[Guest]

Same drill, 5 questions, put your thinking hats on

SHOW ALL WORKING OUT/PROOF!

Enjoy:

1. Prove that, if a, b, c are positive numbers s.t. a < b + c, then: a/(a+1) < b/(b+1) + c/(c+1)

2. Find the sum for: 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(n*(n+1))

3. How many subsets of the set {1, 2, 3, ..., 2011} have an even number of elements?

4. If we have a*b*c = 1, and we also have a + b + c = 1/a + 1/b + 1/c. Prove that at least one of the numbers a, b, or c is equal to 1.

5. A hotel has 20 rooms, and only one storey. The rooms are located along the same side of a common corridor. The hotel owner is a weird fella, he only allows guests to rent his rooms in one of the following two ways:

a) Rent a room for 2 nights in a row, or

b) Rent 2 adjacent rooms for 1 night only.

No other ways of renting the rooms are allowed. Each room costs $1 dollar a night. Now the holiday season lasts 100 days.

It is known that the joke celebrity Justin Beiber stayed in Room 1 on the first night, and Room 20 on the last night of the season. And because he didn't stay according to the rules set out by the hotel owner for renting rooms (i.e. 2 nights in a row or two adjacent rooms for 1 night), he decided to pay the hotel owner an extra $6.50 for each night.

Assuming that no other guests broke the rules of the hotel (i.e. no one else paid extra). Show that the hotel owner took a maximum of $2011 in total throughout the holiday season.

Okay, Ready!! Set!! GOOOOO!!!!!

[superprismatic]

Answer for #3 is 2²⁰¹⁰:

The ith row of Pascal's triangle enumerates the

number of subsets of size j for j=0,1,2,...,i.

(Pascal's triangle starts with the 0th row here).

When i is odd, the row contains an even number

of entries; furthermore, each number in each odd

row is in there twice -- once in an even position

and once in an odd position. Thus, the number

of even subsets of 2011 elements is equal to the

number of odd subsets. Since the total number of

subsets of 2011 elements is 2²⁰¹¹, the number of

even subsets is half of that, which is 2²⁰¹⁰.

[Guest]

Answer for #3 is 2²⁰¹⁰:

The ith row of Pascal's triangle enumerates the

number of subsets of size j for j=0,1,2,...,i.

(Pascal's triangle starts with the 0th row here).

When i is odd, the row contains an even number

of entries; furthermore, each number in each odd

row is in there twice -- once in an even position

and once in an odd position. Thus, the number

of even subsets of 2011 elements is equal to the

number of odd subsets. Since the total number of

subsets of 2011 elements is 2²⁰¹¹, the number of

even subsets is half of that, which is 2²⁰¹⁰.

[spoiler=Indeed Sir ]You are correct. If you think about it this way, it becomes very clear:

Let A be the elements {1, 2, ..., 2011}. If B is a subset of A with an even number of elements, then A \ B (the complement of B in A, i.e. in A but not in B) contains an odd number of elements. So then the even numbered subsets can be paired off one-to-one to the odd numbered subsets. Thus there are just as many even numbered subsets of A as they are odd. So that means exactly half of the subsets has an even number of elements.

We all know that the number of subsets for A is 2^2011. 2^2011 / 2 = 2^2010. Answer given by superprismatic.

I really am amazed at the number of smart people roaming these forums...

[Guest]

Think of the days and rooms as a checkerboard 20 columns across and 100 rows long.

The owner will only allow you to checkout one red and one black square.

The red and black horizontal to each other indicating 2 rooms on the same night,

Red and black vertically indicate the same room on 2 nights.

Our friend JB took room 1 on night one and room 20 on night 100 (both of which are "black squares")

This leaves only 998 black squares left, so only 998 sets of rooms/nights left to purchase.

At $1 a day/room (or $2 per set), we get $1996 max.

Toss in JB's $15 ($1 per night per room + $6.50 sur charge per night per room) and you get 2011.

