Solved Stingers
Fresh puzzles: Stingers
Math Stinger #161
We have three cardboard isosceles right triangles of unequal sizes. No markings of
any kind are allowed. You may overlap the triangles. Is it possible to locate the
midpoint of the hypotenuse of the smallest triangle? How about the midpoint of the
non-hypotenuse side of the smallest triangle? Justify your answers.
Solved by Ian Salomone-Lent, Caitlyn Garrett, Elijah Remington and Andrew Daugherty.
Solution: Use the largest isosceles right triangle to form a perpendicular bisector at the right angle of the smallest triangle.
To locate the mid-point of the non=hypotenuse side of the smallest triangle, just use the middle-size triangle to form a perpendicular bisector at the mid-point of the smallest triangle created above.
Math Stinger #160
A fair coin is tossed eleven times. Find the probability that two heads do not appear
in succession.
Solved by Andrew Daugherty and Ian Salomone-Lent.
Solution: It is related to the Fibonacci numbers! Let Fn be the number of ways to arrange heads or tails in n flips with no two heads in succession.
Then, F1 = 2 for we can have H or T.
F2 = 3 for we can have HT or TH or TT.
F3 = 5 for we can have HTT or HTH or THT or TTH or TTT.
With n coins, the first one can be a head H. But then the second one must be a T. Then the remaining n-2 coins has Fn-2 total different possibilities with no two heads in succession.
With n coins, the first one can be a tail T. Then the remaining n-1 coins has Fn-1 total different possibilities with no two heads in succession.
Therefore, Fn = Fn-1 + Fn-2.
Recursively, one can compute F4 = 3+5 = 8, F5 = 5+8 = 13, F6 = 8+13 = 21, F7 = 13+21 = 34, F8 = 21+34 = 55, F9 = 34+55 = 89, F10 = 55+89 = 144, F11 = 89+144 = 233.
So, the probability is 233 / 211 = 233 / 2048.
Math Stinger #159
Find all positive integers which are one more than the sum of the squares of their
base ten digits. For example, 35 = 1 + 32 + 52.
Solved by Ian Salomone-Lent, Caitlyn Garrett, Andrew Daugherty and William Little.
Solution: Say the number N = a1a2...an has n > 3 digits. Then a1 2 + a22 + ... + an2 + 1 is at most 81n+1 while the number N itself is at least 10n. One can prove by induction that 10n > 81n + 1 for n > 3.
So, N can have 1 or 2 or 3 digits.
N cannot have 1 digit as a1 is clearly not the same as a12+1.
If N has 3 digits, then a12 + a22 + a32 + 1 is at most 81*3+1 = 244. So, the leading digit a1 = 1 or 2 only. But then a12 + a22 + a32 + 1 is at most 4 + 81 + 81 + 1 = 166. So, the leading digit a1 = 1. So N has the form 1ab and we want 1 + a2 + b2 + 1 = 100 + 10a + b or a2 - 10a + b2 - b = 98 which is impossible as a2-10a = a(a - 10) is negative and the left hand side is at most b2 - b = b(b-1) < 10*9 = 90.
So, N has 2 digits. Say N = ab and we want 10a + b = a2 + b2 + 1 or a(10 - a) = b2 - b + 1. Notice that b and b2 are either both odd or both even. So, b2 - b + 1 is an odd number. Hence, a must be odd.
One can go through 1(10 - 1) = 9, 3(10 - 3) = 21, 5(10 - 5) = 25, 7(10 - 7) = 21, 9(10 - 9) = 9. One can check that it is impossible for perfect squares 9 or 25 = b2 - b + 1. So, we must have a = 3 or a = 7.
Notice that 21 = 52 - 5 + 1. Therefore, N = 35 or 75.
Math Stinger #158
Is it possible to place four black unit circles and three white unit circles on the
plane so that the white circles cannot move to other positions while staying on the
plane, if the black circles are fixed in their position? Justify your answer.
Solved by Noah Cruce, Elijah Remington, Andrew Daugherty, Ian Salmone-Lent.
Solution: Yes, there are many possible solutions. One possibility is to place the three white circles next to each other to form an equilateral triangle. Then, one can place three black circles just touching them and form an equilateral triangle. In order for the three white circles to move, it must be a rotation. Then, one simply place the fourth black circle next to the equilateral triangle of white circles to prevent it from rotating.
Math Stinger #157
A strike force is to be selected from a row of eleven agents. It is known that three
of them have special abilities. It is not known who they are, except that they are
evenly spaced in the row. Find, with justification, the minimum number of agents we
need to pick so that at least one agent with special abilities is included.
Solved by Ian Salomone-Lent.
Solution: The answer is 5. Suppose the eleven agents are in a row numbered 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. The 3 special-ability agents are evenly spaced means their numbers form an arithmetic progression.
Then, we pick the agents numbered: 2, 5, 6, 7, 10. Any 3-term arithmetic progression (3 special agents) centered at these position is okay as we already picked the middle one.
For an arithmetic progression centered at 3, it is either 2, 3, 4 or 1, 3, 5. Each one of them has one of the 5 numbers we picked.
For an arithmetic progression centered at 4, it is either 3, 4, 5 or 2, 4, 6 or 1, 4, 7. Again, each one of them has one of the 5 numbers we picked.
By symmetry, arithmetic progressions centered at 8 or 9 are also okay. Hence, 5 agents suffice.
It remains to prove that picking 4 or less numbers is not enough.
First, notice that agent number 6 must be one of the 4 numbers. For otherwise, we have the 5 disjoint arithmetic progressions (except for the middle number) for the special agents: 1,6,11; 2,6,10; 3,6,9; 4,6,8; 5,6,7 which would require picking 5 other numbers to cover these possibilities.
Since 6 is one of the picked number, we can have at most 1 number from 1 to 5 or from 7 to 11.
We cannot pick no number in either range for otherwise we would miss some arithmetic progression.
By symmetry, without loss of generality, say x is the only number picked from 1 to 5.
However, x cannot be 1 or 5 for othewise the arithmetic progression 2,3,4 is not covered.
Similarly, x cannot be 2 or 4 for otherwise the 1,3,5 is not covered. Hence, x = 3.
Then we still need to cover the arithmetic progressions 1,4,7; 2,5,8; 1,5,9 which requires 3 extra numbers. But we only have at most 4 - 1 - 1 = 2 numbers left which is impossible. Hence, 4 or less agents is not enough.
Math Stinger #156
Your calculator is not working properly - it cannot perform multiplication. But it
can add and subtract any two numbers. It can also compute x-1 and x2 of any number x. Can you nevertheless use this defective calculator to multiply two
numbers?
Solved by Ian Salomone-Lent.
Solution: Given two numbers a and b.
Notice that 2ab = (a+b)2 - a2 - b2 which can be done by addition, subtraction, and squaring operations.
How can one get ab then? Observe that (ab)-1 = (2ab)-1 + (2ab)-1. So, we can (ab)-1 by using the reciprocal operation to the quantity 2ab we just found and adding to itself.
Finally, to get ab, one simply takes the reciprocal of (ab)-1.
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.
