2020 AMC 10A Problems/Problem 22
Contents
Problem
For how many positive integers isnot divisible by ? (Recall that is the greatest integer less than or equal to .)
Solution 1
Clearly, fails. Except for the special case of , equals either or . If it equals , this implies that , so their sum is clearly a multiple of , so this will always fail. If it equals , the sum of the three floor terms is , so it is never a multiple of . Thus, we are looking for all such that This implies that either or
Let's analyze the first equation of these two. This equation is equivalent to the statement that there is a positive integer such that Analogously, the second equation implies that So our only that satisfy this condition are that divide or . Using the method to find the number of divisors of a number, we see that has divisors and has divisors. Their only common factor is , so there are positive integers that divide either or . Since the integer is a special case and does not count, we must subtract this from our , so our final answer is
*While this observation may seem strange, it is actually "trivial by intuition" to go straight from to . In fact, "trivial by intuition" is basically a good summary of the solution to this entire problem.
~ihatemath123
Solution 2
Counting down from , , ... we notice that is only not divisible by when n is a divisor of only or ...). Notice how the factors of , and , do not work.
The prime factorization of is , so factors in total.
The prime factorization of is , so factors in total.
However, obviously does not work, so we have to subtract ( is counted twice) from the total. = .
~BLOATED_BAGEL
~Typo fixed by evanhliu2009
Solution 3
First, we notice the following lemma:
: For , if ; and if
: Let , with . If , then . Hence , , and
If , then . Hence , , and
From the lemma and the given equation, we have four possible cases:
Note that cases (2) and (3) are the cases in which the term, is not divisible by . So we only need to count the number of 's for which cases (2) and (3) stand.
Case (2): By the lemma, we have and Hence can be any factor of except for . Since there are possible values of for this case.
Case (3): By the lemma, we have and Hence can be any factor of except for . Since there are possible values of for this case.
So in total, we have total of possible 's.
~mathboywannabe
Solution 4 (Casework)
Note that is a multiple of if lies between two consecutive multiples of .
Let's assume that the above expression does indeed lie betweent two consecutive multiples of . This implies that does not divide either or , meaning that when divided by , none of the quotients are whole. In turn, this also means that they will all have the same whole number part.
If and were to have a different whole number part, then would have to lie within and . If this is confusing, think of why and have the same whole number part ( in this case). Here, . What happens to the whole number part when ?
Since and have identical whole number parts, their floors are identical, since the floor of a number is equal to its whole number part, discarding its fractional component.
Let be the number we get when we floor or if the three of those numbers lie between two consecutive multiples of . Adding them up, we get (due to their floors being the same), which is a mulutiple of . So no matter what, we cannot have and lie between two consecutive multiples of .
What does this mean? It means that there must be some multiple of within this expression (with some restrictions, as we'll see), in order to prevent the violation of the main restriction. We can proceed with casework now.
Let's assume that the multiple of is located at . Luckily, the only prime factors of are and .
We can observe that doesn't work, since and will both round down to when divided by it. However, does work, since it divides both and . The floor of is , , meaning that when we evaluate the given expression for , we will get which isn't a multiple of .
This case works because we have a multiple of at the end of the expression, as well as the beginning, so the whole number parts of and when divided by are not all the same, due to dividing two of these numbers.
The restriction of dividing more than one number within the expression is only valid when we're testing the first number in the given expression (otherwise, the other floors wouldn't round down to the same value as the first).
Case is complete, and we've found that only works.
Now, let's assume that the multiple of is located at . In this case, if divides , it doesn't divide (since two multiples of a number greater than are never consecutive), nor does it divide , for the same reason.
The prime factorization of is , and thus has divisors.
When testing a few values of initially, we observed that causes the expression to be divisible by . Subtracting , we see that there are values of that work for this case.
Finally, let's assume that the multiple of is located at
Our goal is to have the other multiple of below whatever is, so we won't have identical floors throughout the expression.
Once again, any factor of , except for is relatively prime to both and , so when we floor those two numbers, we get an integer that isn't .
factors as , meaning that it has divisors. and either don't work or have already been counted, so there are valid values of for this case.
Adding these three cases, we get values of .
-Benedict T (countmath1)
Video Solutions
Video Solution 1 (Simple)
Education, The Study of Everything
Video Solution 2
https://www.youtube.com/watch?v=_Ej9nnHS07s
~Snore
Video Solution 3
https://www.youtube.com/watch?v=G5UVS5aM-CY&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=4 ~ MathEx
Video Solution 4 (Richard Rusczyk)
https://artofproblemsolving.com/videos/amc/2020amc10a/517
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 |
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