Last update July 2009
Interesting problems
With * marked my own original problems. Hints to solution will be given below (29 Jan 2008).
- * 1.
Prove that for any odd x value x/4 has the
same parity as 3x/4 (here /4
is integer division discarding the modulo).
- * 2. Given a non-reducable fraction. If the numerator and the denominator
increases by 1, the fraction increases by N per cent.
What is the original fraction? (Assume N to be integer 1 to 100)
- 3. You have a balance scale and 13 coins, one of
which is counterfeit. The counterfeit weigh less or more
than the other coins. You can determine the counterfeit
in 3 weighings, but there is a possibility
that after finding the counterfeit you cannot
tell if it is heavier or lighter. What is the most
efficient algorithm of finding the counterfeit
which gives the minimal probability of not knowing
whether the counterfeit is heavier or lighter,
and what is this probability? [E.g. How we have to do our weighings
to find the counterfeit and likely to find out if it is hevier or lighter?]
- * 4. Prove that if in equation x^{2}-y^{2}=z
the sum of the digits of z equal to 10, then y is
divisible by 3, and x gives
a reminder 1 or 8 when divided by 9.
- * 5. What base of a number system must be to have such
property for number N that if the sum of digits of some number
is divisible by N then the number itself is divisible by N as well?
- * 6. Why a bus tilts left while turning right, and
a motorcycle tilts right while turning right? (You undersand it, but can
you express it in clear logic words?)
- * 7. Prove that any calendar year has Friday 13th.
- 8. Why in the mirror the left
swaps with the right, but the top does not swap with the bottom?
- 9. What is the next number 2,4,6,8,...?
Given that it is not 10, and the function is analytic.
- 10. A number ends with 2. If you
move 2 to the first place then the
number gets twice bigger. What is that
number?
- 11. I was born on k-th day in n-th
month. 12k+31n=545. When I was born?
- 12. John's son is as many weeks old as John's grandson
days old is. And the grandson is as many months old as
John years old is. Three of them are 100 years old.
How old is John?
- 13. Why the sum of subsequent odd numbers is a
perfect square?
- 14. A number gives remainder 1 when divided by 2,
it gives remainder 2 when divided by 3, 3 when by 4,
4 when by 5, 5 when by 6, and 0 when by 7. What are
2 minimal such numbers?
- 15. What is the probability that the first digit is 7 in the
infinite sequense 1, 2, 4, 8, 16, 32, 64, ...?
- 16. Two men talking.
- How many kids you have?
- 3
- How old are they?
- The product of their ages gives 36.
- Okay, tell me something else so I can guess.
- Do you see the number on that building? The sum of their ages gives that number.
- I still cannot figure out.
- Well, the older son has red hair.
- Oh, now I got it.
Question: what is the number on the building?
- 17. A man came to a hotel and paid $15 for the night. In the morning the receptionist
had found that the room was only $10 and gave $5 to a boy to return back to the man.
The boy gave back $3 to the man and left $2 to himself (stole).
Question: since the man received back $3 he paid $12, plus $2 the boy hid, makes $14, not $15;
where $1 had gone?
- 18. In town A all residents always tell lie. In town B all residents allways tell the truth.
Find a simple question (with yes/no answer) that will tell you in which of 2 towns
you are if asked the first encoutered person. Residents of A and B visit each other, so
when asking you do not know whether it is a visitor or a resident.
- 19. On one islind there live philosophers with green, brown, and grey eyes. If any
of philosophers guess his own colour of eyes he must leave the islind within one night.
One day one of the philisophers said "I see at least one person with green eyes".
Later some philosophers have to leave the island. Why?
- 20. 2 circles with radii 8sm and 10sm have 3 tangent lines: 2 internal and 1 external.
The internal tangent lines cross at 90 degrees. Find the area of triangle made by these
3 lines. (Must be simple solution with integer answer)
- 21. In a triangle ABC point D is on AB and point F is on AC. Angle BAC is 20 degrees.
Angles ACB=ABC=80. Angles FBC=50 and DCB=20. Find angle CDF.
- 22. In a game "guess the prize" a player is given 4 boxes. He selects a box and then
another box is open and is shown to be empty. In the rest 3 boxes the probability
to find the prize now is not equal. Find these probabilities.
- 23. One person deals 5 random cards from 52 card deck. He takes 1 card aside and puts
the rest 4 cards on the table face up. The other person tells what card is the one set aside.
Find encoding algorithm that 4 cards put on the table give enough information about
one card.
- * 24. The same problem as above but 6 cards are dealt instead of 5 and all 6 are put
on the table, 2 face down. Another person tells what cards are face down and which one is which.
Find the algorithm to encode this information.
- 25. Two men are dining together in a restaraunt. The waiter asks them to play the following
game. They took out their wallets and count money. The person who has least amount in
his wallet takes all the money. Both men think that the game is fair and profitable
because everyone wins more then can lose. Find the solution to the paradox.
- 26. One man traveled exactly 1km north, 1km east, 1km south, 1km west and ended in exactly
the same place where he started from. Tell where is this man.
