Puzzles Page 2

 Find all sets of consecutive integers that add up to 1000.AnswerThere are total 8 such series: 1. Sum of 2000 numbers starting from -999 i.e. summation of numbers from -999 to 1000. (-999) + (-998) + (-997) + ..... + (-1) + 0 + 1 + 2 + ..... + 997 + 998 + 999 + 1000 = 1000 2. Sum of 400 numbers starting from -197 i.e. summation of numbers from -197 to 202. (-197) + (-196) + (-195) + ..... + (-1) + 0 + 1 + 2 + ..... + 199 + 200 + 201 + 202 = 1000 3. Sum of 125 numbers starting from -54 i.e. summation of numbers from -54 to 70. (-54) + (-53) + (-52) + ..... + (-1) + 0 + 1 + 2 + ..... + 68 + 69 + 70 = 1000 4. Sum of 80 numbers starting from -27 i.e. summation of numbers from -27 to 52. (-27) + (-26) + (-25) + ..... + (-1) + 0 + 1 + 2 + ..... + 50 + 51 + 52 = 1000

5. Sum of 25 numbers starting from 28 i.e. summation of numbers from 28 to 52.
28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 = 1000
6. Sum of 16 numbers starting from 55 i.e. summation of numbers from 55 to 70.
55 + 56 + 57 + 58 + 59 +60 + 61 + 62 + 63 + 64 + 65 + 66 + 67 + 68 + 69 + 70 = 1000
7. Sum of 5 numbers starting from 198 i.e. summation of numbers from 198 to 202.
198 + 199 + 200 +201 + 202 = 1000

Find all sets of consecutive integers that add up to 1000.

There are total 8 such series:
1. Sum of 2000 numbers starting from -999 i.e. summation of numbers from -999 to 1000.
(-999) + (-998) + (-997) + ..... + (-1) + 0 + 1 + 2 + ..... + 997 + 998 + 999 + 1000 = 1000
2. Sum of 400 numbers starting from -197 i.e. summation of numbers from -197 to 202.
(-197) + (-196) + (-195) + ..... + (-1) + 0 + 1 + 2 + ..... + 199 + 200 + 201 + 202 = 1000
3. Sum of 125 numbers starting from -54 i.e. summation of numbers from -54 to 70.
(-54) + (-53) + (-52) + ..... + (-1) + 0 + 1 + 2 + ..... + 68 + 69 + 70 = 1000
4. Sum of 80 numbers starting from -27 i.e. summation of numbers from -27 to 52.
(-27) + (-26) + (-25) + ..... + (-1) + 0 + 1 + 2 + ..... + 50 + 51 + 52 = 1000
5. Sum of 25 numbers starting from 28 i.e. summation of numbers from 28 to 52.
28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 = 1000
6. Sum of 16 numbers starting from 55 i.e. summation of numbers from 55 to 70.
55 + 56 + 57 + 58 + 59 +60 + 61 + 62 + 63 + 64 + 65 + 66 + 67 + 68 + 69 + 70 = 1000
7. Sum of 5 numbers starting from 198 i.e. summation of numbers from 198 to 202.
198 + 199 + 200 +201 + 202 = 1000
8. Sum of 1 number starting from 1000.
1000 = 1000

There is a 4-character code, with 2 of them being letters and the other 2 being numbers.

How many maximum attempts would be necessary to find the correct code? Note that the code is case-sensitive.

The maximum number of attempts required are 16,22,400

There are 52 possible letters - a to z and A to Z, and 10 possible numbers - 0 to 9. Now, 4 characters - 2 letters and 2 numbers, can be selected in 52*52*10*10 ways. These 4 characters can be arranged in 4C2 i.e. 6 different ways - the number of unique patterns that can be formed by lining up 4 objects of which 2 are distinguished one way (i.e. they must be letters) and the other 2 are distinguished another way (i.e. they must be numbers).

