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Okay, so time for Round III of Talent Search Questions :) Last round was quite easy, all questions got answered in a night, so well done to everyone.

Let's see how you guys cope with these: :P

1. What are the last 2 digits of 6^2011 + 7^2011

2. Prove that a 10 x 10 chessboard can not be split up into 25 4x1 strips.

3. How many zeros are there at the end of 2011! ?

4. A company has 5 directors. Regulations of the company that any majority (3 or more) must be able to open its strongroom, but any minority (2 or less) of directors cannot. It is proposed that the door to the strongroom be equipped with 10 locks, so that when all ten locks are present, the door can be unlocked. Now each director is given a set of keys to exactly n doors, find all possible values of n such that there is a way to allocate the keys according to the regulations of the company.

5. Okay, this question is a mother of a question :D If anyone get's it correctly on their first attempt, I will be amazed at your amazingnessnessness :D (KK NO GOOGLING!!!)

Here goes: A magician has one hundred cards named 1 to 100. He divides them into 3 groups, and puts each group in separate boxes, boxes are coloured red, white and blue. Each box has at least one card in it.

A member of the audience then chooses 2 boxes, picks a card from each and announces the sum of the two cards. Given this sum, the magician identifies the box where no cards were taken.

Now the question is: How many ways are there to put all cards in the boxes so that this trick always works?

(Two ways are considered different if at least one card is put in a different box)

Hope these aren't too hard :) If you get stuck, here's a quick joke: :D

A statistician refuses to fly after reading that alarmingly high probabilities that of a bomb being on any given plane. But then a few weeks later he reads that the probability that 2 bombs are on any given plane is very low, so now when he flies, he carries a bomb with him at all times.

Edited by Twinhelix
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Paint rows 1, 3, 5, 7, and 9 white and the other rows black. There are now 50 squares of each color (2 modulo 4). Each vertical strip covers exactly two white squares, while a horizontal strip covers either 0 or 4 white squares (always 0 modulo 4). So we need an odd number of vertical strips to cover exactly 50 white squares. But for symmetrical reasons we also need an odd number of horizontal strips. Since we should have 25 strips total, this is impossible.

The number of zeros is the smallest of the exponents of 2 and 5 when factoring into primes. Since 2 obviously has a higher exponent in 2011!, we only need to know the exponent of 5.

In the set (1,2,3,...,2011), there are 402 numbers divisible by 5, 80 numbers divisible by 25, 16 numbers divisible by 125, 3 numbers divisible by 625, and no numbers divisible by 3125.

So the exponent of 5 is 402+80+16+3=521. Hence there are 521 zeros at the end of 2011!.

Edited by shakingdavid
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Paint rows 1, 3, 5, 7, and 9 white and the other rows black. There are now 50 squares of each color (2 modulo 4). Each vertical strip covers exactly two white squares, while a horizontal strip covers either 0 or 4 white squares (always 0 modulo 4). So we need an odd number of vertical strips to cover exactly 50 white squares. But for symmetrical reasons we also need an odd number of horizontal strips. Since we should have 25 strips total, this is impossible.

The number of zeros is the smallest of the exponents of 2 and 5 when factoring into primes. Since 2 obviously has a higher exponent in 2011!, we only need to know the exponent of 5.

In the set (1,2,3,...,2011), there are 402 numbers divisible by 5, 80 numbers divisible by 25, 16 numbers divisible by 125, 3 numbers divisible by 625, and no numbers divisible by 3125.

So the exponent of 5 is 402+80+16+3=521. Hence there are 521 zeros at the end of 2011!.

Yup your methods are correct!! :) just double check your adding for number 3 though :)

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n = 6. I had trouble just naming the keys until I decided to go backwards and only list the keys each director is missing. Then it was fairly easy to solve. Below are the 4 keys each director does not have, they have the other 6. Each lock corresponds to a key from 0 to 9 for 10 keys total.

missing keys per director

1234

1567

2589

3680

4790

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10,20,30...2010.

a1=0+10*1

an=0+201*10. therefore couting 201 zeros.

only 5*2 produce 10.

5 set

5,15,25,2005.

a0=5 + 10*0

an=5+10*200

the ending 5 is 201.

therefore total 0 zeros 201+201=402 zeros

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5 set

5,15,25,2005.

a0=5 + 10*0

an=5+10*200

the ending 5 is 201.

last digit is 0

10,20,30...2010.

a1=0+10*1

an=0+201*10. therefore couting 201 zeros.