[Guest]

a/(a+1) < b/(b+1) + c/(c+1)

changed to

a(b+1)(c+1) < b(a+1)(c+1) + c(a+1)(b+1)

equal

abc + ab + ac + a < abc + ba + bc + b + abc + ca + cb + c

eqal

a < abc + 2bc + b + c.

it is ture

because

a < b + c

0 < abc + 2bc

[spoiler=Indeed Sir ]You are correct. If you think about it this way, it becomes very clear:

Let A be the elements {1, 2, ..., 2011}. If B is a subset of A with an even number of elements, then A \ B (the complement of B in A, i.e. in A but not in B) contains an odd number of elements. So then the even numbered subsets can be paired off one-to-one to the odd numbered subsets. Thus there are just as many even numbered subsets of A as they are odd. So that means exactly half of the subsets has an even number of elements.

We all know that the number of subsets for A is 2^2011. 2^2011 / 2 = 2^2010. Answer given by superprismatic.

I really am amazed at the number of smart people roaming these forums...

[Guest]

1-1/(n+1)

Same drill, 5 questions, put your thinking hats on

SHOW ALL WORKING OUT/PROOF!

Enjoy:

1. Prove that, if a, b, c are positive numbers s.t. a < b + c, then: a/(a+1) < b/(b+1) + c/(c+1)

2. Find the sum for: 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(n*(n+1))

3. How many subsets of the set {1, 2, 3, ..., 2011} have an even number of elements?

4. If we have a*b*c = 1, and we also have a + b + c = 1/a + 1/b + 1/c. Prove that at least one of the numbers a, b, or c is equal to 1.

5. A hotel has 20 rooms, and only one storey. The rooms are located along the same side of a common corridor. The hotel owner is a weird fella, he only allows guests to rent his rooms in one of the following two ways:

a) Rent a room for 2 nights in a row, or

b) Rent 2 adjacent rooms for 1 night only.

No other ways of renting the rooms are allowed. Each room costs $1 dollar a night. Now the holiday season lasts 100 days.

It is known that the joke celebrity Justin Beiber stayed in Room 1 on the first night, and Room 20 on the last night of the season. And because he didn't stay according to the rules set out by the hotel owner for renting rooms (i.e. 2 nights in a row or two adjacent rooms for 1 night), he decided to pay the hotel owner an extra $6.50 for each night.

Assuming that no other guests broke the rules of the hotel (i.e. no one else paid extra). Show that the hotel owner took a maximum of $2011 in total throughout the holiday season.

Okay, Ready!! Set!! GOOOOO!!!!!

[Guest]

a/(a+1) < b/(b+1) + c/(c+1)

changed to

a(b+1)(c+1) < b(a+1)(c+1) + c(a+1)(b+1)

equal

abc + ab + ac + a < abc + ba + bc + b + abc + ca + cb + c

eqal

a < abc + 2bc + b + c.

it is ture

because

a < b + c

0 < abc + 2bc

Yup! Nice and simple one to start things off well done

[Guest]

the total number of subsets including odd and even is

(2011 chooose 0) + (2011 choose 1) +....(2011 choose 2010) + (2011 choose 2011) = (1 + 1) ^ 2011 = 2^2011

Because there is 2011 numbers. 2011 = odd + even. Any specific odd subset is along with a specific even subset. therefore , the number of odd and the number of even is the same.

Therefore the answer is 2010

Same drill, 5 questions, put your thinking hats on

SHOW ALL WORKING OUT/PROOF!

Enjoy:

1. Prove that, if a, b, c are positive numbers s.t. a < b + c, then: a/(a+1) < b/(b+1) + c/(c+1)

2. Find the sum for: 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(n*(n+1))

3. How many subsets of the set {1, 2, 3, ..., 2011} have an even number of elements?

4. If we have a*b*c = 1, and we also have a + b + c = 1/a + 1/b + 1/c. Prove that at least one of the numbers a, b, or c is equal to 1.

5. A hotel has 20 rooms, and only one storey. The rooms are located along the same side of a common corridor. The hotel owner is a weird fella, he only allows guests to rent his rooms in one of the following two ways:

a) Rent a room for 2 nights in a row, or

b) Rent 2 adjacent rooms for 1 night only.

No other ways of renting the rooms are allowed. Each room costs $1 dollar a night. Now the holiday season lasts 100 days.

It is known that the joke celebrity Justin Beiber stayed in Room 1 on the first night, and Room 20 on the last night of the season. And because he didn't stay according to the rules set out by the hotel owner for renting rooms (i.e. 2 nights in a row or two adjacent rooms for 1 night), he decided to pay the hotel owner an extra $6.50 for each night.

Assuming that no other guests broke the rules of the hotel (i.e. no one else paid extra). Show that the hotel owner took a maximum of $2011 in total throughout the holiday season.

