Find Next Number

Puzzle:

Find the next number in the series.

• 1211
• 122111
• 312211
• 12221131
• ?

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Solution:

Answer is 1113213211.

Just reverse numbers 2nd and 4th (alternative) and look at that series.

• 1211
• 111221
  
• 312211
• 13112221
  

So after starting the sequence with 1211, each term in the sequence consists of groups of two numbers based on the previous term - the first being the quantity and the second specifying which digit. For example, first term has "one 1", "one 2" and "two 1s". So the second numbers becomes 111221. Now the 5th number is 1113213111. Since we reverse only even terms, the answer is 1113213211.

Grandfather's Age

Puzzle:

My grandson is about as many days as my son in weeks, and my grandson is as many months as I am in years. My grandson, my son and I together are 120 years. Can you tell me my age in years?

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Solution:

I am 72 years old.

Let m be my age in years. If s is my son's age in years, then my son is 52s weeks old. If g is my grandson's age in years, then my grandson is 365g days old. Thus,

365g = 52s.
Since my grandson is 12g months old,
12g = m.
Since my grandson, my son and I together are 120 years,
g + s + m = 120.
The above system of 3 equations in 3 unknowns (g, s and m) can be solved as follows.
m / 12 + 365 m / (52 x 12) + m = 120 or
52 m + 365 m + 624 m = 624 x 120 or
m = 624 x 120 / 1041 = 72.
So, I am 72 years old.

Hourglasses

Puzzle:

You have two hourglasses: a 7 minute one and an 11 minute one. Using just these hourglasses, how can you accurately measure time 15 minutes?

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Solution:

1. Start both the 7 minute hour glass & 11 minute hour glass.
2. Wait till the 7 minute hour glass times out. Time is 7 minute!

3. Restart the 7 minute hour glass. At this time 11 minute hour glass will have 4 minutes left to time out.

4. As soon as 11 minute glass times out invert the 7 minute hour glass. Total time now is 11 minutes.

5. After inverting 7 minute hour glass, it will now have 4 minutes left for time out.

6. After these 4 minutes times out, the total time is 15 minutes.

Self Referencing Number

Puzzle:

ABCDEFGHI is a nine digit number. A is the count of number of 0's in the number, B is the number of 1's, C is number of 2's and so on. Find the number.

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Solution:

The self referencing nine digit number is 521001000.

The Six Doors

Puzzle:

You are in a room that has six doors. You must pass through all six doors, each door once only, in the correct order. Three are entrances only and three are exits only. Door A must be followed by door B or E; D by A or F; B by C or E; C by D or F; E by B or D; and F by C or D. Also A, F are entrance and B is exit. In what order you must pass through the doors?

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Solution:

The order is CFDABE.

Doors A, F are entrance and B is exit. We must alternate exits and entrances. Let the unknown entrance be X. With a "_" denoting an exit door the possible orders are:
(a) _A_F_X ; (b) _F_A_X ; (c) _X_A_F ; (d) _X_F_A ; (e) _A_X_F ; (f) _F_X_A

Neither (a) nor (c) are possible since A is followed by B or E and neither leads to F. Since A can only be preceded by D, (e) and (f) are possible as DA_X_F and _F_XDA, respectively. With D used, door F can only be preceded by C to give DA_XCF and CF_XDA, respectively. Door B must occupy the remaining space in each possibility (third door is exit) so X=E but C cannot follow E in the former and for the latter, B cannot follow F. This eliminates (e) and (f). Since only D precedes A, (b) is possible as _FDA_X and (d) as _X_FDA. For (d), the exit B must occupy a "_" but cannot be followed by F, and if it is the first door, neither the C or E that must follow as X can be followed by the remaining E or C, respectively. This only leaves (b) and only C (of the possible B,C,E) can precede F as CFDA_X and B must occupy the final "_" as an exit leaving E=X. Notice that the condition that E is followed by B or D cannot be satisfied.

