Clever High School Math Problems

[Repeekthgil]

I teach primarily high school freshman Algebra I and Geometry, maybe Calc next year (fingers crossed)

Do any of you remember a favorite problem posed in high school? The sort of problem that stopped you and made you think.

My favorite is: A mother and child weigh x, the father and child weigh y, and the mother and father weigh z. Find each weight (assume they are the same m/f/c.)

Also, for geometry, any problem with a bunch of lines with angles and shapes formed that you can reason through with the transversal rules, int. angle sum, etc...

How about you?

[dudiobugtron]

I quite like "A mother's age is ____ times her daughter's age. The sum of their ages is ______".
Easy if you have a good understanding of algebra, but in practice a lot of kids get really tripped up when trying to write an equation for it. I don't know what level 'Freshman high school' corresponds to, though, so it might be too easy for them!

I also really like this one (possibly not appropriate for your students though):

A mother is 21 years older than her child. 6 years from now, the mother will be 5 times as old as the child. Where is the father?

[Plasma_Wolf]

I've started my master course of mathematics this year (algebra), but I'm interested in most branches of math (especially in the high school teaching part).

One of the final exam questions of this year at the highest Dutch high school level was this:

Consider the following function: [imath]f_p(x)=2x^2-px^4[/imath] (the graph is symmetric in the y-axis). This function has three vertices, the points O=(0,0), A and B, where the coordinates of the latter two are dependent on the value of p. there is one value of p for which the length of OA equals the length of AB.

The exercise: determine this value of p.

[gmalivuk]

I also really like this one (possibly not appropriate for your students though):

A mother is 21 years older than her child. 6 years from now, the mother will be 5 times as old as the child. Where is the father?

That only works if you say it's a child from the moment of conception.

Better to slide the first part, since it's not like it changes: In ten years, a mother will be 21 years older than her child. In six years, the mother will be 5 times as old as the child. Where is the child's father now? (or "What is the mother doing now?")

[Repeekthgil]

Having sex is the answer, right? It might be a little inappropriate for the US market...

That first type, the sum of 2 numbers (ages) is this, their product that (though we'd been doing differences, b/c it makes for easier systems) and it still gave some of them troubles. Real world applications, though perhaps trivial, but the best questions often are...

[Farabor]

My favorite problem was posed to me after 5th grade as part of an entrance exam to an advanced math program

Find a 10 digit number such that the first digit tells how many 0s there are in the number, the second digit tells how many 1s there are in the number, the third how many 2's, all the way up to the 10th tell how many 9s.
There's only one such number, and simple algebra or a bit of trial and error will find it, but you have to conceptualize the problem right first.

[Repeekthgil]

That's a great one. I'm going to file that away in pack problems.

[Derek]

My favorite problem was posed to me after 5th grade as part of an entrance exam to an advanced math program

Find a 10 digit number such that the first digit tells how many 0s there are in the number, the second digit tells how many 1s there are in the number, the third how many 2's, all the way up to the 10th tell how many 9s.
There's only one such number, and simple algebra or a bit of trial and error will find it, but you have to conceptualize the problem right first.

6210001000

I just got this by starting with 9000000000 and correcting digits until the result converged.

[doogly]

That is rather neat. Is there a way to prove uniqueness?

[sparkyb]

That is rather neat. Is there a way to prove uniqueness?

I was able to solve it, but then I also tried to prove. I gave up and tried Googling, and found this solution which I think is a valid proof of the answer's uniqueness. However it didn't satisfy me. The problem statement said it could be solved with simple algebra if you express the problem correctly. That solution is a proof, but not really an algebraic one, so I wonder if one really does exist. I got at far as recognizing that the sum of the digits (call them A through J) A+B+...+J = 10, but also, since each is a count of digits with that value somewhere else in the number B+2C+3D+...+9J = 10. I got stuck there, though, because A=2,B=6,C=2 is another valid solution to just those 2 equations, but I couldn't think of how to algebraicly express the missing constraints.

While Googling for the answer to that one, I found a similar and interesting problem:

Find a 10 digit number that uses all the digits 0-9 and:
the first digit is divisible by 1,
the first 2 digits (taken as a 2-digit number) are divisible by two,
the first 3 digits (taken as a 3-digit number) are divisible by three, etc.
(there is only one such number)

[jestingrabbit]

Subtracting your two equations and rearranging, you end up with A= C + 2D + 3E +...+ 8J. So, you need to balance this, and its a very strong statement.

