# Mind Blowing Puzzles

CAT is over and its time for some recreation. Here’s the first one *On the recent TV reality show “Losing Big,” five celebrities competed to shed pounds during the eight-week contest. Each of the five did lose weight on the show, with no two …

CAT is over and its time for some recreation. Here's the first one

On the recent TV reality show "Losing Big," five celebrities competed to shed pounds during the eight-week contest. Each of the five did lose weight on the show, with no two losing the same number of pounds. Given the clues below, can you find each star's full name (one surname is Brown), the field of entertainment in which he or she is famous, and the amount of weight he or she dropped during the contest?

1. The most any of the five lost during the eight weeks is 60 lbs., while the least any contestant lost is 15 lbs.
2. Jerry lost twice as much weight as Hart.
3. Tom lost 5 more lbs. than West, who shed 10 more lbs. than the cable network news anchor.
4. Willa dropped twice as much weight as the tennis player.
5. King and the rock star made a side bet on which of them would lose the most weight.
6. Willa, Stone, and the comedian all placed their weight-loss hopes on extreme exercise.
7. Alec isn't the new anchor.
8. Mia lost half as many pounds as the talk show host.

and before I forget its better to post it

There's a bank which held a week-long contest to provide its tellers incentive to convince their regular customers to open new accounts. All five tellers participated, and the winner received two days off with pay. You have to find out each teller's first and last name (one first name is Trudy; one surname is West) and how
many new accounts each one acquired using the following eight clues. (1) The five brought in a total of 132 new accounts; (2) The winner brought in 36 accounts, but no teller brought in fewer than 15; (3) Billings got six more new accounts than Meyer; (4) Ian didn't get the fewest accounts; (5) Russo got an odd number of accounts; (6) Jessica got five more accounts than Robert; (7) Ellen got three more accounts than Kurtz; ( 8 ) The two men together got a total of 44 accounts.

and before I forget its better to post it

There's a bank which held a week-long contest to provide its tellers incentive to convince their regular customers to open new accounts. All five tellers participated, and the winner received two days off with pay. You have to find out each teller's first and last name (one first name is Trudy; one surname is West) and how
many new accounts each one acquired using the following eight clues. (1) The five brought in a total of 132 new accounts; (2) The winner brought in 36 accounts, but no teller brought in fewer than 15; (3) Billings got six more new accounts than Meyer; (4) Ian didn't get the fewest accounts; (5) Russo got an odd number of accounts; (6) Jessica got five more accounts than Robert; (7) Ellen got three more accounts than Kurtz; ( 8 ) The two men together got a total of 44 accounts.

Name.....Surname....Accounts

Trendy ....Billing......36
Jessica.....Heyer.......30
Robert....Russo........25
Eleen.......West........22
Ian.........Kurtz..........19

I have taken Ian and Robert as males and rest as item log.

Here 1) The five brought in a total of 132 new accounts; (2) The winner brought in 36 accounts, but no teller brought in fewer than 15;
( 8 ) The two men together got a total of 44 accounts. forms the starting point. see two men have 44 and one has score of 36. so these two cant have 36 as otherwise one will be 8. now we have scores of three ( 36 and two have 44) = 80 so we have 52 as score of other two.

now try out P&C; most of them get elimintaed due to condition no 2nd

and yes that's my 1800. Didnt want to waste it by posting on some other forum

Name.....Surname....Accounts

Trendy ....Billing......36
Jessica.....Heyer.......30
Robert....Russo........25
Eleen.......West........22
Ian.........Kurtz..........19

I have taken Ian and Robert as males and rest as item log.

