Difference between revisions of "2011 AMC 10B Problems/Problem 23"
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<math>\begin{aligned} 11^{2011} &= (1+10)^{2011} \\ &= 1 + \dbinom{2011}{1} \cdot 10 + \dbinom{2011}{2} \cdot 10^2 + \cdots \end{aligned}</math> | <math>\begin{aligned} 11^{2011} &= (1+10)^{2011} \\ &= 1 + \dbinom{2011}{1} \cdot 10 + \dbinom{2011}{2} \cdot 10^2 + \cdots \end{aligned}</math> | ||
− | In all of the other terms, the power of <math>10</math> is greater than <math> | + | In all of the other terms, the power of <math>10</math> is greater than <math>2</math> and so is equivalent to <math>0</math> modulo <math>1000,</math> which means we can ignore it. We have: |
<math>\begin{aligned}11^{2011} &\equiv 1 + 2011\cdot 10 + \dfrac{2011 \cdot 2010}{2} \cdot 100 \\ &\equiv 1+20110 + \dfrac{11\cdot 10}{2} \cdot 100\\ &= 1 + 20110 + 5500\\ &\equiv 1 + 110 + 500\\&=611 \pmod{1000} \end{aligned}</math> | <math>\begin{aligned}11^{2011} &\equiv 1 + 2011\cdot 10 + \dfrac{2011 \cdot 2010}{2} \cdot 100 \\ &\equiv 1+20110 + \dfrac{11\cdot 10}{2} \cdot 100\\ &= 1 + 20110 + 5500\\ &\equiv 1 + 110 + 500\\&=611 \pmod{1000} \end{aligned}</math> | ||
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Therefore, the hundreds digit is <math>\boxed{\textbf{(D) } 6}.</math> | Therefore, the hundreds digit is <math>\boxed{\textbf{(D) } 6}.</math> | ||
− | + | Side note: By [[Euler's Totient Theorem]], <math>a^{\phi (1000)} \equiv 1 \pmod{1000}</math> for any <math>a</math> relatively prime with 1000, so <math>a^{400} \equiv 1 \pmod{1000}</math> and <math>11^{2011} \equiv 11^{11} \pmod{1000}</math>. We can then proceed using the clever application of the Binomial Theorem, or we can just proceed with solution 6 from here. | |
== Solution 2 == | == Solution 2 == | ||
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-jackshi2006 | -jackshi2006 | ||
− | ==Solution 4 (naive solution | + | ==Solution 4 (naive solution) == |
Since we are only looking at the last 3 digits and <math>2011^2</math> has the same last 3 digits as <math>11^2</math>, we can find <math>11^{2011}</math> instead. | Since we are only looking at the last 3 digits and <math>2011^2</math> has the same last 3 digits as <math>11^2</math>, we can find <math>11^{2011}</math> instead. | ||
Line 44: | Line 44: | ||
From this information, we can figure out <math>11^{11}</math> and <math>11^{61}</math> end in 611. Adding various multiples of 50 to the exponent gives us the fact that <math>11^{2011}</math>'s last digits are 611. We get <math>\boxed{\textbf{(D) } 6}</math>. | From this information, we can figure out <math>11^{11}</math> and <math>11^{61}</math> end in 611. Adding various multiples of 50 to the exponent gives us the fact that <math>11^{2011}</math>'s last digits are 611. We get <math>\boxed{\textbf{(D) } 6}</math>. | ||
+ | -ThisUsernameIsTaken | ||
+ | |||
+ | ==Solution 5 (modular bash)== | ||
+ | We know that the hundreds digit of <math>2011^{2011}</math> is just the hundreds digit of <math>11^{2011} \equiv 011 \pmod{1000}</math>. If we actually take a look at the powers of <math>11 \pmod{1000}</math>, we notice a pattern: | ||
+ | |||
+ | <cmath>11^{1} \equiv 011 \pmod{1000}</cmath> | ||
+ | <cmath>11^{2} \equiv 121 \pmod{1000}</cmath> | ||
+ | <cmath>11^{3} \equiv 331 \pmod{1000}</cmath> | ||
+ | <cmath>11^{4} \equiv 641 \pmod{1000}</cmath> | ||
+ | <cmath>11^{5} \equiv 051 \pmod{1000}</cmath> | ||
+ | <cmath>11^{6} \equiv 561 \pmod{1000}</cmath> | ||
+ | Notice how the units digit is always one, the tens digit is always the previous tens digit plus the ones digit (or one) and the hundreds digit is always the previous hundreds digit plus the previous tens digit. Knowing this, we confidently repeat this pattern without actually multiply the previous term by <math>11</math> out (if you generally multiply a few numbers by <math>11</math>, you can see why this pattern holds). | ||
+ | |||
+ | <cmath>11^{7} \equiv 171 \pmod{1000}</cmath> | ||
+ | <cmath>11^{8} \equiv 881 \pmod{1000}</cmath> | ||
+ | <cmath>11^{9} \equiv 691 \pmod{1000}</cmath> | ||
+ | <cmath>11^{10} \equiv 601 \pmod{1000}</cmath> | ||
