Difference between revisions of "2018 AMC 10B Problems/Problem 13"
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− | ==Solution== | + | == Solution 1 == |
− | Note that <math>10^{2k}+1</math> for some odd <math>k</math> will suffice <math>\mod {101}</math>. Each <math>2k \in \{2,4, | + | 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,-10,1)\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>\dfrac{2020}{4}=\boxed{\textbf{(C) } 505}</math> numbers in that list. | ||
+ | <br> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Note that <math>10^{2k}+1</math> for some odd <math>k</math> will suffice <math>\mod {101}</math>. Each <math>2k \in \{2,6,10,\dots,2018\}</math>, so the answer is <math>\boxed{\textbf{(C) } 505}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | If we divide each number by <math>101</math>, we see a pattern occuring in every 4 numbers. <math>101, 1000001, 10000000001, \dots</math>. We divide <math>2018</math> by <math>4</math> to get <math>504</math> with <math>2</math> left over. Looking at our pattern of four numbers from above, the first number is divisible by <math>101</math>. This means that the first of the <math>2</math> left over will be divisible by <math>101</math>, so our answer is <math>\boxed{\textbf{(C) } 505}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | Note that <math>909</math> is divisible by <math>101</math>, and thus <math>9999</math> is too. We know that <math>101</math> is divisible and <math>1001</math> isn't so let us start from <math>10001</math>. We subtract <math>9999</math> to get 2. Likewise from <math>100001</math> we subtract, but we instead subtract <math>9999</math> times <math>10</math> or <math>99990</math> to get <math>11</math>. We do it again and multiply the 9's by <math>10</math> to get <math>101</math>. Following the same knowledge, we can use mod <math>101</math> to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is <math>{0,-9,-99 ( 2),11, 0, ...}</math>. Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide <math>2017</math> by four to get <math>504</math> remainder <math>1</math>. Thus the answer is <math>504</math> plus the 1st term or <math>\boxed{\textbf{(C) } 505}</math>. | ||
+ | ==Solution 5== | ||
+ | Note that <math>101=x^2+1</math> and <math>100...0001=x^n+1</math>, where <math>x=10</math>. We have that <math>\frac{x^n+1}{x^2+1}</math> must have a remainder of <math>0</math>. By the remainder theorem, the roots of <math>x^2+1</math> must also be roots of <math>x^n+1</math>. Plugging in <math>i,-i</math> to <math>x^n+1</math> yields that <math>n\equiv2\mod{4}</math>. Because the sequence starts with <math>10^2+1</math>, the answer is <math>\lceil 2018/4 \rceil=\boxed{\textbf{(C) } 505}</math> | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2018|ab=B|num-b=12|num-a=14}} | ||
+ | {{MAA Notice}} |
Latest revision as of 22:53, 3 August 2024
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.
Solution 2
Note that for some odd
will suffice
. Each
, so the answer is
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
.
Solution 5
Note that and
, where
. We have that
must have a remainder of
. By the remainder theorem, the roots of
must also be roots of
. Plugging in
to
yields that
. Because the sequence starts with
, the answer is
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.