Here 1) The five brought in a total of 132 new accounts; (2) The winner brought in 36 accounts, but no teller brought in fewer than 15;
( 8 ) The two men together got a total of 44 accounts. forms the starting point. see two men have 44 and one has score of 36. so these two cant have 36 as otherwise one will be 8. now we have scores of three ( 36 and two have 44) = 80 so we have 52 as score of other two.

now try out P&C; most of them get elimintaed due to condition no 2nd

Congo on 1800 ... my patience with puzzles has gone down drastically after cat mann ... give m some more puzzles to get back on track
siddhesh_j Says
Congo on 1800 ... my patience with puzzles has gone down drastically after cat mann ... give m some more puzzles to get back on track

sir jee aap kaho hum post naa kare. ye le next year ka cat problem (paper leaked - mein hoo don ranjit don )

Cloe had just come in from her lacrosse game when her twin brother Tyler telephoned. "Officer O'Henry wants to come over. It seems he has some unreliable eyewitnesses to deal with," Tyler told her after she picked up.
A familiar figure to these twelve-year-olds, O'Henry had come to rely on their help on tricky cases. Being a smart and serious student of puzzles himself, he characteristically started his briefing by giving the twins a small history lesson.

"There are many classic puzzles in which people either always tell the truth or always lie," he began. "You ask them questions and either you determine who is the liar or you discover some fact without explicitly determining anyone's honesty.

"For example, here is what I consider to be the queen of the classic liar problems. You are in unfamiliar country walking on a path. You encounter a single armed warrior at a fork in the road. Each branch of the fork leads to a village. You know that one road leads to a group of peaceful people who always tell the truth and will give you food. The other leads to a village of congenital liars who will kill you. The people from the two villages look alike. You are permitted one yes-or-no question to the warrior and then you must choose a fork and meet your fate. What question should you ask?"

Here's why: Suppose you are pointing toward the liar's village. The liar will deny that you are pointing to his village and the truthteller will, too. If you point toward the truthtellers' village, both will say they come from there. In either case, you know which way you should go.
O'Henry continued, "The trouble is that the liars most of us know in real life are only occasional liars. Sometimes they choose to tell the truth and sometimes they lie. That might seem to make things easier, but it doesn't. If the warrior in the example were an occasional liar, he might tell you to go to his real village in response to the question.
"In fact if there were at most one occasional liar in a group of warriors while the rest told the truth, you would need at least three warriors to determine what to do using this question. The majority would tell you the truth.

"My problem concerns a crime scene, but I can't tell you the details because they are confidential, so we'll make it a game. Suppose that there is an object in an opaque box. You know it is one object and that it can be either red or black, either large or small, and either a shoe or a sock. You want to discover what's in the box.
"Three people have seen the object. At most one is an occasional liar. The others are honest."

Problem 1. Suppose you may ask each person at most three yes-or-no questions provided you ask a total of no more than seven questions. Can you figure out the size, color and type of object?
"Now for the more difficult problem: Six people have seen the object in the box. At most two of these people are occasional liars. The others are honest."
Problem 2. Can you ask each person at most two yes-or-no questions and figure out what is in the box?
Problem 3. How much better could you do if socks could be only black and shoes could be only large?
Craggy Prime is a hermit. He gets his first name because he likes to climb rocks. The twin sleuths Cloe and Eli gave him his second name because of his remarkable ability with numbers, especially primes.
The twins were walking on a trail near Dog Creek when they first met him. The local police had just told the twins that they were looking for a major drug shipment, possibly arriving on a numbered bus.

Something strange about the number on a bus I saw coming through here," said Craggy without introduction and quite coincidentally. "This bus number has two digits and the sum of those digits equals the sum of the bus number's prime factors."

Problem 1. Cloe and Eli quickly figured out what the number on the bus must be. Can you?
Hint: Consider the number 45. The prime factors are 3, 3 and 5. So the sum of the prime factors is 3 + 3 + 5 = 11. The sum of the digits is 4 + 5 = 9. So, 45 is not the answer.
The twins reported the bus number to the police, who found the bus and traces of drugs. The police were convinced, however, there were other buses involved in the scheme. The twins went looking and found Craggy perched atop the volcanic plug called French's Dome. "Yes, I found another strange one," he said as he awoke from his reverie. "Also two digits. The sum of the digits is half the sum of its prime factors."
"That's not enough information," Cloe protested.
"Right," said Craggy. "The bus number in question also has only two prime factors."
"I've got it," said Eli.

