2011 AMC 10B Problems/Problem 23
What is the hundreds digit of
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.
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
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.
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 are looking for Rewrite as and we are looking for the of this, which is Ignore the digits past the hundreds place and bash it out. We do this by substituting each result of the previous power of 2 and multiplying in . Next, we use the expression and substitute the values for the different powers in and simplify to as we go (by ignoring the thousands digits and beyond during multiplication) to get , and the hundreds digit is .
Solution 6 (Super Easy)
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 .
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 .
Video Solution by TheBeautyofMath
Video Solution by OmegaLearn
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