Consider an example : Let's assume that @ represents letter and # represents number. the 6 possible ways of arranging them are : @@##, @#@#, @##@, #@@#, #@#@, ##@@

Hence, the required answer is
= 52*52*10*10*6
= 16,22,400 attempts
= 1.6 million approx.

Thanks to Tim Sanders for opening BrainVista's brain !!!
How many possible combinations are there in a 3x3x3 rubics cube?

In other words, if you wanted to solve the rubics cube by trying different combinations, how many might it take you (worst case senerio)?

How many for a 4x4x4 cube?
Submitted

There are 4.3252 * 10^19 possible combinations for 3x3x3 Rubics and 7.4012 * 10^45 possible combinations for 4x4x4 Rubics.

Let's consider 3x3x3 Rubics first.

There are 8 corner cubes, which can be arranged in 8! ways.
Each of these 8 cubes can be turned in 3 different directions, so there are 3^8 orientations altogether. But if you get all but one of the corner cube into chosen positions and orientations, only one of 3 orientations of the final corner cube is possible. Thus, total ways corner cubes can be placed = (8!) * (3^8)/8 = (8!) * (3^7)

Similarly, 12 edge cubes can be arranged in 12! ways.
Each of these 12 cubes can be turned in 2 different directions, so there are 2^12 orientations altogether. But if you get all but one of the edge cube into chosen positions and orientations, only one of 2 orientations of the final edge cube is possible. Thus, total ways edge cubes can be placed = (12!) * (2^12)/2 = (12!) * (2^11)

Here, we have essentially pulled the cubes apart and stuck cubes back in place wherever we please. In reality, we can only move cubes around by turning the faces of the cubes. It turns out that you can't turn the faces in such a way as to switch the positions of two cubes while returning all the others to their original positions. Thus if you get all but two cubes in place, there is only one attainable choice for them (not 2!). Hence, we must divide by 2.

Total different possible combinations are
= [(8!) * (3^7)] * [(12!) * (2^11)] / 2
= (8!) * (3^7) * (12!) * (2^10)
= 4.3252 * 10^19

Similarly, for 4x4x4 Rubics total different possible combinations are
= [(8!) * (3^7)] * [(24!)] * [(24!) / (4!^6)] / 24
= 7.4011968 * 10^45

Note that there are 24 edge cubes, which you can not turn in 2 orientations (hence no 2^24 / 2). Also, there are 4 center cubes per face i.e. (24!) / (4!^6). You can switch 2 cubes without affecting the rest of the combination as 4*4*4 has even dimensions (hence no division by 2). But pattern on one side is rotated in 4 directions over 6 faces, hence divide by 24.
Substitute digits for the letters to make the following relation true.
N  E  V  E  R

L  E  A  V  E

+           M  E

-----------------

A  L  O  N  E
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be 3.

A tough one!!!

Since R + E + E = 10 + E, it is clear that R + E = 10 and neither R nor E is equal to 0 or 5. This is the only entry point to

solve it. Now use trial-n-error method.

N  E  V  E  R               2  1  4  1  9

L  E  A  V  E               3  1  5  4  1

+           M  E            +           6  1

-----------------           -----------------

A  L  O  N  E               5  3  0  2  1

One of the four people - Mr. Clinton, his wife Monika, their son Mandy and their daughter Cindy - is a singer and another is a dancer. Mr. Clinton is older than his wife and Mady is older than his sister.
1. If the singer and the dancer are the same sex, then the dancer is older than the singer.
2. If neither the singer nor the dancer is the parent of the other, then the singer is older than the dancer.
3. If the singer is a man, then the singer and the dancer are the same age.
4. If the singer and the dancer are of opposite sex then the man is older than the woman.
5. If the dancer is a woman, then the dancer is older than the singer.
Whose occupation do you know? And what is his/her occupation?

Cindy is the Singer. Mr. Clinton or Monika is the Dancer.

From (1) and (3), the singer and the dancer, both can not be a man. From (3) and (4), if the singer is a man, then the dancer must be a man. Hence, the singer must be a woman.