Problem 2. Which bus number did Eli calculate?
"You are very clever," Craggy said with his rough smoker's chuckle. "There's another two-digit bus number where the sum of the digits is greater than the sum of its prime factors and the number of prime factors is not 5."
"That's not enough information either," said Eli.
"Ok, I'll nail it down by telling you how many prime factors it has," said Craggy.
"No need," Cloe interrupted. "I know what it is."

Problem 3. What number did Cloe say? How did she know?

Craggy Prime is a hermit. He gets his first name because he likes to climb rocks. The twin sleuths Cloe and Eli gave him his second name because of his remarkable ability with numbers, especially primes.
The twins were walking on a trail near Dog Creek when they first met him. The local police had just told the twins that they were looking for a major drug shipment, possibly arriving on a numbered bus.

Something strange about the number on a bus I saw coming through here," said Craggy without introduction and quite coincidentally. "This bus number has two digits and the sum of those digits equals the sum of the bus number's prime factors."

Problem 1. Cloe and Eli quickly figured out what the number on the bus must be. Can you?
Hint: Consider the number 45. The prime factors are 3, 3 and 5. So the sum of the prime factors is 3 + 3 + 5 = 11. The sum of the digits is 4 + 5 = 9. So, 45 is not the answer.
The twins reported the bus number to the police, who found the bus and traces of drugs. The police were convinced, however, there were other buses involved in the scheme. The twins went looking and found Craggy perched atop the volcanic plug called French's Dome. "Yes, I found another strange one," he said as he awoke from his reverie. "Also two digits. The sum of the digits is half the sum of its prime factors."
"That's not enough information," Cloe protested.
"Right," said Craggy. "The bus number in question also has only two prime factors."
"I've got it," said Eli.

Problem 2. Which bus number did Eli calculate?
"You are very clever," Craggy said with his rough smoker's chuckle. "There's another two-digit bus number where the sum of the digits is greater than the sum of its prime factors and the number of prime factors is not 5."
"That's not enough information either," said Eli.
"Ok, I'll nail it down by telling you how many prime factors it has," said Craggy.
"No need," Cloe interrupted. "I know what it is."

Problem 3. What number did Cloe say? How did she know?

1. 27
Sum of digits,S= 2+7= 9
Sum of prime factors,P= 3+3+3= 9

2. 91
Sum of digits, S= 9+1= 10
No. of prime factors= 2
Sum of prime factors,P= 13+7= 20= 2S

3. 18
As it is the only two digit no. having no. of prime factors not equal to 5 and
for which S>P.

@Shashank Shekha,

Is there any easy way to calculate other than trial and error??

@warrior

Warrior bhai, sahi ja rahe ho!! Awesome puzzles. aapne yeah puzzle kaha se liya? Hume bhee source batana please

1. 27
Sum of digits,S= 2+7= 9
Sum of prime factors,P= 3+3+3= 9

2. 91
Sum of digits, S= 9+1= 10
No. of prime factors= 2
Sum of prime factors,P= 13+7= 20= 2S

3. 18
As it is the only two digit no. having no. of prime factors not equal to 5 and
for which S>P.

watson.............

From MS 17th DEC 2006

I'm not particularly a dog person, I've generally ended up keeping a fox, a bat, a couple of crows, some snakes, whole bunches of squirrels, three or four tortoises and too many cats to remember. Seven in fact on last count, the last one being a fishing feline that fell off the eighth floor of my Calcutta home while trying to take a flying leap at a pigeon for lunch from the outer window ledge it was sunning itself on. It died on the spot. So what's the big deal that cats always land on their feet? This is what I found out. Seems like these guys can survive falls from the first to third floors and after that from the ninth to the highest record being the 71st