Spoiler:

Let the kth digit be the last positive digit, and z be the number of zeroes. I claim

k>= z >= (k-1)*(the number of k's) >= k-1 and z >= (9-k)

the first inequality is from the fact that the zth digit is positive, the second and third are from the new equation, and the fourth is from the fact that the last 9-k digits must be 0.

From this, we have that z >= 4, which is the bound we get when k = 5 (any other k gives us a necessarily larger z). Using the first inequality, we know k is at least 4 too, and that the kth digit is 1, so B>0.

If z = k-1, then the only way to balance the new equation is to have B = 9-z. Only 3 digits are nonzero, so z = 7, k = 8, and we have a contradiction (digit sum too high).

Now, if z = k, using the equation we introduced at the start, we must have that C=1, B non zero, and the zth and 0th digit is the only other non zero digit. So, z = 6, B=2, C=1 and the 6th digit is a 1.

[Derek]

While Googling for the answer to that one, I found a similar and interesting problem:

Find a 10 digit number that uses all the digits 0-9 and:
the first digit is divisible by 1,
the first 2 digits (taken as a 2-digit number) are divisible by two,
the first 3 digits (taken as a 3-digit number) are divisible by three, etc.
(there is only one such number)

The first digit is our choice.
The second digit is one of {0, 2, 4, 6, 8}.
The third digit depends on the first two (the first three must sum to a multiple of 3), but in particular we can choose the first digit so that any third digit is possible.
The fourth digit is one of {0, 4, 8} if the third digit is {0, 2, 4, 6, 8}, or the fourth digit is {2, 6} if the third digit is {1, 3, 5, 7, 9}.
The fifth digit is {0, 5}.
The sixth digit is {0, 2, 4, 6, 8}, and the fourth, fifth, and sixth digits must sum to a multiple of 3.
The seventh digit is fuck seven, but the previous numbers will restrict you to one or two choices.
The eight digit is like the fourth but the pattern is more complicated. It is dependent on the sixth and seventh, and you'll have one or two choices.
The ninth digit depends on the previous, they must sum to a multiple of 9.
The tenth digit is 0.

Now add to the consideration that we must use each digit exactly once.

Therefore the fifth digit is 5.
The ninth digit will be completely determined by the first 8 (because the digits 1 to 9 add to 9).
There are five even digits, and the even positions must be even, so the odd positions must be odd.
Since the third digit is {1, 3, 7, 9}, the fourth digit is {2, 6}
The fourth + fifth digit is {25, 65}, so the sixth digit is {8, 4}, respectively, so the fourth digit determines the sixth.
Since the sixth digit is always even, the eighth digit depends only on the seventh digit. For {1, 9} we have {6}, for {3, 7} we have {2}. (This is because 8 divides 200)
Thus the seventh digit determines the eighth digit, which determines the fourth digit.
And the second is {8, 4}, so the second and sixth digits determine each other. So Seventh -> eighth -> fourth -> sixth -> second.
This leaves the first, third, and seventh digits to be determined. All have possibilities {1, 3, 7, 9}, so we have 4*3*2 = 24 possibilities. Actually the second digit will put some restrictions on the first and third, so it's even better. We can work this by hand.

Possible seventh digits, and consequential other digits:
{_4_25816_0, _8_65432_0, _8_65472_0, _4_25896_0}
{1896543270, 9816543270, 1836547290, 1896547230, 3816547290, 9816547230, 1472589630, 7412589630}
Checking for divisibility by 7 gives us 3816547290.

[Copper Bezel]

The fourth + fifth digit is {25, 65}, so the sixth digit is {8, 4}, respectively, so the fourth digit determines the sixth.
Since the sixth digit is always even, the eighth digit depends only on the seventh digit. For {1, 9} we have {6}, for {3, 7} we have {2}. (This is because 8 divides 200)

Argh. That should have been obvious, that having two digits where the truncated number is divisible by 4, each following an odd digit, means that those digits have to be 2 and 6, so digits 2 and 6 must then be 4 and 8, for all of four possible ways to arrange the even digits. I missed that, and kinda made a tree starting with the patterns that worked for the first four (and consequently the first five) digits ... It took me a good bit longer to get to 3816547290.