2nd digit is 0 with last digit is 0

100,200,300,...2000. 20 counts

2nd digit is 5 with last digit is 0.

50,150,....1950.

a0=50+100*0

an=50+100*19. counts = 20.

3rd digits is 0. with last, 2nd is 0

1000,2000. 2

3rd digist is 5

500,1500. 2.

total 201+201+20+20+4=446.

no correct where is wrong?

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99 reasoning: raising 7 to the 10th power will give you 49 in the last 2 digits, since the last 2 digits of a product are only affected by the last 2 digits of the factors, then the last 2 digits of 7^100 are the same as the last 2 digits of 49^10, which are 01. similarly, the last 2 digits of 7^2000 are the same as the last 2 digits of 01^2. the only thing left os to multiply these last 2 digits (ie 01) by 7^11 to find the last 2 digits of 7^2011. the last 2 digits of 7^11 are 43. the last 2 digits of 6^2011 are 56: 6^10 has 76 as last 2 digits, 76^2 has also 76 as last digits which are the last 2 digits of 6^200, now 76^10 gives the same last 2 digits as 6^200, which is also 76. now we only have to find the last 2 digits of 76*6^11, this is 56.. so the last 2 digits of 6^2011 + 7^2011 are: 56 + 43 = 99

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1 in red box, 100 in blue box, 2 thru 99 in white box

1 in red box, 100 in white box, 2 thru 99 in blue box

1 in blue box, 100 in red box, 2 thru 99 in white box

1 in blue box, 100 in white box, 2 thru 99 in red box

1 in white box, 100 in blue box, 2 thru 99 in red box

1 in white box, 100 in red box, 2 thru 99 in blue box

Are these considered six different ways or one way? The split up is the same in all. They all work, as well.

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Superprismatic. Your solution omits possibilities

Not the full solution but I noticed immediately that the following combination also works. By placing every third card in each box, i.e. Red box cards 1,4,7,10,13 ...., White box 2,5,8,11..., and Blue box 3,6,9,12....the box not selected can easily be determined by dividing sum of two cards by three and looking at remainder. ) remainder = Blue box, 1 remainder =White box and 2 remainder - Blue box.

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Paint rows 1, 3, 5, 7, and 9 white and the other rows black. There are now 50 squares of each color (2 modulo 4). Each vertical strip covers exactly two white squares, while a horizontal strip covers either 0 or 4 white squares (always 0 modulo 4). So we need an odd number of vertical strips to cover exactly 50 white squares. But for symmetrical reasons we also need an odd number of horizontal strips. Since we should have 25 strips total, this is impossible.

The number of zeros is the smallest of the exponents of 2 and 5 when factoring into primes. Since 2 obviously has a higher exponent in 2011!, we only need to know the exponent of 5.

In the set (1,2,3,...,2011), there are 402 numbers divisible by 5, 80 numbers divisible by 25, 16 numbers divisible by 125, 3 numbers divisible by 625, and no numbers divisible by 3125.

So the exponent of 5 is 402+80+16+3=521. Hence there are 521 zeros at the end of 2011!.

Haha wonderful adding there, got an extra 20, if only I could make that work with my money :P of course it's 501

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This is basically an old school question that i remember in particular because of the ease of solution. It discusses of 2 probabilities. 1. 3 set of cards, with 1st set containing card 1, 2nd set containing cards upto 99 and the third containing the 100th card. Depending upon the sum now it can be said that if sum is 100 or less, set no. 1 and 2 have been chosen, if its 101, set 1 and 3 are chosen or else set 2 and 3 are chosen. 2. Another ways is to arrange cards in 3 sets with 1st set containing multiples of 3, 2nd containing modulo 1 (remainder is 1 after dividing by 3), 3rd containing modulo 2 cards. Now if the sum is multiple of 3 then set 2 and 3 are chosen, if sum is modulo 1 then set 1 and 2 are chosen or else set 1 and 3 are chosen. These are the only 2 probabilities (2nd case is not formulated for any other number except 3 because the no. of boxes are 3). Now the 2 cases can be used for 3 different boxes in 3! ways, i.e. 6 ways, for each case. So the total possibilities are 6 +6 = 12..