floor. But they usually end up in those happy mouse hunting grounds on cloud ninety-nine when they fall from the fourth to the eighth floors. Bet you don't know why. There's a cool reason though.
So as I was continuing, this guy has a collection of five groups of animals in his private zoo comprising mammals, birds, snakes, spiders and insects. Now assume that usually animals have one head, mammals have four legs, birds have two legs, snakes have zero legs, spiders have eight legs and insects have six legs. However, our dude has picked up some freak beasts like in a mammal with three legs, a bird with two heads, a snake with three heads, a spider with seven legs and an insect with four legs. Now given the following information you have
to determine how many of each group of animals the weirdo has in his kept.
(1) There are a total of 100 heads and 376 legs; (2) Each group has a different quantity of animals; (3) The most populous group has 10 more members than the least populous group; (4) There are twice as many insect legs as there are bird legs; (5) There are as many snake heads as there are spider heads.

From MS 17th DEC 2006

I'm not particularly a dog person, I've generally ended up keeping a fox, a bat, a couple of crows, some snakes, whole bunches of squirrels, three or four tortoises and too many cats to remember. Seven in fact on last count, the last one being a fishing feline that fell off the eighth floor of my Calcutta home while trying to take a flying leap at a pigeon for lunch from the outer window ledge it was sunning itself on. It died on the spot. So what's the big deal that cats always land on their feet? This is what I found out. Seems like these guys can survive falls from the first to third floors and after that from the ninth to the highest record being the 71st

floor. But they usually end up in those happy mouse hunting grounds on cloud ninety-nine when they fall from the fourth to the eighth floors. Bet you don't know why. There's a cool reason though.
So as I was continuing, this guy has a collection of five groups of animals in his private zoo comprising mammals, birds, snakes, spiders and insects. Now assume that usually animals have one head, mammals have four legs, birds have two legs, snakes have zero legs, spiders have eight legs and insects have six legs. However, our dude has picked up some freak beasts like in a mammal with three legs, a bird with two heads, a snake with three heads, a spider with seven legs and an insect with four legs. Now given the following information you have
to determine how many of each group of animals the weirdo has in his kept.
(1) There are a total of 100 heads and 376 legs; (2) Each group has a different quantity of animals; (3) The most populous group has 10 more members than the least populous group; (4) There are twice as many insect legs as there are bird legs; (5) There are as many snake heads as there are spider heads.

my ans 15 mammals, 17 insects, 19 snakes, 21 spiders and 25 birds.

you will have 97 different mamamls - just minus the extra heads from the 5 abnormal animals

so u have 92 normal animals

if the normal animals are a,b,c,d,e

we have a+b+c+d+e = 92 ( eq on heads)

similary 4a+2b+8d+6e = 360 ( = 376- 3-2-0-7-4)

also its given that 6e+ 4 = 2(2b+2) => 2b =3e ( condition 4)

now because we have diff of 10 betn max and min assume they are g, g+2, g+4, g+6, g+10

sub these in a+b+c+d+e = 92 we get g= 14

i.e the animals are 14, 16, 18,20, 24

this seems to fit 2b =3e ( b= 24 and e= 16)