[Derek]

The fourth + fifth digit is {25, 65}, so the sixth digit is {8, 4}, respectively, so the fourth digit determines the sixth.
Since the sixth digit is always even, the eighth digit depends only on the seventh digit. For {1, 9} we have {6}, for {3, 7} we have {2}. (This is because 8 divides 200)

Argh. That should have been obvious, that having two digits where the truncated number is divisible by 4, each following an odd digit, means that those digits have to be 2 and 6, so digits 2 and 6 must then be 4 and 8, for all of four possible ways to arrange the even digits. I missed that, and kinda made a tree starting with the patterns that worked for the first four (and consequently the first five) digits ... It took me a good bit longer to get to 3816547290.

So in fairness I first solved it with a similar approach to your. I started with all valid permutations of the first three digits and using the first few obvious deductions I built out the rest of the string from there, getting the correct answer. While doing this I saw a few more patterns, so then I went back and made this more mathematical solution.

[Copper Bezel]

Ah. I feel slightly less silly now. And yeah, it sounds like the same process. I had little sets of possible digits 1 and 3, eliminated based on possible 2s, then 4s, and so on. I ended up just happy that filtering them out based on digit 7 could come last. = )

I guess it's a lesson in looking to identify all of the constraints first. = )

[mfb]

That puzzle can be extended to other bases, with other nice solutions.

Spoiler:

The base has to be even, otherwise there is no solution

It would be interesting to see if there is a solution for a base larger than 14 (there is one with 14). I wrote a hack to search for solutions, but that is not quick enough to search beyond base 16.

[Lopsidation]

Nothing works for bases 16 through 36. Unfortunately, 38 looks like it would take way too long to run.

I doubt it's ever possible above 14, but I don't see any way to prove it either. The conditions are all too global.

[dudiobugtron]

Edit: Ooops, I was solving the wrong problem...

Old post here:

Spoiler:

Nothing works for bases 16 through 36. Unfortunately, 38 looks like it would take way too long to run.

I doubt it's ever possible above 14, but I don't see any way to prove it either. The conditions are all too global.

For some base b, we have digits d₀, d₁, ... , d_b-1.
Also, each d_n tells us how many of the d_i 's equal n.

There are two constraints.

Firstly, all of the digits must add up to b:
(1) d₀ + d₁ + ... + d_b-1 = b

Secondly, all of the digits multiplied by their position number must add up to b as well.

(2) d₀ x 0 + d₁ x 1 + ... + d_b-1 x (n-1) = b

This puts huge limitations on it. Basically, you can only put a number in a later column if you have enough zeros make up for the loss.

-----------

However, it also illustrates that the solution Derek posted has some pretty obvious extensions to higher bases - basically, d₀ will be b-4, d₁ will be 2, d₂ will be 1, and d_b-4 will be 1 as well. (As long as b-4 is greater than 2.)

This satisfies condition (1), as b-4 + 2 + 1 + 1 = b
and also condition (2), since (b-4) x 0 + 2 x 1 + 1 x 2 + 1 x (b-4) = b as well.

I can't see why this won't work for any base greater than 6. For example, for 16:

12, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0

Is a solution. Isn't it?

And in general:
b-4, 2, 1, 0, 0, ... , 0, 1, 0, 0, 0

[americablanco]

The exercise: determine this value of p.

Every (real) number?

Do you mean where |OA| = |OB| = |AB| ?

[LaserGuy]

A two-digit number has the property that the square of its tens digit plus ten times its units digit equals the square of its units digit plus ten times its tens digit. Determine all two-digit numbers which have this property, and are prime numbers.

Came off a math contest. Not too difficult, but requires a bit of thought to work it out.

[gmalivuk]

Do you mean where |OA| = |OB| = |AB| ?

Yes, since that's what is implied by

there is one value of p for which the length of OA equals the length of AB.

together with

(the graph is symmetric in the y-axis)

[brenok]

N = 10x + y
x^2 + 10y = y^2 + 10x
x^2 - y^2 = 10x - 10y

This is true when x=y.
When x!=y:

x^2 - y^2 = 10(x - y)
(x^2 - y^2) / (x - y) = 10
x + y = 10

Solutions: 11, 19, 22, 28, 33, 37, 44, 46, 55, 64, 65, 73, 77, 82, 88, 91, 99, primes underlined.

