Stingers

ALERT: THE MORE CORRECT ANSWERS YOU SUBMIT, THE MORE ENTRIES YOU HAVE IN EACH SEMESTER'S RAFFLE!

Math Stinger #150
Let us assume that a given pair of people either know each other or are strangers.  If six people enter a room, show that there must be either three people who know each other pairwise or three people who are pairwise strangers.
Please send solutions to shollid4 at kennesaw dot edu.

Math Stinger #148
For two given positive integers n and k, how many different sequences of positive integers a1 ≤ a2 ≤ a3 ≤ ... ≤ ak are there in which ak ≤ n?
Please send solutions to shollid4 at kennesaw dot edu.

Math Stinger #146
Two functions of x are differentiable and not identically equal to zero.  FInd an example of two such functions having the property that the derivative of their quotient is the quotient of their derivatives.
Please send solutions to shollid4 at kennesaw dot edu.

 

Spring 2022 solvers and number of Raffle entries:
JS .... 1
SC .... 1
JE .... 2
AM .... 1

Math Stinger #149
The factorial function is useful in many areas of mathematics. It is defined by: n! = n(n-1)(n-2)(n-3)...(3)(2)(1). So, for example, 6! = (6)(5)(4)(3)(2)(1) = 720  and 10! = (10)(9)(8)(7)(6)(5)(4)(3)(2)(1) = 3628800. 
Determine how many zeros occur at the end of 10000!
Please send solutions to shollid4 at kennesaw dot edu.

First solution: JE Spring 2022

So, I first wanted to figure out a way to evaluate the amount of trailing 0s in 1000! by not calculating it, ideally by just evaluating 1000. 2*5 is 10, so maybe find the amount of numbers between 1 and 1000 are divisible by 5, as well as the ones divisible by 2. Since more numbers between 1-1000 are divisible by 2 than 5, I can just evaluate it by dividing by 5. 


However, I realized that 5 wouldn't cover all trailing zeroes due to the presence of powers of 5. However, I realized that the power of 5 correlates with the maximum amount of trailing zeroes it will contribute. That makes sense - 10 raised to the nth power equal to (2*5) raised to the same power. So, dividing 1000 by increasing powers of 5, then adding all these quotients, will cover all additions of trailing 0s to 1000!


1000/5 = 200
1000/25 = 40
1000/125 = 8
1000/625 = 1.6 (will be counted as 1 due to the nature of the problem)


200+40+8+1 = 249


Therefore, there are 249 trailing zeroes in 1000!

Older solutions published to: Solved Stingers

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