Find the Number

Puzzle:

There is a five digit number ABCDE (A, B, C, D & E are 5 different single digit numbers). If you multiply this number with 4, you get reverse of the number. Find A, B, C, D & E respectively.

ABCDE * 4 = EDCBA


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Solution:

21978 * 4 = 87912

Look at the A in ABCDE. A can be only 1 or 2 because E can be 9 at the most. On the other hand, we look at the A in EDCBA. We found that A must be an even number, which should be 2. Since E x 4 produces a number within the one's place as 2, we know E must be 8. Since there is no carry from B x 4 to E's place, we know B must be 1. Since B = 1, we know D x 4 will produce a number within one's place as 8 so that 8 + 3 (carry) = 11. The only choice for D is 7. C can only be 8 or 9 to make the product of BC x 4 be close to DC. Since E is 8 already, C must be 9.

So ABCDE=21978

25 August 2007

Puzzle:

At a fruit stall in Puzzle land, an Apple costs Rs.20, an Orange costs Rs.24, and a Pear costs Rs.16.

What is the cost of a mango?

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Solution:

Cost of mango is Rs.20.

1. Each letter in the word costs Rs.4 Since the word mango has 5 words, its cost is Rs20.

2. Assign A = 1; B = 2; C = 3.... and so on upto Z = 26. Now find the score of the fruits.
APPLE - 1+16+16+12+5= 50
PEAR - 16+5+1+18 =40
ORANGE - 15+18+1+14+7+5 = 60
Multiply the score with Rs.0.4 to get the cost of the fruit.
So mango - 13+1+14+7+15 = 50, 50*0.4=Rs.20

24 August 2007

Puzzles:

There are five competitor A, B, C, D, and E and they enter a running race that awards gold, silver, and bronze medals. Each of the following compound statements about the race is false, although one of the two clauses in each may be true.

• A didn't win the gold, and B didn't win the silver.
• D didn't win the silver, and E didn't win the bronze.
• C won a medal, and D didn't.
• A won a medal, and C didn't.
• D and E both won medals.

Who won each of the medals?

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Solution:

A won the Gold medal.
D won the Silver medal.
C won the Bronze medal.

23 August 2007

Puzzle:

You have three boxes of fruit. First box contains just Apples, second contain just Oranges and the third box contains a mixture of both. Each box is labeled -- one says "Apples", one says "Oranges", and one says "Apples and Oranges." However, it is known that none of the boxes are labeled correctly. How can you label the boxes correctly if you are only allowed to take and look at just one piece of fruit from just one of the boxes?

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Solution:

Take a piece of fruit from the box marked "apples and oranges." Suppose the fruit you take is an apple. Then that box must be the box containing just apples. Therefore, the box marked "oranges" can't be the box containing just apples, and it can't be the box containing just oranges either -- so it must be the box containing apples and oranges. The remaining box is therefore the box containing just oranges.

If the fruit you take out is an orange, the solution is derived in a similar fashion: the box marked "apples and oranges" is the box containing just oranges; the box marked "apples" is the box containing both apples and oranges; and the box marked "oranges" is the one containing just apples.

22 August 2007

Puzzle:

Find next number in the series.

6 ..25 .. 64 ..81 ..32 ..?


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Solution:

The answer is 1.
Well, you can write every number as a power of another number:
6 equals 6^1,
25 equals 5^2,
64 equals 4^3,
81 equals 3^4,
32 equals 2^5.

So the next number in the series is 1^6 that equals 1.

21 August 2007

Puzzle:

Find the next letter in this sequence.

F S S E N T T T T T T T T T T T T T T T T T T T T ?



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Solution:

Answer is F. The sequence is the first letter of numbers starting at fifteen, as far as thirty-nine. The next letter would be F for forty.

20 August 2007

Puzzle:

You have a 12 liter jug, an 8 liter jug, and a 5 liter jug. The 12 liter jug is full, and the other two are empty. How can you divide the 12 liters of water equally (i.e., so two of the jugs have exactly 6 liters of water in them, and the third is empty)?