Okay, Ready!! Set!! GOOOOO!!!!!

[araver]

4. If we have a*b*c = 1, and we also have a + b + c = 1/a + 1/b + 1/c. Prove that at least one of the numbers a, b, or c is equal to 1.

abc=1 =>

1/a=bc

1/b=ac

1/c=ab.

Hence a + b + c = 1/a + 1/b + 1/c = bc + ac + ab

Computing (1-a)(1-b)(1-c)=1-a-b-c+ab+ac+ca-abc = (1-abc) +(bc+ac+ab-a-b-c)=0

So, 1-a=0 OR 1-b=0 OR 1-c=0.

[Guest]

a+b+c=1/a+1/b+1/c

=

a+b+c = bc + ac + ab

times a both side

a^2 + ab + ac = abc + a^2c+ a^2b

a^2+a(b+c) = 1 + a^2(b+c)

a^2-1 = (b+c)(a^2-1)

assume a,b,c not equal 1, therfore a^2-1 not 0

=> b + c = 1

same

a+ b = 1

a + c = 1

from abc=1

a(1-a)(1-a)=1

equal a(a-1)^2=1

therefore a > 0.

also

b(b-1)^=1

c(b-1)^2=1

the eqn x(x-1)^2=1.

has only 1 real root.

so assumption failed.

therefore must some a or b or c = 1

Same drill, 5 questions, put your thinking hats on

SHOW ALL WORKING OUT/PROOF!

Enjoy:

1. Prove that, if a, b, c are positive numbers s.t. a < b + c, then: a/(a+1) < b/(b+1) + c/(c+1)

2. Find the sum for: 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(n*(n+1))

3. How many subsets of the set {1, 2, 3, ..., 2011} have an even number of elements?

4. If we have a*b*c = 1, and we also have a + b + c = 1/a + 1/b + 1/c. Prove that at least one of the numbers a, b, or c is equal to 1.

5. A hotel has 20 rooms, and only one storey. The rooms are located along the same side of a common corridor. The hotel owner is a weird fella, he only allows guests to rent his rooms in one of the following two ways:

a) Rent a room for 2 nights in a row, or

b) Rent 2 adjacent rooms for 1 night only.

No other ways of renting the rooms are allowed. Each room costs $1 dollar a night. Now the holiday season lasts 100 days.

It is known that the joke celebrity Justin Beiber stayed in Room 1 on the first night, and Room 20 on the last night of the season. And because he didn't stay according to the rules set out by the hotel owner for renting rooms (i.e. 2 nights in a row or two adjacent rooms for 1 night), he decided to pay the hotel owner an extra $6.50 for each night.

Assuming that no other guests broke the rules of the hotel (i.e. no one else paid extra). Show that the hotel owner took a maximum of $2011 in total throughout the holiday season.

Okay, Ready!! Set!! GOOOOO!!!!!

[Guest]

abc=1 =>

1/a=bc

1/b=ac

1/c=ab.

Hence a + b + c = 1/a + 1/b + 1/c = bc + ac + ab

Computing (1-a)(1-b)(1-c)=1-a-b-c+ab+ac+ca-abc = (1-abc) +(bc+ac+ab-a-b-c)=0

So, 1-a=0 OR 1-b=0 OR 1-c=0.

your solution is better.

[Guest]

Think of the days and rooms as a checkerboard 20 columns across and 100 rows long.

The owner will only allow you to checkout one red and one black square.

The red and black horizontal to each other indicating 2 rooms on the same night,

Red and black vertically indicate the same room on 2 nights.

Our friend JB took room 1 on night one and room 20 on night 100 (both of which are "black squares")

This leaves only 998 black squares left, so only 998 sets of rooms/nights left to purchase.

At $1 a day/room (or $2 per set), we get $1996 max.

Toss in JB's $15 ($1 per night per room + $6.50 sur charge per night per room) and you get 2011.

wow very smart

[Guest]

#5 the owner can make $2013 DURING the season

in the last night, there will be 2, non adjacent, vacant rooms. the owner can rent them each for 2 consecutive nights, the money he gets from renting them is ($4). $2 are for the night spent during the last day of the holiday season, and $2 are for the first day after the holiday, that's assuming that the person renting the room is willing to stay 1 night after the holiday, so the maximum the owner can POSSIBLY get is $2013 from the holiday, not counting the $2 from the first night after the holiday.