CASE I : Singer is a woman and Dancer is also a woman
Then, the dancer is Monika and the singer is Cindy.

CASE II : Singer is a woman and Dancer is also a man
Then, the dancer is Mr. Clinton and the singer is Cindy.

In both the cases, we know that Cindy is the Singer. And either Mr. Clinton or Monika is the Dancer.
There are 20 people in your applicant pool, including 5 pairs of identical twins.

If you hire 5 people randomly, what are the chances you will hire at least 1 pair of identical twins? (Needless to say, this could cause trouble ;))
Submitted

The probability to hire 5 people with at least 1 pair of identical twins is 25.28%

5 people from the 20 people can be hired in 20C5 = 15504 ways.

Now, divide 20 people into two groups of 10 people each :
G1 - with all twins
G2 - with all people other than twins

Let's find out all possible ways to hire 5 people without a single pair of indentical twins.
People from G1    People from G2    No of ways to hire G1 without a single pair of indentical twins    No of ways to hire G2    Total ways
0    5    10C0    10C5    252
1    4    10C1    10C4    2100
2    3    10C2 * 8/9    10C3    4800
3    2    10C3 * 8/9 * 6/8    10C2    3600
4    1    10C4 * 8/9 * 6/8 * 4/7    10C1    800
5    0    10C5 * 8/9 * 6/8 * 4/7 * 2/6    10C0    32
Total    11584

Thus, total possible ways to hire 5 people without a single pair of indentical twins = 11584 ways

So, total possible ways to hire 5 people with at least a single pair of indentical twins = 15504 - 11584 = 3920 ways

Hence, the probability to hire 5 people with at least a single pair of indentical twins
= 3920/15504
= 245/969
= 0.2528
= 25.28%
In a hotel, rooms are numbered from 101 to 550. A room is chosen at random. What is the probability that room number starts with 1, 2 or 3 and ends with 4, 5 or 6?

There are total 450 rooms.

Out of which 299 room number starts with either 1, 2 or 3. (as room number 100 is not there) Now out of those 299 rooms only 90 room numbers end with 4, 5 or 6

So the probability is 90/450 i.e. 1/5 or 0.20
Draw 9 dots on a page, in the shape of three rows of three dots to form a square. Now place your pen on the page, draw 4 straight lines and try and cover all the dots.

You're not allowed to lift your pen.

Note: Don't be confined by the dimensions of the square.
Submitted

There are 3 persons X, Y and Z. On some day, X lent tractors to Y and Z as many as they had. After a month Y gave as many tractors to X and Z as many as they have. After a month Z did the same thing. At the end of this transaction each one of them had 24.

Find the tractors each originally had?

One way to solve it is by making 3 equations and solve them simultaneously. But there is rather easier way to solve it using Backtracing.

It's given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X & Y as many as they had. It means that after getting tractors from Z their tractors got doubled. So before Z gave them tractors, they had 12 tractors each and Z had 48 tractors. (12, 12, 48)

Similarly, before Y gave tractors to X & Z, they had 6 & 24 tractors respectively and Y had 42 tractors i.e. (6, 42, 24)

Again, before X gave tractors to Y & Z, they had 21 & 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12)

Hence, initially X had 39 tractors, Y had 21 tractors and Z had 12 tractors.

There is a 50m long army platoon marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position.

In the mean time the whole platoon has moved ahead by 50m.

The question is how much distance did the last person cover in that time. Assuming that he ran the whole distance with uniform speed.
Submitted

The last person covered 120.71 meters.

It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered - while person moving forward and backword - are equal.

Let's assume that when the last person reached the first person, the platoon moved X meters forward.

Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.

Similarly, while moving back the last person moved [50-(50-X)] X meters whereas the platoon moved (50-X) meters.

Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X

Solving, X=35.355 meters

Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters

Note that at first glance, one might think that the total distance covered by the last person is 100 meters, as he ran the total lenght of the platoon (50 meters) twice. TRUE, but that's the relative distance covered by the last person i.e. assuming that the platoon is stationary.
Assume that you have enough coins of 1, 5, 10, 25 and 50 cents.

How many ways are there to make change for a dollar? Do explain your answer.

There are 292 ways to make change for a dollar using coins of 1, 5, 10, 25 and 50 cents.

Let's generalised the teaser and make a table as shown above.

If you wish to make change for 75 cents using only 1, 5, 10 and 25 cent coins, go to the .25 row and the 75 column to obtain 121 ways to do this.

The table can be created from left-to-right and top-to-bottom. Start with the top left i.e. 1 cent row. There is exactly one way to make change for every amount. Then calculate the 5 cents row by adding the number of ways to make change for the amount using 1 cent coins plus the number of ways to make change for 5 cents less using 1 and 5 cent coins.

Let's take an example:
To get change for 50 cents using 1, 5 and 10 cent coins.
* 50 cents change using 1 and 5 cent coins = 11 ways
* (50-10) 40 cents change using 1, 5 and 10 cent coins = 25 ways
* 50 cents change using 1, 5 and 10 cent coins = 11+25 = 36 ways

Let's take another example:
To get change for 75 cents using all coins up to 50 cent i.e. 1, 5, 10, 25 and 50 cents coins.
* 75 cents change using coins upto 25 cent = 121 ways
* (75-50) 25 cents change using coins upto 50 cent = 13 ways
* 75 cents change using coins upto 50 cent = 121+13 = 134 ways

For people who don't want to tease their brain and love to do computer programming, there is a simple way. Write a small multi-loop program to solve the equation: A + 5B + 10C + 25D + 50E = 100
where,
A = 0 to 100
B = 0 to 20
C = 0 to 10
D = 0 to 4
E = 0 to 2

The program should output all the possible values of A, B, C, D and E for which the equation is satisfied.

In a Road Race, one of the three bikers was doing 15km less than the first and 3km more than the third. He also finished the race 12 minutes after the first and 3 minutes before the third.

Can you find out the speed of each biker, the time taken by each biker to finish the race and the length of the course?

Assume that there were no stops in the race and also they were driving with constant speeds through out the

Let us assume that
Speed of First biker = V1 km/min
Speed of Second biker = V2 km/min
Speed of Third biker = V3 km/min
Total time take by first biker = T1 min
Total distance = S km

Now as per the data given in the teaser, at a time T min
X1 = V1 * T         ----> 1

X1 - 15 = V2 * T    ----> 2

X1 - 18 = V3 * T    ----> 3

At a Distance S Km.
S = V1 * T1         ----> 4

S = V2 * (T1 + 12)  ----> 5

S = V3 * (T1 + 15)  ----> 6

Thus there are 6 equations and 7 unknown data that means it has infinite number of solutions.

By solving above 6 equations we get,
Time taken by first biker, T1 = 60 Min.
Time taken by Second biker, T2 = 72 Min.
Time taken by first biker, T3 = 75 Min.

Also, we get
Speed of first biker, V1 = 90/T km/min
Speed of second biker, V2 = (5/6)V1 = 75/T km/min
Speed of third biker, V3 = (4/5)V1 = 72/T km/min

Also, the length of the course, S = 5400/T km

Thus, for the data given, only the time taken by each biker can be found i.e. 60, 72 and 75 minutes. For other quantities, one more independent datum is required i.e. either T or V1 or V2 or V3

Thanks to Theertham Srinivas for the answer !!!
What is the four-digit number in which the first digit is 1/3 of the second, the third is the sum of the first and second, and the last is three times the second?

The 4 digit number is 1349.

It is given that the first digit is 1/3 of the second. There are 3 such possibilities.
1. 1 and 3
2. 2 and 6
3. 3 and 9
Now, the third digit is the sum of the first and second digits.
1. 1 + 3 = 4
2. 2 + 6 = 8
3. 3 + 9 = 12
It is clear that option 3 is not possible. So we are left with only two options. Also, the last digit is three times the second, which rules out the second option. Hence, the answer is 1349.

