Difference between revisions of "2011 AMC 10B Problems/Problem 23"
Williamgolly (talk | contribs) |
m (→Solution 4 (naive solution, EXTREMELY bashy)) |
||
(13 intermediate revisions by 7 users not shown) | |||
Line 19: | Line 19: | ||
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. | |
== Solution 2 == | == Solution 2 == | ||
− | We need to compute <math>2011^{2011} \pmod{1000}.</math> By the Chinese Remainder Theorem, it suffices to compute <math>2011^{2011} \pmod{8}</math> and <math>2011^{2011} \pmod{125}.</math> | + | We need to compute <math>2011^{2011} \pmod{1000}.</math> By the [[Chinese Remainder Theorem]], it suffices to compute <math>2011^{2011} \pmod{8}</math> and <math>2011^{2011} \pmod{125}.</math> |
In modulo <math>8,</math> we have <math>2011^4 \equiv 1 \pmod{8}</math> by Euler's Theorem, and also <math>2011 \equiv 3 \pmod{8},</math> so we have <cmath>2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.</cmath> | In modulo <math>8,</math> we have <math>2011^4 \equiv 1 \pmod{8}</math> by Euler's Theorem, and also <math>2011 \equiv 3 \pmod{8},</math> so we have <cmath>2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.</cmath> | ||
Line 33: | Line 33: | ||
== Solution 3 == | == Solution 3 == | ||
− | Notice that the hundreds digit of <math>2011^{2011}</math> won't be affected by <math>2000</math>. Essentially we could solve the problem by finding the hundreds digit of <math>11^{2011}</math>. Powers of <math>11</math> are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of <math>11</math> 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 <math>1, 5, 10, 10, 5,</math> and <math>1</math>. Adding all numbers from right to left, we get <math>161051</math>, which is also <math>11^5</math>. In other words, each number is <math>10^n</math> steps from the right side of the row. The hundreds digit is <math>0</math>. We can do the same for <math>11^2011</math>, but we only need to find the <math>3</math> digits from the right. Observing, every <math>3</math> number from the right is <math>1 + 2 + 3... + n</math>. So to find the third number from the right on the row of <math>11^{2011}</math>, < | + | Notice that the hundreds digit of <math>2011^{2011}</math> won't be affected by <math>2000</math>. Essentially we could solve the problem by finding the hundreds digit of <math>11^{2011}</math>. Powers of <math>11</math> are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of <math>11</math> 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 <math>1, 5, 10, 10, 5,</math> and <math>1</math>. Adding all numbers from right to left, we get <math>161051</math>, which is also <math>11^5</math>. In other words, each number is <math>10^n</math> steps from the right side of the row. The hundreds digit is <math>0</math>. We can do the same for <math>11^{2011}</math>, but we only need to find the <math>3</math> digits from the right. Observing, every <math>3</math> number from the right is <math>1 + 2 + 3... + n</math>. So to find the third number from the right on the row of <math>11^{2011}</math>, <cmath>f(11^n) = 1 + 2 + 3... + (n-1),</cmath> or <math>\frac{(2010 \cdot 2011)}{2}</math>, or <math>2021055</math>. 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 <math>2011</math>. We must carry the <math>1</math> in <math>2011</math>'s tens digit to the <math>5</math> in <math>2021055</math>'s unit digit to get <math>\boxed{\textbf{(D) } 6}</math>. The one at the very end of the row doesn't affect anything, so we can leave it alone. |
-jackshi2006 | -jackshi2006 | ||
+ | ==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. | ||
+ | After this, we can repeatedly multiply the last 3 digits by 11 and take the last 3 digits of that product. We discover that <math>11^{51}</math>'s last 2 digits are -11, the same as <math>11^1</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 | ||
==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 19:22, 23 January 2021
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.
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
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.