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Solution:

Fill the 8 liter jug with the 12 liter jug, leaving 4 liters remaining. Fill the 5 liter jug with the 8 liter jug, leaving 3 liters remaining. Empty the 5 liter jug into the 12 liter jug. Now there are 9 liters in the 12 liter jug and 3 liters in the 8 liter jug. Pour the 3 liters from the 8 liter jug into the 5 liter jug. Now fill the 8 liter jug with water from the 12 liter jug, leaving 1 liter in the 12 liter jug. Fill the 5 liter jug (which already has 3 liters in it) from the 8 liter jug, leaving 6 liters in the 8 liter jug. Empty the 5 liter jug into the 12 liter jug. Now there are 6 liters in the 12 liter jug, 6 liters in the 8 liter jug, and the 5 liter jug is empty.

19 August 2007

Puzzle:

Take six 6s and place any mathematical operator in the equation to make it equal to 100.

6 6 6 6 6 6 = 100

There are multiple ways of doing this. One of them is given below.

(666-66)/6 = 100

Find other ways.

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Solution:

1. ((6 + 6) / 6)^6 + 6 * 6 = 100

2. 6! / (6 * 6) * (6 - (6 / 6)) = 100

3. ((6!-(6!/6))/6)*(6/6) = 100

18 August 2007

Puzzle:

1. Ravi is as old as Ram will be when Ravi is twice as old as Ram was when Ravi's age was half the sum of their present ages.

2. Ram is as old as Ravi was when Ram was half the age he will be 10 years from now.

How old are Ravi and Ram?

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Solution:

Ravi is 40 years old and Ram is 30 years old.


Let Ravi's age be a and let Ram's age be b.
The first sentence tells us that "Ravi is as old as Ram will be when Ravi is twice as old as Ram was when Ravi's age was half the sum of their present ages".
"Half the sum of their present ages" is: 0.5*(a+b)
So "Ravi's age was half the sum of their present ages", the following number of years ago: a - (0.5*(a+b)) = 0.5*a-0.5*b
At that moment, "Ram's age" was: b - (0.5*a-0.5*b) = 1.5*b-0.5*a.
So "twice that age" is: 3*b-a
"Ravi's age will be twice that age", in the following number of years: (3*b-a) - a = 3*b-2*a
At that moment, Ram will be: b + (3*b-2*a) = 4*b-2*a
And we are told that "that's Ravi's age", so: 3*a = 4*b.

The second sentence tells us that "Ram is as old as Ravi was when Ram was half the age he will be 10 years from now.".
"10 years from now", Ram will be: b+10
"Half that age" is: 0.5*b+5
"Ram was that age", the following number of years ago: b - (0.5*b+5) = 0.5*b-5
At that time, "Ravi was", the following number of years old: a - (0.5*b-5) = a-0.5*b+5
And that "is Ram's current age", so: b = a-0.5*b+5 or a = 1.5*b-5.

Now we have two equations:
3*a = 4*b and a = 1.5*b-5.
Solve these equations. So Ravi's age is 40, and Ram's age is 30.

17 August 2007

Puzzle:

Take a look at the clock below. Break the clock into exactly five pieces such that sums of all the numbers on each piece are 8, 10, 12, 14 and 16.

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Solution:

1+2+...+11+12 = 78 and 8+10+12+14+16 = 60. So you probably should try to decrease the total on the clock. How could you decrease the total without throwing away something? Have a good look at the numbers 10, 11 and 12. If you break the clock between the 1 and 2 of twelve, you loose 12 in the total but you gain 1 and 2. So overall the total would decrease by 9. Breaking between the 1 and 1 of 11 and between the 1 and 0 of 10 also decreases the total by 9. So if you break between two of the numbers 10, 11 and 12, the total will become 78-9-9 = 60.