[skeptical scientist]

One of my favorites is the two-trains puzzle:

Two trains are on the same track a distance 100 km apart heading towards one another, each at a speed of 50 km/h. A fly starting out at the front of one train, flies towards the other at a speed of 75 km/h. Each time the fly reaches a train, the fly turns around and returns to the train it just left left. How many kilometers does the fly travel before getting squashed in the collision of the two trains?

[PM 2Ring]

One of my favorites is the two-trains puzzle:

Two trains are on the same track a distance 100 km apart heading towards one another, each at a speed of 50 km/h. A fly starting out at the front of one train, flies towards the other at a speed of 75 km/h. Each time the fly reaches a train, the fly turns around and returns to the train it just left left. How many kilometers does the fly travel before getting squashed in the collision of the two trains?

To clarify: the fly travels at 75 km/h relative to the ground.

I like this one because you can do it the long way, using infinite series, or you can notice that there's a simpler way. Allegedly, John von Neumann solved this problem in a few moments in his head, doing it the long way.

[dudiobugtron]

One of my favorites is the two-trains puzzle:

Two trains are on the same track a distance 100 km apart heading towards one another, each at a speed of 50 km/h. A fly starting out at the front of one train, flies towards the other at a speed of 75 km/h. Each time the fly reaches a train, the fly turns around and returns to the train it just left left. How many kilometers does the fly travel before getting squashed in the collision of the two trains?

To clarify: the fly travels at 75 km/h relative to the ground.

I like this one because you can do it the long way, using infinite series, or you can notice that there's a simpler way. Allegedly, John von Neumann solved this problem in a few moments in his head, doing it the long way.

I also really like this puzzle.

However, while it's not not *that* hard to get a good approximation in your head by summing the first few terms of the infinite series (especially if you spot the pattern), even working out the first term of the series is pretty difficult for high school students! So basically they'll come to the conclusion that they can't do it, unless they figure out the trick.

[doogly]

And it's nice to have a way to check if you have the next von Neumann in the room, so you can give them a cookie.

[DeGuerre]

One of my favourites: On Saturday, a climber starts at the bottom of a mountain at 9am, and reaches the summit at 5pm. After camping for the night, the climber starts the descent on Sunday at 9am, and reaches the bottom at 5pm. Make no assumptions about the speed of the ascent or descent; in fact, the climber may even have descended slightly during the ascent and vice versa.

Show that there must be some time of day where the climber was at the exact same height from the ground both on Saturday and on Sunday.

For people who know some maths, this is of course a simple consequence of the intermediate value theorem. Proving that rigorously is quite hard, but it should be possible for any high school student to understand why it should be true.

[donbright]

I found it interesting to try to trisect an angle using an edge and compass. In youth, I was so sure that if I just tried hard enough, I could figure it out! How can something so simple be impossible, and how did someone prove it was impossible?

[skeptical scientist]

How can something so simple be impossible, and how did someone prove it was impossible?

By using algebra.

[Tirian]

I was just rereading Benjamin Bold's Famous Problems of Geometry and How to Solve Them, which is a very accessible guide to showing how high school math can be used to tackle the impossibility of angle-trisection with compass and straightedge, the Delian problem, and Gauss' construction of the heptadecagon.

[LaserGuy]

There are a number of interesting intro calculus type problems that I recall being quite enjoyable. Something like:

-A man is standing at the origin, and begins walking at speed v in the x direction. His dog is standing on the y axis at y = y0 and runs toward the man at speed Av where A > 1. If the direction of the dog's running is always toward the man, at what position will the dog catch up, and what path will the dog take?

-A silo of radius 1m sits in the center of a field. A cow is staked to one end of the silo on a 1m tether. Calculate the total grazing area of the cow.

[Jrthedawg]

One fun high school project:

Draw any polygon whose sides are all 90 degree angles (if the polygon is not convex, there can be many sides). Pretend as if this polygon is a building from birds-eye view, and that a cow is attached to some part of the building with a leash of a certain length. What is the cows grazing area?

You can have students draw their own buildings.