[Guest]

Given a < b + c

Therefore, dividing both sides by (a+1),

a/(a+1) < b/(a+1) + c/(a+1)

As both b and c are < a, this inequality will also hold if we replace a by b or c in the denominators of the RHS

Therefore, a/(a+1) < b/(b+1) + c/(c+1)

QED

[Guest]

Think of the days and rooms as a checkerboard 20 columns across and 100 rows long.

The owner will only allow you to checkout one red and one black square.

The red and black horizontal to each other indicating 2 rooms on the same night,

Red and black vertically indicate the same room on 2 nights.

Our friend JB took room 1 on night one and room 20 on night 100 (both of which are "black squares")

This leaves only 998 black squares left, so only 998 sets of rooms/nights left to purchase.

At $1 a day/room (or $2 per set), we get $1996 max.

Toss in JB's $15 ($1 per night per room + $6.50 sur charge per night per room) and you get 2011.

Yup that solution is perfect! Damnit I was hoping to trick some people with that one haha

[Guest]

Given a < b + c

Therefore, dividing both sides by (a+1),

a/(a+1) < b/(a+1) + c/(a+1)

As both b and c are < a, this inequality will also hold if we replace a by b or c in the denominators of the RHS

Therefore, a/(a+1) < b/(b+1) + c/(c+1)

QED

"As both b and c are < a" <----- this is not true?

i.e. let a = 1, b = 2, c = 2. Then 1 < 2+2. yet both b and c are >a?

[Guest]

#5 the owner can make $2013 DURING the season

in the last night, there will be 2, non adjacent, vacant rooms. the owner can rent them each for 2 consecutive nights, the money he gets from renting them is ($4). $2 are for the night spent during the last day of the holiday season, and $2 are for the first day after the holiday, that's assuming that the person renting the room is willing to stay 1 night after the holiday, so the maximum the owner can POSSIBLY get is $2013 from the holiday, not counting the $2 from the first night after the holiday.

Okay I should have clarified. Hotel was only open during the holiday season. Otherwise your answer would be correct yes, but that wouldn't have been a hard problem.

[Guest]

Still number 2 up for grabs! And also remember number 4 from still lacks a valid proof.

[Guest]

For #2, reformulate as (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1)) and notice that the second value in the bracket of each term is annihilated by the first value in the next term. This leaves 1/1 - 1/(n+1) or n/(n+1)

[Guest]

For #4

Substitute 1/a for bc to get b^2c^c - b^2c - bc^2 + b +c - 1 = 0 and factorise this to get (c-1)(b-1)(bc - 1) = 0

This gives that either b = 1 or c = 1 (or bc = 1). If bc = 1, this implies, from abc = 1, that a = 1. Therefore either a = 1 or b = 1 or c = 1. Hence rewording says that at least one of a, b, c must be 1.

[Guest]

Sorry, typo. Last answer should read:

For #4

Substitute 1/a for bc to get b^2c^2 - b^2c - bc^2 + b +c - 1 = 0 and factorise this to get (c-1)(b-1)(bc - 1) = 0

This gives that either b = 1 or c = 1 (or bc = 1). If bc = 1, this implies, from abc = 1, that a = 1. Therefore either a = 1 or b = 1 or c = 1. Hence rewording says that at least one of a, b, c must be 1.

[Guest]

Answer for 2:

sum of 1/(1*2) + 1/ (2*3)+....1/(n*(n+1))= n^2/n*(n=1)

proof :

1/(1*2) + 1/(2*3) = 4/6 = n^2/n*(n=1) for n= 2 ..... (1)

same as above + 1 / (3*4) = 9/12 = n^2 /(n*(n=1)..... (2)

(2) + 1/ (4*5) = 16/20 = n^2/ (n*(n+1)

by deduction sum

= n^2 / n*(n=1)

[dark_magician_92]

its actually summation 1/n(n+1)

which is = to 1/n - 1/n+1, so we get on expanding from 1 to n

= 1-1/2+1/2-1/3+1/3-1/4 ........... 1/n - 1/n+1 = 1- 1/n+1 = n/n+1

[dark_magician_92]

#3

if null set is included then 2^2010, it is

2011 C O + 2011C2+.......+2011C2010 = 2^2011 /2 =2^2010