16 August 2007

Puzzle:

You have 5 boxes and each box has 5 marbles in it. Marbles in 4 bags weight 10g each and in one bag weight 9g each. You don't know which bag has 9g marbles and you are given a chance to use a weight scale for one time. How can you find the bag with 9g marbles?

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Solution:

Number the bags from 1 through 5. Take 1 marble from bag 1, 2 marbles from bag 2, 3 marbles from bag 3, 4 marbles from bag 4 and 5 marbles from bag 5. Now put them on the weightscale. Now there are 5 possible measurements. If all marbles would weigh 10g, the total would be 150g. But one or more marbles weigh 9g. So if you took one 9g marble, the total would be 149g. If you took two 9g marbles, the total would be 148g. And so on.

Therefore, 150-(Measured Weight) gives the bag number which contain 9g marbles.
(Ex: If measured weight is 147g, 150-147=3 is the bag number as you have taken three 9g marbles from bag number 3)

15 August 2007

I Wish To All Indians
A
Happy Independence Day

Puzzle:

Crack the code!
A message is hidden inside the code. Can you decode it?

L4A1Q10P1Q6.. P3M2A3Q3P1P8Z6L7Q3M2Z3P8.. L7A1P5

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Solution:

The message is "HAPPY INDEPENDENCE DAY".

The message can be easily decoded using a QWERTY keyboard. Each two characters (except Q10 for P and P10 for Q, which are three characters) represent a letter. The first character of a letter is always Q, A, Z, P, L or M, specifying which row on a QWERTY keyboard the letter is on. The number preceding the letter is the position of the letter on a QWERTY keyboard from left to right (in case of Q, A and Z) or from right to left (in case of P, L and M) along the row. For example, F can be represented as A4 or L6, E can be represented as Q3 or P8.

14 August 2007

Puzzle:

After numbers, today we will try letter series.

Can you find the next letter in the series?

S - T - I - L - N - T - F - Y - ?



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Solution:

Just go through the question once again.
"Can you find the next letter in the series?"
Write down all the first letters of each word and reverse the order.

S - T - I - L - N - T - F - Y - C

That's the series. The answer is C.

13 August 2007

Puzzle:

Find the next number in the series.

3 - 7 - 10 - 11 - 12 - 17 - ?



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Solution:

Stop looking at numbers. Write them in words and try again.

The only vowel they contain, when written in words is the letter E. So next number that has this characteristic is 20 (Twenty).

12 August 2007

Puzzle:

I have classified all the 26 English letters into two groups.

• W R T Y U I S F H J K L Z X C V N M
• Q E O P A D G B
  

On what criteria did I classified these letters?

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Solution:

The letters are classified depending on how many enclosed areas there are within the character. Sounds crazy for the letters R, E and G placed in the wrong group? No. Write them all in small letters and check once again.

• w r t y u i s f h j k l z x c v n m
• q e o p a d g b

The first group have zero enclosed space where as second group have one or more than one enclosed spaces.

11 August 2007

Puzzle:

At a family get-together function, the following people are present: one grandfather, one grandmother, four sons, three daughters, three fathers, three mothers, seven children, five grandchildren, five brothers, four sisters, one son-in-law and one daughter-in-law. But not these many people has attended as it sounds. Find how many people has attended the function.

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Solution:

Eleven family members attended the function. We have a grandfather, grandmother, their two children who were brother and sister and they got married to another sister and brother and three sons and two daughters to them. Thus totaling of eleven family members.
(Let A, B be grandfather and grandmother. They have two children C and D, C is son and D is daughter. C is married to E and D is married to F and E, F are sister and brother. C and E have three sons and D and F have two daughters)

10 August 2007

Puzzle:

An explorer wishes to cross a barren desert that requires 6 days to cross, but one man can only carry enough food for 4 days. What is the fewest number of other men required to help carry enough food for him to cross?

