Solved Stingers
Fresh puzzles: Stingers
Math Stinger #155
In a math class, 21% of the students get an A and 81% of those A students are part-time students. Among part-time students, 31% of them do not get an A. What percentage of the whole class are part-time students?
Solved by Faye Le, Nha Phan, and Andrew Daugherty.
Solution: Let x be the proportion of part-time students in the whole class.
Then, x - 0.31x = 0.69x gives the proportion of part-time students getting an A.
Hence, 0.69x = 0.21 * 0.81 which gives x = (0.21 * 0.81) / (0.69) = 1701/6900 = 24.65217... %.
Note: Faye Le gave a nice solution using conditional probability.
Math Stinger #152
Without using any calculator or computer, determine (with justification) which of the following two quantities 1.0000000001 + 1 / 1.0000000001 or 0.9999999999 + 1 / 0.9999999999 is bigger?
Solved by Andrew Daugherty, Ian Salomone-Lent, Tsui Yi Cheung, and William Little.
Solution: Let x = 0.0000000001. Then, we want to compare (1+x) + 1/(1+x) and (1-x) + 1/(1-x).
Subtract the two quantities and use common denominator,
(1+x) + 1/(1+x) - (1-x) - 1/(1-x) = 2x + (1-x)/[(1+x)(1-x)] - (1+x)/[(1-x)(1+x)] = 2x - 2x/(1 - x2) = 2x[1 - 1/(1 - x2)].
Note that 0 < x2 < 1. We also have 0 < 1 - x2 < 1.
Hence, 1/(1 - x2) > 1/1 = 1 and the difference 2x[1 - 1/(1 - x2)] is negative.
Therefore, the second quantity 0.9999999999 + 1 / 0.9999999999 is bigger
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.
First solution: IS Fall 2022
Solution: Set f(x) = 1/x + 1 and g(x) = -1/x. Then (f/g)' = (-1 - x)' = -1 and f'/g' = (-1/x2)/(1/x2) = -1.
Method: The initial plan was to see if there was an easy looking choice for one
function, to see if that would lead to the second. Rewriting the question via quotient
rule led to the equation:
f'/g' = (g*f' - g'*f)/g2
This equation looked pretty solvable as long as g was such that its derivative equaled
its square. I came up with g = -1/x by brute force, but in retrospect you could solve
the ODE g' = g2 if you'd like. This led to the equation:
f' = -f'/x - f/x2 => f'(1 + 1/x) = -f/x2 => f' = -f/(x2
+ x)
From here, we only need to solve a separable ODE, whose solution is f = 1/x +
1, assuming C = 0.
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!
Math Stinger #145
What is the number of k-tuples chosen from 1,..., n containing no two consecutive
integers?
Please send solutions to shollid4 at kennesaw dot edu.
First correct solution: AM Spring 2022
(n-k+1) ways
k
Math Stinger #147
Often a problem can appear difficult until it is looked at in just the right way.
This is just such a problem. The answer is "obvious" after you find it.
Find two numbers containing no zeros that have a product of 1,000,000,000.
After solving this problem you may wish to tackle the following variation:
Explain why it is not possible to find two numbers containing no zeros that have a
product of 1,000,000,000,000.
The above two problems indicate that powers of ten can be classified according to whether or not they can be factored into the product of two numbers that contain no zeros. As a final puzzle, find as many powers of ten as you can that can be factored into two numbers that contain no zeros.
First correct solution to JE Spring 2022:
1,000,000,000 is the product of 512 and 1953125.
Since powers of 10 only have the prime factors of 2 and 5, the only two numbers that could be multiplied to get that number had to be 5 raised to the same power and 2 raised to the same power. This is the only way to not have one of the values have a factor of 10, which would result in having a 0 as the last digit. So, for 10 raised to the nth power, it’s the product of 5 to the nth power and 2 to the nth power. However, if either 2 to the nth power or 5 to the nth power has a 0 in it, that power of 10 cannot be expressed as the product of two numbers with no zeroes in them.
Both 5 to the 12th power and 2 to the 12th power have 0’s in them, either can be used to show that 1,000,000,000,000 cannot be expressed as the product of two values that contain no 0’s.
As for the powers of 10 that can be expressed, I’ve found 0-7, 9, and 18.
Math Stinger #144
What is the distance between the two poles?
Note: The above picture is certainly not to scale.
SOLVED! First solver: JS, second solver SC. Spring 2022
Solution: the poles must be 0m apart. This "problem" is often attributed to a large corporation's interview process.