so solvew for other and get the ans
To raise travel funds to go to the Midget Stanley Cup, hockey players on the Summerset Sticks sold Tim Horton's doughnuts throughout the area. Each player sold a different number of dozen doughnuts, with the top five salesmen, each of whom plays a different position on the team, selling a total of 2,000 dozen among them. Given the sales figures below, can you solve this Challenger Logic Puzzle by finding the full name (one first name is Alex and one last name is Orr) of each of the top five Tim Horton's salesmen, the position he plays for the Summerset Sticks, and the number of dozen doughnuts he sold?
1. Eric, who isn't Lemieux, isn't the left wing.
2. Howe sold half as many dozen as the Sticks center did.
3. Tony, who isn't the defenseman, and Esposito are team co-captains.
4. The top salesman isn't the boy who plays right wing for the team.
5. Eric sold twice as many doughnuts as Esposito.
6. Lemieux and the center competed in selling doughnuts at the entrance to the local Food Giant.
7. Top-five salesman Clarke and the boy who plays goalie each ate a dozen doughnuts while they were out selling.
8. Chad, who didn't finish in fifth place in number of dozens sold, isn't the top-five salesman who plays right wing.
9. Lemieux sold 100 dozen more Tim Horton's doughnuts than Peter, who sold 200 dozen more than the defenseman.
10. The fifth-place salesman sold 200 dozen doughnuts.
To raise travel funds to go to the Midget Stanley Cup, hockey players on the Summerset Sticks sold Tim Horton's doughnuts throughout the area. Each player sold a different number of dozen doughnuts, with the top five salesmen, each of whom plays a different position on the team, selling a total of 2,000 dozen among them. Given the sales figures below, can you solve this Challenger Logic Puzzle by finding the full name (one first name is Alex and one last name is Orr) of each of the top five Tim Horton's salesmen, the position he plays for the Summerset Sticks, and the number of dozen doughnuts he sold?
1. Eric, who isn't Lemieux, isn't the left wing.
2. Howe sold half as many dozen as the Sticks center did.
3. Tony, who isn't the defenseman, and Esposito are team co-captains.
4. The top salesman isn't the boy who plays right wing for the team.
5. Eric sold twice as many doughnuts as Esposito.
6. Lemieux and the center competed in selling doughnuts at the entrance to the local Food Giant.
7. Top-five salesman Clarke and the boy who plays goalie each ate a dozen doughnuts while they were out selling.
8. Chad, who didn't finish in fifth place in number of dozens sold, isn't the top-five salesman who plays right wing.
9. Lemieux sold 100 dozen more Tim Horton's doughnuts than Peter, who sold 200 dozen more than the defenseman.
10. The fifth-place salesman sold 200 dozen doughnuts.

Well solved it, at the expense of a few hrs which goes into the timesheet as self study Well I must admit that I was incredibly lucky. I will explain why. Neways

The answer: I know its gotta be right cos it satisfies all the conditions

Full Name----------Position---------Units sold
Alex Howe----------Defenseman------200
Peter Clarke -------Center----------- 400
Tony Lemieux ------Rightwinger------- 500
Eric Orr ------------Goalie -----------600

The method:
First I came up with the combination of units that are sold by the individual. Concluded that this is the only combination possible with the given conditions.
This wouldve been really simple if the first names and the last names were mentioned explicitly This is where the my method has a major flaw. Though not explicitly mentioned, I assumed and bifurcated the names into first and last respectively and this is where I was lucky:biggrin: I mean just read those names agn. It was a sorta intuition
Then I drew 3 matrices. First name v/s Position, Last name v/s position and First name v/s Last name, jus to keep track of who is not what and in 15 minutes it was done.

Ofcourse by elimination and multiple-connect interpretations, the answer couldve been obtained, wouldve taken a lot of time and concentration though.

Moral: Dont always go for the tried and tested option

here's today's puzzle

A Number Pyramid is composed of the 10 different numbers 0-9 with a top row of 1 number resting on a second row of 2 sitting on a third row of 3 supported by a bottom row of 4. For example, a Number Pyramid could be:

0
1 2
3 4 5
6 7 8 9

Given the clues below, can you determine the composition of Number Pyramid ?