Difference between Bholu's and Molu's age is 2 years and the difference between Molu's and Kolu's age is 5 years.

What is the maximum possible value of the sum of the difference in their ages, taken two at a time?

The maximum possible value of the sum of the difference in their ages - taken two at a time - is 14 years.

It is given that -
"Difference between Bholu's and Molu's age is 2 years"
"Difference between Molu's and Kolu's age is 5 years"

Now, to get the maximum possible value, the difference between Bholu's and Kolu's age should be maximum i.e. Molu's age should be in between Bholu's and Kolu's age. Then, the difference between Bholu's and Kolu's age is 7 years.

Hence, the maximum possible value of the sum of the difference in their ages - taken two at a time - is (2 + 5 + 7) 14 years.
If it is given that:
25 - 2 = 3
100 x 2 = 20
36 / 3 = 2

What is 144 - 3 = ?
Submitted

There are 3 possible answers to it.

Answer 1 : 9
Simply replace the first number by its square root.
(25) 5 - 2 = 3
(100) 10 x 2 = 20
(36) 6 / 3 = 2
(144) 12 - 3 = 9

Answer 2 : 11
Drop the digit in the tens position from the first number.
(2) 5 - 2 = 3
1 (0) 0 x 2 = 20
(3) 6 / 3 = 2
1 (4) 4 - 3 = 11

You will get the same answer on removing left and right digit alternatively from the first number i.e remove left digit from first (2), right digit from second (0), left digit from third (3) and right digit from forth (4).
(2) 5 - 2 = 3
10 (0) x 2 = 20
(3) 6 / 3 = 2
14 (4) - 3 = 11

Answer 3 : 14
Drop left and right digit alternatively from the actual answer.
25 - 2 = (2) 3 (drop left digit i.e. 2)
100 * 2 = 20 (0) (drop right digit i.e. 0)
36 / 3 = (1) 2 (drop left digit i.e. 1)
144 - 3 = 14 (1) (drop right digit i.e. 1)

A 3 digit number is such that it's unit digit is equal to the product of the other two digits which are prime. Also, the difference between it's reverse and itself is 396.

What is the sum of the three digits?

The required number is 236 and the sum is 11.

It is given that the first two digits of the required number are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third digit is the multiplication of the first two digits. Thus, first two digits must be either 2 or 3 i.e. 22, 23, 32 or 33 which means that there are four possible numbers - 224, 236, 326 and 339.

Now, it is also given that - the difference between it's reverse and itself is 396. Only 236 satisfies this condition. Hence, the sum of the three digits is 11.

There are 4 mugs placed upturned on the table. Each mug have the same number of marbles and a statement about the number of marbles in it. The statements are: Two or Three, One or Four, Three or One, One or Two.

Only one of the statement is correct. How many marbles are there under each mug?

A simple one.

As it is given that only one of the four statement is correct, the correct number can not appear in more than one statement. If it appears in more than one statement, then more than one statement will be correct.

Hence, there are 4 marbles under each mug.
At University of Probability, there are 375 freshmen, 293 sophomores, 187 juniors, & 126 seniors. One student will randomly be chosen to receive an award.

What percent chance is there that it will be a junior? Round to the nearest whole percent

19%

This puzzle is easy. Divide the number of juniors (187) by the total number of students (981), & then multiply the number by 100 to convert to a percentage.

Hence the answer is (187/981)*100 = 19%

If you were to dial any 7 digits on a telephone in random order, what is the probability that you will dial your own phone number?

Assume that your telephone number is 7-digits.

1 in 10,000,000

There are 10 digits i.e. 0-9. First digit can be dialed in 10 ways. Second digit can be dialed in 10 ways. Third digit can be dialed in 10 ways. And so on.....

