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 .
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