+ | <cmath>11^{11} \equiv 611 \pmod{1000}</cmath> | ||
+ | Since <math>11^{11}</math> ends in _<math>11</math>, the pattern "cycles" every <math>10</math> times. <math>\frac{2011}{10}=201</math> remainder <math>1</math> , meaning that <math>2011^{2011} \equiv 11^{2011} \equiv 611 \pmod{1000}</math>, giving us a hundreds digit of <math>\boxed{\textbf{(D) } 6}</math>. | ||
+ | |||
+ | ==Solution 6 (Super Easy) (Wrong)== | ||
+ | We know that the hundreds digit of <math>2011^{11}</math> is just the hundreds digit of <math>11^{11}</math> (mod 1000). If we actually take a look at the powers of 11, we notice a pattern: | ||
+ | |||
+ | <cmath>11^{1} = 11</cmath> | ||
+ | |||
+ | <cmath>11^{2} = 121</cmath> | ||
+ | |||
+ | <cmath>11^{3} = 1331</cmath> | ||
+ | |||
+ | <cmath>11^{4} = 14641</cmath> | ||
+ | |||
+ | |||
+ | We notice that the hundreds digit just increases by 1 starting from 1(the hundreds digit of <math>11^{1}</math>). All we have to do now is add all the numbers from 1 to 11 (since the hundreds digit increases by 1 with every power of 11 that comes after) to get the hundreds digit of <math>11^{11}</math>. | ||
+ | |||
+ | <math>1 + 2 + 3 + 4.... + 11 = 66</math>. Thus the answer is <math>\boxed{\textbf{(D) } 6}</math>. | ||
+ | |||
+ | -TheOfficalPro | ||
+ | |||
+ | Ilovebender I believe this solution is wrong, as I can see that the hundreds digit doesn't increase by 1, it goes from 0 to 1 to 3 to 6. A more accurate solution would be the hundreds digit are triangular numbers. | ||
+ | |||
+ | -Extremelysupercooldude: In your solution, you find the hundreds digit of <math>2011^{11}</math>, but the problem asks for the hundreds digit of <math>2011^{2011}</math>, but you never cite way the two hundreds digits are the same. | ||
+ | <!-- Latex edits by MrThinker :D --> | ||
+ | |||
+ | -Wesserwes7254: I think this is a more revised version of this solution: | ||
+ | We know that the hundreds digit of <math>2011^{2011}</math> is just the hundreds digit of <math>11^{2011} \equiv 011 \pmod{1000}</math>. If we actually take a look at the powers of <math>11 \pmod{1000}</math>, we notice a pattern: | ||
+ | |||
+ | <cmath>11^{1} \equiv 011 \pmod{1000}</cmath> | ||
+ | |||
+ | <cmath>11^{2} \equiv 121 \pmod{1000}</cmath> | ||
+ | |||
+ | <cmath>11^{3} \equiv 331 \pmod{1000}</cmath> | ||
+ | |||
+ | <cmath>11^{4} \equiv 641 \pmod{1000}</cmath> | ||
+ | |||
+ | <cmath>11^{5} \equiv 051 \pmod{1000}</cmath> | ||
+ | |||
+ | <cmath>11^{6} \equiv 561 \pmod{1000}</cmath> | ||
+ | |||
+ | Notice how the units digit is always one, the tens digit is always the previous tens digit plus the ones digit (or one) and the hundreds digit is always the previous hundreds digit plus the previous tens digit. Knowing this, we confidently repeat this pattern without actually multiply the previous term by <math>11</math> out (if you generally multiply a few numbers by <math>11</math>, you can see why this pattern holds). | ||
+ | |||
+ | <cmath>11^{7} \equiv 171 \pmod{1000}</cmath> | ||
+ | |||
+ | <cmath>11^{8} \equiv 881 \pmod{1000}</cmath> | ||
+ | |||
+ | <cmath>11^{9} \equiv 691 \pmod{1000}</cmath> | ||
+ | |||
+ | <cmath>11^{10} \equiv 601 \pmod{1000}</cmath> | ||
+ | |||
+ | <cmath>11^{11} \equiv 611 \pmod{1000}</cmath> | ||
+ | |||
+ | Since <math>11^{11}</math> ends in _<math>11</math>, the pattern "cycles" every <math>10</math> times. <math>\frac{2011}{10}=201</math> remainder <math>1</math> , meaning that | ||
+ | <math>2011^{2011} \equiv 11^{2011} \equiv 611 \pmod{1000}</math>, giving us a hundreds digit of <math>\boxed{\textbf{(D) } 6}</math>. | ||
+ | |||
+ | -Solution 1's sidenote can start this solution and make it correct. | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/7rosJnqLhs8?t=323 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/-H4n-QplQew?t=1423 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2011|ab=B|num-a=24|num-b=22}} | {{AMC10 box|year=2011|ab=B|num-a=24|num-b=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:47, 13 June 2024
Contents
Problem
What is the hundreds digit of
Solution 1
Since we know that
To compute this, we use a clever application of the binomial theorem.