:)
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For any solution, we realize that every pair of directors must in fact lack at least one key in common. Otherwise they could open the door together. But no two pairs can lack the same key, because then we would have a triplet lacking that key. And that triplet wouldn't be able to open the door. So a mapping from the pairs of directors to the keys they lack in common would look something like this:

Directors 1,2: don't have key 1

Directors 1,3: don't have key 2

Directors 1,4: don't have key 3

Directors 1,5: don't have key 4

Directors 2,3: don't have key 5

Directors 2,4: don't have key 6

Directors 2,5: don't have key 7

Directors 3,4: don't have key 8

Directors 3,5: don't have key 9

Directors 4,5: don't have key 10

Since every pair must be mapped to at least one key, and no key can appear twice, we must assign exactly one missing key for each pair (there are 10 pairs and 10 keys). So where do we end up? Each director is missing exactly 4 keys, one for each pair containing that director. That makes n=6 the only possibility.

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For any solution, we realize that every pair of directors must in fact lack at least one key in common. Otherwise they could open the door together. But no two pairs can lack the same key, because then we would have a triplet lacking that key. And that triplet wouldn't be able to open the door. So a mapping from the pairs of directors to the keys they lack in common would look something like this:

Directors 1,2: don't have key 1

Directors 1,3: don't have key 2

Directors 1,4: don't have key 3

Directors 1,5: don't have key 4

Directors 2,3: don't have key 5

Directors 2,4: don't have key 6

Directors 2,5: don't have key 7

Directors 3,4: don't have key 8

Directors 3,5: don't have key 9

Directors 4,5: don't have key 10

Since every pair must be mapped to at least one key, and no key can appear twice, we must assign exactly one missing key for each pair (there are 10 pairs and 10 keys). So where do we end up? Each director is missing exactly 4 keys, one for each pair containing that director. That makes n=6 the only possibility.

Yup that's perfect! There are 5C2 pairs of directors, i.e. 10 pairs, so each must lack a key :) well done

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This is basically an old school question that i remember in particular because of the ease of solution. It discusses of 2 probabilities. 1. 3 set of cards, with 1st set containing card 1, 2nd set containing cards upto 99 and the third containing the 100th card. Depending upon the sum now it can be said that if sum is 100 or less, set no. 1 and 2 have been chosen, if its 101, set 1 and 3 are chosen or else set 2 and 3 are chosen. 2. Another ways is to arrange cards in 3 sets with 1st set containing multiples of 3, 2nd containing modulo 1 (remainder is 1 after dividing by 3), 3rd containing modulo 2 cards. Now if the sum is multiple of 3 then set 2 and 3 are chosen, if sum is modulo 1 then set 1 and 2 are chosen or else set 1 and 3 are chosen. These are the only 2 probabilities (2nd case is not formulated for any other number except 3 because the no. of boxes are 3). Now the 2 cases can be used for 3 different boxes in 3! ways, i.e. 6 ways, for each case. So the total possibilities are 6 +6 = 12..

:)

extend the first technique to use {1, 99, rest} and {1,98,rest}. The sum-splitter would be 100 and 99 respectively (instead of 101).

That would add 12 more possibilities.

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extend the first technique to use {1, 99, rest} and {1,98,rest}. The sum-splitter would be 100 and 99 respectively (instead of 101).

That would add 12 more possibilities.

Sorry, please ignore previous post.

extend the first technique to use

{ [1,2], [99,100], [rest] }

{ [1,2,3], [98,99,100], [rest] }

...

{ [1,2...48,49], [52,53,...99,100], [50,51] }

The same test would apply (checking if sum is >,=, or < 101).

This would add 48 new unique permutations, and thus 48*6=288 more possibilities, bringing the total to an even 300.

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Sorry, please ignore previous post.

extend the first technique to use

{ [1,2], [99,100], [rest] }

{ [1,2,3], [98,99,100], [rest] }

...

{ [1,2...48,49], [52,53,...99,100], [50,51] }

The same test would apply (checking if sum is >,=, or < 101).

This would add 48 new unique permutations, and thus 48*6=288 more possibilities, bringing the total to an even 300.

I don't think so. Let's look at { [1,2], [99,100], [rest] }. You can make 100 in two ways: 98 from [rest] and 2 from [1,2] AND 1 from [1,2] and 99 from [99,100].

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