# A problem you find interesting

Let's say your friend knows nothing about olympiad math problems, and as an example you want to give him or her one that shows the idea of problem solving but at the same time isn't overly complex so they can find it intellectually interesting and understand the problem. What problem would you give them? (looking for a specific one)

Note by Michael Tong
4 years, 9 months ago

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I like this one - it's about mid-late AIME, and I wrote it!

Call a number huge if the sum of its digits is at least $$23.$$ There are exactly $$35$$ huge three-digit numbers, and there are exactly $$2205$$ huge four-digit numbers. Find the number of five-digit huge numbers.

- 4 years, 9 months ago

I like it.. but what I first thought is that these American Michael Ts are messing with my brain!

- 4 years, 9 months ago

He's michael tAng, i'm michael tOng :D

- 4 years, 9 months ago

Show that at a party there are always at least 2 people who greet the same number of participants.

- 4 years, 9 months ago

I know it's an easy problem but here's the solution if someone needs it ;)

Let $$n$$ be the number of participants.

$$[1]$$ The number of participants someone could greet is a number between $$1$$ and $$n-1$$ (if someone greets $$0$$ people than no one can greet $$n-1$$ people and the number is between $$0$$ and $$n-2$$).

Let's suppose that each participant has a card with a number corresponding to the "number of participants greeted by him": in consequence of what was said in $$[1]$$, there are $$n-1$$ possible values for this number. But the participants are $$n$$. Therefore there are at least $$2$$ participants who have a card with the same number (box principle): this shows that there are at least $$2$$ people who greet the same number of participants.

- 4 years, 9 months ago

This one is actually given from me to a friend that is curious what "math olympiad" is. Which is also a well-known problem, but hey why not.

A standard ($$8 \times 8$$) chessboard has two opposite corners removed. We want to place 31 dominoes (there are 62 squares remaining), each covering two adjacent squares, on the board, such that all dominoes lie on the board and all squares are covered. Can we do it?

It's a fairly simple statement, but with a nontrivial answer. Exactly as the question asks for?

Also, if you want a problem that is easy to understand, always try giving a combinatorics question; second contender would be some kind of geometry that doesn't involve naming points. People are easily shunned by variables. Combinatorics wins against geometry just because problems in combinatorics are more "real-life" like; they can imagine the problem simpler. But that's my opinion, and yours might differ.

- 4 years, 9 months ago

1)Every road in Sikinia is one-way.Every pair of cities is connected exactly by one direct road.Show that there exists a city which can be reached from every city directly or via at most one other city.

2)One of the $$n+1$$ numbers from {$$1,2,3,\cdots ,2n$$} is divisible by another.

- 4 years, 9 months ago

Find $$i^i$$, where $$i=\sqrt{-1}$$.

That's cool :P

i^i = e^(-pi/2) which is real.

- 4 years, 9 months ago

An old manuscript describing the location of a buried treasure: There are only two trees, A and B, in a flat terrain, and a bed of tomatoes. "A" is a hose, and B an appletree. From the center K of the tomatoes bed, measure the distance in a straight line to the hose. Turn 90 degrees to the left and go through the same distance to point C. Go back to the tomatoes bed. Measure the straight line distance to the appletree. Turn 90 degrees to the right and go through the same distance to the point D. The treasure is at the midpoint T of the segment CD. An adventurer found the manuscript, identified the trees, but as the tomatoes bed disappeared over time, could not locate it, and gave up the search. The student Brilli the Ant, under same conditions, says it would be able to locate the treasure. Show you how to solve the problem, ie, give the coordinates of point T as a function of the coordinates of A = (5, 3) and B = (8, 2).

Comment deleted Sep 25, 2013

616

- 4 years, 9 months ago

What is the last digit of $$2^{2013}$$?

How many positive divisors does $$120$$ have?

- 4 years, 9 months ago

Something more complex than that though. I don't think anybody would find a question like that "interesting"

- 4 years, 9 months ago

Well, you didn't specify the problemsolving-level of this friend. While these problems might not be that hard, the math behind them is quite interesting! Furthermore, alot of advanced questions boil down to these simple questions.

- 4 years, 9 months ago

last digit=[2^10]^2012^3=248=192 last digit =2

120=2^3 35 hence total no of divisor=(3+1)(1+1)(1+1)=16 since[ 2^10]^odd=last two digits are 24 and in case of even it is 76

- 4 years, 9 months ago

120=2^3 35 hence total no of divisor=(3+1)(1+1)(1+1)=16

- 4 years, 9 months ago

last digit=[2^10]^2012^3=248=192 last digit =2

since[ 2^10]^odd=last two digits are 24 and in case of even it is 76

- 4 years, 9 months ago

Why does the ASS congruence not hold?

- 4 years, 9 months ago

What is $$\infty^\infty$$?

- 4 years, 9 months ago