Difference between revisions of "2011 AMC 10B Problems/Problem 23"
Jackshi2006 (talk | contribs) (→Solution 2) |
Jackshi2006 (talk | contribs) (→Solution 3) |
||
Line 33: | Line 33: | ||
== Solution 3 == | == Solution 3 == | ||
− | Notice that the hundreds digit of 2011^2011 won't be affected by 2000. Essentially we could solve the problem by finding the hundreds digit of 11^2011. | + | Notice that the hundreds digit of 2011^2011 won't be affected by 2000. Essentially we could solve the problem by finding the hundreds digit of 11^2011. Powers of 11 are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of 11 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 1, 5, 10, 10, 5, and 1. Adding all numbers from right to left, we get 161051, which is also 11^5. In other words, each number is 10^n steps from the right side of the row. The hundreds digit is 0. We can do the same for 11^2011, but we only need to find the 3 digits from the right. Observing, every 3 number from the right is 1 + 2 + 3... + n. So to find the third number from the right on the row of 11^2011, f(11^n) = 1 + 2 + 3... + (n-1), or (2010 * 2011)/2, or 2021055. 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 2011. We must carry the 1 in 2011's tens digit to the 5 in 2021055's unit digit to get 6. The one at the very end of the row doesn't affect anything, so we can leave it alone. |
==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}} |
Revision as of 17:21, 28 November 2019
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
Sidenote: By Euler's Totient Theorem, for any , 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 2011^2011 won't be affected by 2000. Essentially we could solve the problem by finding the hundreds digit of 11^2011. Powers of 11 are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of 11 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 1, 5, 10, 10, 5, and 1. Adding all numbers from right to left, we get 161051, which is also 11^5. In other words, each number is 10^n steps from the right side of the row. The hundreds digit is 0. We can do the same for 11^2011, but we only need to find the 3 digits from the right. Observing, every 3 number from the right is 1 + 2 + 3... + n. So to find the third number from the right on the row of 11^2011, f(11^n) = 1 + 2 + 3... + (n-1), or (2010 * 2011)/2, or 2021055. 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 2011. We must carry the 1 in 2011's tens digit to the 5 in 2021055's unit digit to get 6. The one at the very end of the row doesn't affect anything, so we can leave it alone.
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.