We do it again and multiply the 9's by to get . Following the same knowledge, we can use mod to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is . Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide by four to get remainder . Thus the answer is plus the 1st term or .
-googleghosh
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.