2020 AMC 10B Problems/Problem 19
In a certain card game, a player is dealt a hand of cards from a deck of distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as . What is the digit ?
We're looking for the amount of ways we can get cards from a deck of , which is represented by .
We need to get rid of the multiples of , which will subsequently get rid of the multiples of (if we didn't, the zeroes would mess with the equation since you can't divide by 0)
, , leaves us with 17.
Converting these into, we have
Since this number is divisible by but not , the last digits must be divisible by but the last digits cannot be divisible by . This narrows the options down to and .
Also, the number cannot be divisible by . Adding up the digits, we get . If , then the expression equals , a multiple of . This would mean that the entire number would be divisible by , which is not what we want. Therefore, the only option is -PCChess
It is not hard to check that divides the number, As , using we have . Thus , implying so the answer is .
Solution 5 (Very Factor Bashy CRT)
We note that: Let . This will help us find the last two digits modulo and modulo . It is obvious that . Also (although this not so obvious), Therefore, . Thus , implying that .
As in Solution 2, we see that
which contains no factors of Therefore, the sum of the digits must not be a multiple of This sum is
It follows that cannot be a multiple of ruling out choices and Therefore, our possibilities are and Now, notice that is divisible by Therefore, we can plug each possible value of into and test for divisibility by Conveniently, we see that the first value, works. Thus, the answer is (To make our argument more rigorous, we can also test divisibility by for and to show that these values do not work.)
The total number of ways to choose from is
Using divisibility rules, we have that A is not a multiple of . Then, divide this equation by 10. This implies that the new number is divisible by but not . This means that is either or . However, is a multiple of , meaning has to be
Solution 8 (Very time consuming)
As stated in previous solutions, the number of ways to choose from is
Canceling out common factors , you get this -
When you multiply the remaining numbers, you get the product as . From this product, we can then determine that is equal to
~Puck_0 (Minor LaTeX)
Video Solution 1
Video Solution 2
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