2014 AMC 10B Problems/Problem 17
What is the greatest power of that is a factor of ?
We begin by factoring the out. This leaves us with .
We factor the difference of squares, leaving us with . We note that all even powers of more than two end in .... Also, all odd powers of five more than end in .... Thus, would end in ... and thus would contribute one power of two to the answer, but not more.
We can continue to factor as a difference of cubes, leaving us with times an odd number (Notice that the other number is . The powers of end in , so the two powers of will end with . Adding will make it end in . Thus, this is an odd number). ends in ..., contributing two powers of two to the final result.
Or we can see that ends in , and is divisible by only. Still that's powers of .
Adding these extra powers of two to the original factored out, we obtain the final answer of .
First, we can write the expression in a more primitive form which will allow us to start factoring. Now, we can factor out . This leaves us with . Call this number . Thus, our final answer will be , where is the largest power of that divides . Now we can consider , since by the answer choices.
Note that The powers of cycle in with a period of . Thus, This means that is divisible by but not , so and our answer is .
Convert . We can factor out to get that . Using the adjusted Lifting The Exponent lemma ( for all even and odd ), we get that the answer is
Factor out to get . Since , but , has 3 factors of 2. Hence is the largest power of two which divides the given number
Like Solution 1, factor out to get . Using engineer's induction, we observe that for any positive integer (where is an odd positive integer), it appears that the least even numbers directly above and below in value must contain a maximum multiple of and a maximum multiple of . Hence, the answer is which is .
For all integers where where n is an odd integer, must end in Thus, we find that and respectively end in and
We know that this number takes the form where is an integer that ends in . Because is a multiple of times an even number while is , we find that must be where is an odd number
We know that this number ends. Because it is more than the number , which is a multiple of , we find which is an even number that is not divisible by . Thus, it must have a maximum of multiple of .
This means that for any number being in the form where is an odd integer, must have a maximum of factors of while must have a maximum of factor of .
Using difference of squares, we get . Factoring a out, we get , and grouping like terms give .
Then, you would go ahead and innocently choose , right? No! Note that , where is any odd integer greater than or equal to , it always ends in . So, ends in and ends in , so they add up to an extra three 's. Therefore, the answer is actually .
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