In all of the other terms, the power of is greater than
and so is equivalent to
modulo
which means we can ignore it. We have:
Therefore, the hundreds digit is
Side note: By Euler's Totient Theorem, for any
relatively prime with 1000, so
and
. We can then proceed using the clever application of the Binomial Theorem, or we can just proceed with solution 6 from here.
Solution 2
We need to compute By the Chinese Remainder Theorem, it suffices to compute
and
In modulo we have
by Euler's Theorem, and also
so we have
In modulo we have
by Euler's Theorem, and also
Therefore, we have
After finding the solution we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is
Solution 3
Notice that the hundreds digit of won't be affected by
. Essentially we could solve the problem by finding the hundreds digit of
. Powers of
are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of
can be found by reading rows of the triangle and adding extra numbers up. [add source] For example, the sixth row of the triangle is
and
. Adding all numbers from right to left, we get
, which is also
. In other words, each number is
steps from the right side of the row. The hundreds digit is
. We can do the same for
, but we only need to find the
digits from the right. Observing, every
number from the right is
. So to find the third number from the right on the row of
,
or
, or
. The last digit is five, but we must remember to add the number on the right of it, which, by observing other rows is obviously
. We must carry the
in
's tens digit to the
in
's unit digit to get
. The one at the very end of the row doesn't affect anything, so we can leave it alone.
-jackshi2006
Solution 4 (naive solution)
Since we are only looking at the last 3 digits and has the same last 3 digits as
, we can find
instead.
After this, we can repeatedly multiply the last 3 digits by 11 and take the last 3 digits of that product. We discover that
's last 2 digits are -11, the same as
.
From this information, we can figure out and
end in 611. Adding various multiples of 50 to the exponent gives us the fact that
's last digits are 611. We get
.
-ThisUsernameIsTaken
Solution 5 (modular bash)
We know that the hundreds digit of is just the hundreds digit of
. If we actually take a look at the powers of
, we notice a pattern:
Notice how the units digit is always one, the tens digit is always the previous tens digit plus the ones digit (or one) and the hundreds digit is always the previous hundreds digit plus the previous tens digit. Knowing this, we confidently repeat this pattern without actually multiply the previous term by
out (if you generally multiply a few numbers by
, you can see why this pattern holds).
Since
ends in _
, the pattern "cycles" every
times.
remainder
, meaning that
, giving us a hundreds digit of
.
Solution 6 (Super Easy) (Wrong)
We know that the hundreds digit of is just the hundreds digit of
(mod 1000). If we actually take a look at the powers of 11, we notice a pattern:
We notice that the hundreds digit just increases by 1 starting from 1(the hundreds digit of ). All we have to do now is add all the numbers from 1 to 11 (since the hundreds digit increases by 1 with every power of 11 that comes after) to get the hundreds digit of
.
. Thus the answer is
.
-TheOfficalPro
Ilovebender I believe this solution is wrong, as I can see that the hundreds digit doesn't increase by 1, it goes from 0 to 1 to 3 to 6. A more accurate solution would be the hundreds digit are triangular numbers.
-Extremelysupercooldude: In your solution, you find the hundreds digit of , but the problem asks for the hundreds digit of
, but you never cite way the two hundreds digits are the same.
-Wesserwes7254: I think this is a more revised version of this solution:
We know that the hundreds digit of is just the hundreds digit of
. If we actually take a look at the powers of
, we notice a pattern:
Notice how the units digit is always one, the tens digit is always the previous tens digit plus the ones digit (or one) and the hundreds digit is always the previous hundreds digit plus the previous tens digit. Knowing this, we confidently repeat this pattern without actually multiply the previous term by out (if you generally multiply a few numbers by
, you can see why this pattern holds).
Since ends in _
, the pattern "cycles" every
times.
remainder
, meaning that
, giving us a hundreds digit of
.
-Solution 1's sidenote can start this solution and make it correct.
Video Solution by TheBeautyofMath
https://youtu.be/7rosJnqLhs8?t=323
~IceMatrix
Video Solution by OmegaLearn
https://youtu.be/-H4n-QplQew?t=1423
~ pi_is_3.14
See Also
2011 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.