1. The four numbers in row 4, none of which is 6, sum to 12.
2. The leftmost number in row 3 minus the rightmost number in row 3 equals 7.
3. The number at the apex of Number Pyramid minus the rightmost number in the bottom row equals 6.
4. The number at the apex plus the two numbers in the second row sum to 17.
5. The rightmost numbers in the four rows of Number Pyramid add to 15.
6. The leftmost number in row 4 minus the second leftmost number in row 4 equals 1.
here's today's puzzle

A Number Pyramid is composed of the 10 different numbers 0-9 with a top row of 1 number resting on a second row of 2 sitting on a third row of 3 supported by a bottom row of 4. For example, a Number Pyramid could be:

0
1 2
3 4 5
6 7 8 9

Given the clues below, can you determine the composition of Number Pyramid ?
1. The four numbers in row 4, none of which is 6, sum to 12.
2. The leftmost number in row 3 minus the rightmost number in row 3 equals 7.
3. The number at the apex of Number Pyramid minus the rightmost number in the bottom row equals 6.
4. The number at the apex plus the two numbers in the second row sum to 17.
5. The rightmost numbers in the four rows of Number Pyramid add to 15.
6. The leftmost number in row 4 minus the second leftmost number in row 4 equals 1.

8
3 6
7 9 0
5 4 2 1
8
3 6
7 9 0
5 4 2 1

not correct dude it dosent satisfy that apex minus bottom right should be 6
Well solved it, at the expense of a few hrs which goes into the timesheet as self study Well I must admit that I was incredibly lucky. I will explain why. Neways

The answer: I know its gotta be right cos it satisfies all the conditions

Full Name----------Position---------Units sold
Alex Howe----------Defenseman------200
Peter Clarke -------Center----------- 400
Tony Lemieux ------Rightwinger------- 500
Eric Orr ------------Goalie -----------600

The method:
First I came up with the combination of units that are sold by the individual. Concluded that this is the only combination possible with the given conditions.
This wouldve been really simple if the first names and the last names were mentioned explicitly This is where the my method has a major flaw. Though not explicitly mentioned, I assumed and bifurcated the names into first and last respectively and this is where I was lucky:biggrin: I mean just read those names agn. It was a sorta intuition
Then I drew 3 matrices. First name v/s Position, Last name v/s position and First name v/s Last name, jus to keep track of who is not what and in 15 minutes it was done.

Ofcourse by elimination and multiple-connect interpretations, the answer couldve been obtained, wouldve taken a lot of time and concentration though.

Moral: Dont always go for the tried and tested option

correct boss

the 9 th and 10th option will give that doughnuts are 200,300,400,500,600

The fifth-place salesman sold 200 dozen doughnuts. -- series start from 200

Lemieux sold 100 dozen more Tim Horton's doughnuts than Peter, who sold 200 dozen more than the defenseman -- the terms differe by 100
and it should have multiples of 100 as its mentioned twice / half somewhere

after that for Eric its either 400 or 600. choose one of them amd solve by hit and trial
amrutesh666 Says
Well solved it, at the expense of a few hrs which goes into the timesheet as self study

now solve this also

One of downtown Summerset's most popular attractions is Artisan Alley, a narrow throughfare of eight shops, each featuring a different local artist or craftsman practicing his or her craft and offering his or her work for sale. The shops are at at 1, 3, 5, and 7 Artisan Alley on one side and at 2, 4, 6, and 8 Artisan Alley on the other side, with the shop at 1 Artisan Alley directly across from the one at 2 Artisan Alley, the shop at 3 Artisan Alley directly opposite the one at 4 Artisan Alley, etc. From the map data in the clues below, can you locate the artist and type of art at each Artisan Alley address?
1. Mary Moses isn't the glassblower.
2. Perry Picasso is between and next door to Wayne Wyeth and the jewelry designer.
3. The glassblower isn't Greta Greco.
4. April Adams is next door to the water color artist.
5. Ronald Rodin isn't the jewelry designer.
6. Vince Vermeer isn't the artist at 7 Artisan Alley.
7. The sculptor isn't Wayne Wyeth.
8. Ronald Rodin is between and adjacent to the goldsmith and the glassblower.
9. The photographer and the sculptor are in adjacent studios.
10. Cary Cassatt is directly opposite the weaver; neither is at 8 Artisan Alley.
11. Perry Picasso is neither the sculptor nor the oil painter.
12. Greta Greco is directly across the street from the photographer, who isn't Ronald Rodin.
Its my 2100