- * 27. (Sydney puzzle) Find the smallest number for which the following procedure
does not give 9.
1. Take any number
2. Multiply by 3
3. Add all digits
4. Multiply by 3
5. Add all digit
6. The result is 9
For example, 5 -> 5*3=15 -> 1+5=6 -> 6*3=18 -> 1+8=9
- * 28. Prove that 1*1!+2*2!+3*3!+...+n*n!=(n+1)!-1
- * 29.
This is a math trick I show with my daughter.
I take 7 defined playing cards from a deck. Shuffle them.
My daughter closes itself in another room. Then I put 4 (out of those 7)
random cards on the table face up. The rest 3 are aside face down.
Then I look at the cards on the table and turn down one card out of 4.
Then anyone speaks loud 3 cards which are left face up. And my daughter
in the other room says which card was turned down by me. The trick is
that selecting a particular card and leaving particular 3 others I pass
the information about the card turned down. The question: find a simple
rule (algorithm) allowing this. Hint: my part involves a bit of counting
but simple enough to make it quick on the fly; were as my daughter
is doing it with a piece of paper (I can do it in my head but for
her it is more difficult) - it is less prone to mistake.
- * 30.
What is the probability of getting 2 subsequent numbers when selecting and removing
numbers at random from a list 1 to N? In other words, how many permutations in
the list of numbers 1 ro N have at least two numbers in order?
Hints/solutions
- 1. x=4k+1 or x=4k+3. In the first case (4k+1)/4 -> k
and 3(4k+1)/4 -> 3k. In the second (4k+3)/4 -> k and
3(4k+3)/4 -> 3k+2.
- 2. Let x and y be relatively prime.
Then (x+1)/(y+1)=(100+N)x/100y, then x(Ny+100+N)=100y.
Now x must divide 100 because x and y are relatively prime.
That is x can be only 1,2,4,5,10,20,25,or 50.
Substituting x we get 100+N={100-N,50-N,25-N,20-N,10-N,5-N,4-N,2-N}y,
that is one of A-N numbers must divide 100+N
- N=1
101={99,49,24,19,9,4,3,1}y
101=1y.
x=50, y=101
- N=2
102={98,48,23,18,8,3,2}y
102=2y or 102=3y.
The later gives x=20, which is not prime to y.
x=25, y=51
- N=3 103 is prime, hence 103={1}y
25/103
- N=4 No solution (15/39 and 20/104 not good)
- N=5 10/21
- 3. ...
- 4. The condition is x^{2}=y^{2}+1 mod 9.
Squares in mod 9 are 0,1,4,7, which means that only 1 and 0 match the equation.
Hence y^{2}=0 mod 9 and x^{2}=1 mod 9, i.e.
y=0 mod 3 and x=(1 or -1) mod 9.
- 5. A = X^{n}a_{n} + ... + X^{2}a_{2}
+ Xa_{1} + a_{0} = Σa_{i}
+ (X^{n}-1)a_{n} + ... + (X^{2}-1)a_{2}
+ (X-1)a_{1} = Σa_{i} + (X-1)
( (...)a_{n} + ... + (X^{2}+1)a_{2} + a_{1} )
. This means that X must be 1+kN to
make A=Σa_{i} mod N.
- 6. Try this or
this
- 7. Each year (regardless leap of not) has at least one month starting at any day
of the week. Hence, there is a month starting on Sunday, and 13th is Friday. The first
statement can be proven by considering reminders of division by 7 for a non-leap year
and leap year.
- 8. This is very difficult question. I do not know its answer.
- 9. One of solutions is 34. Let us say F(x)=2x+f(x), where
f(x)=0 for x=1,2,3,4. Obviously the simplest function is
f(x)=(x-1)(x-2)(x-3)(x-4). Id est F(5)=2*5+4*3*2*1=34
- 10. The answer is 105263157894736842
- 11.
12k+31n=545
k=(545-31n)/12=45-2n+(5-7n)/12
Let a=(5-7n)/12 then -8<a<1 because 0<n<13
n=(5-12a)/7=-a+(5-5a)/7
Let b=(5-5a)/7 then 0<b<6
Now since b has to be divisible by 5 it must be only 5
a=-6 and n=11
- 12. Assuming that they all have their birthday in the same day we get:
12x+7x+x=100 where x is grandsons age.
- 13. One simple proof is a square made of little squares: look at squares of
different size with the same bottom left corner.
- 14. ...
- 15. 2^{n}=a*10^{x}
log_{10}(a*10^{x})=x+b where 0<b<1
For big n b will cover uniformly the region (0,1) bacause
each next b gets log_{10}2 added minus the whole part (ergodicity).
The probability that a starts with 7 is the length of region between
b=log_{10}7 and b=log_{10}8, because this
corresponds to 7<a<8. Hence probability is
log_{10}8-log_{10}7=log_{10}(8/7)
- 16. The only cases which gives the same sum are 2,2,9 and 6,6,1. Any other
case of product of 36 give different sum.
- 17. This is old well known puzzle.
- 18. Are you a visitor?
- 19. To be continued ...
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