Thus, 7-digit can be dialed in 10*10*10*10*10*10*10 (=10,000,000) ways. And, you have just one telephone number. Hence, the possibility that you will dial your own number is 1 in 10,000,000.

Note that 0123456 may not be a valid 7-digit telephone number. But while dialing in random order, that is one of the possible 7-digit number which you may dial.
An anthropologist discovers an isolated tribe whose written alphabet contains only six letters (call the letters A, B, C, D, E and F). The tribe has a taboo against using the same letter twice in the same word. It's never done.

If each different sequence of letters constitues a different word in the language, what is the maximum number of six-letter words that the language can employ?
Submitted

The language can employ maximum of 720 six-letter words.

It is a simple permutation problem of arranging 6 letters to get different six-letter words. And it can be done in in 6! ways i.e. 720 ways.

In otherwords, the first letter can be any of the given 6 letters (A through F). Then, whatever the first letter is, the second letter will always be from the remaining 5 letters (as same letter can not be used twice), and the third letter always be from the remaining 4 letters, and so on. Thus, the different possible six-letter words are 6*5*4*3*2*1 = 720

Kate, Demi, Madona, Sharon, Britney and Nicole decided to lunch together in a restaurant. The waiter led them to a round table with six chairs.

How many different ways can they seat?

There are 120 different possible seating arrangments.

Note that on a round table ABCDEF and BCDEFA is the same.

The first person can sit on any one of the seats. Now, for the second person there are 5 options, for the third person there are 4 options, for the forth person there are 3 options, for the fifth person there are 2 options and for the last person there is just one option.

Thus, total different possible seating arrangements are
= 5 * 4 * 3 * 2 * 1
= 120
3 blocks are chosen randomly on a chessboard. What is the probability that they are in the same diagonal?

There are total of 64 blocks on a chessboard. So 3 blocks can be chosen out of 64 in 64C3 ways.
So the sample space is = 41664

There are 2 diagonal on chessboard each one having 8 blocks. Consider one of them.
3 blocks out of 8 blocks in diagonal can be chosen in 8C3 ways.
But there are 2 such diagonals, hence favourables = 2 * 8C3 = 2 * 56 = 112

The require probability is
= 112 / 41664
= 1 / 372
= 0.002688

What is the area of the triangle ABC with A(e,p) B(2e,3p) and C(3e,5p)?

where p = PI (3.141592654)

A tricky ONE.

Given 3 points are colinear. Hence, it is a straight line.

Hence area of triangle is 0.

Silu and Meenu were walking on the road.

Silu said, "I weigh 51 Kgs. How much do you weigh?"

Meenu replied that she wouldn't reveal her weight directly as she is overweight. But she said, "I weigh 29 Kgs plus half of my weight."

How much does Meenu weigh?

Meenu weighs 58 Kgs.

It is given that Meenu weighs 29 Kgs plus half of her own weight. It means that 29 Kgs is the other half. So she weighs 58 Kgs.

Solving mathematically, let's assume that her weight is X Kgs.
X = 29 + X/2
2*X = 58 + X
X = 58 Kgs
Consider the sum: ABC + DEF + GHI = JJJ

If different letters represent different digits, and there are no leading zeros, what does J represent?

The value of J must be 9.

Since there are no leading zeros, J must be 7, 8, or 9. (JJJ = ABC + DEF + GHI = 14? + 25? + 36? = 7??)

Now, the remainder left after dividing any number by 9 is the same as the remainder left after dividing the sum of the digits of that number by 9. Also, note that 0 + 1 + ... + 9 has a remainder of 0 after dividing by 9 and JJJ has a remainder of 0, 3, or 6.

The number 9 is the only number from 7, 8 and 9 that leaves a remainder of 0, 3, or 6 if you remove it from the sum 0 + 1 + ... + 9. Hence, it follows that J must be 9.
A man has Ten Horses and nine stables as shown here.
[] [] [] [] [] [] [] [] []
The man wants to fit Ten Horses into nine stables. How can he fit Ten horses into nine stables?
Submitted