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Solution:

The answer is clearly that only two other men are required. The first helper only goes one day into the desert. He feeds the other two men during the first day, so that at the beginning of the second day, he only has one day rations left. So he goes back to camp. On the second day, the second helper feeds himself and the explorer. On the beginning of the third day the helper now has two days rations left so he heads back. The explorer is two days into the journey and still has all four days of his food left, so he continues on alone.

09 August 2007

Puzzle:

A man needs to pay his rent and he was out of money. He found that his rent was worth about one gold link on his chain per day. What is the fewest number of cuts he can make in his 23-link chain to pay the rent for up to 23 days?

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Solution:

Considering that the chain is not closed, it requires only two links to be cut. Cut link number 4 and link number 11 counting from the same beginning link. He then has 2 pieces of length 1 (the cut links), and one of 3, 6, and 12. He can then pay the rent as follows. One each of the first two days he can give a cut link. On the third day he gives the chain of 3 and gets his two cut links back. He uses them on days 4 and 5, and then trades all given so far and gives the 6-link chain on day 6. He then again repeats the first steps for days 7-11. On day 12 he gets all those links back and gives the 12-link chain. The then repeats the actions of the first 11 days to go all the way though day 23. For those knowing numbering systems, it will be noticed that this is basically a trinary numbering scheme.

08 August 2007

Puzzle:

You have six match sticks as shown in figure.
Can you rearrange them to make nothing!!!

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Solution:

Rearrange the matchsticks to form the word "NIL" as shown below...

07 August 2007

Puzzle:

Ravi is a strange liar. He lies on six days of the week, but on the seventh day he always tells the truth. He made the following statements on three successive days:

Day 1: "I lie on Monday and Tuesday."
Day 2: "Today, it's Thursday, Saturday, or Sunday."
Day 3: "I lie on Wednesday and Friday."

On which day does Ravi tell the truth?

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Solution:

We know that Ravi tells the truth on only a single day of the week. If the statement on day 1 is untrue, this means that he tells the truth on Monday or Tuesday. If the statement on day 3 is untrue, this means that he tells the truth on Wednesday or Friday. Since Ravi tells the truth on only one day, these statements cannot both be untrue. So, exactly one of these statements must be true, and the statement on day 2 must be untrue.

Assume that the statement on day 1 is true. Then the statement on day 3 must be untrue, from which follows that Ravi tells the truth on Wednesday or Friday. So, day 1 is a Wednesday or a Friday. Therefore, day 2 is a Thursday or a Saturday. However, this would imply that the statement on day 2 is true, which is impossible. From this we can conclude that the statement on day 1 must be untrue.

This means that Ravi told the truth on day 3 and that this day is a Monday or a Tuesday. So day 2 is a Sunday or a Monday. Because the statement on day 2 must be untrue, we can conclude that day 2 is a Monday.

So day 3 is a Tuesday. Therefore, the day on which Ravi tells the truth is Tuesday.

06 August 2007

Puzzle:

Assume that you have a number of long fuses, of which you only know that they burn for exactly one hour after you lighted them at one end. However, you don't know whether they burn with constant speed, so the first half of the fuse can be burnt in only ten minutes while the rest takes the other fifty minutes to burn completely. Also assume that you have a lighter.

How can you measure exactly three quarters of an hour with these fuses?

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Solution:

With only two fuses that burn exactly one hour, one can measure three quarters of an hour accurately, by lighting the first fuse at both ends and the other fuse at one end simultaneously. When the first fuse is burnt out after exactly half an hour you know that the second fuse still has exactly half an hour to go before it will be burnt completely, but we won't wait for that. We will now also light the other end of the second fuse. This means that the second fuse will now be burnt completely after another quarter of an hour, which adds up to exactly three quarters of an hour since we started lighting the first fuse!

05 August 2007

Puzzle:

The investigator asked Jaclyn about her children. Jaclyn said "I have 3 daughters, Alice, Betty, and Cindy. The product of their ages is 36. The sum of their ages is the same as the street number of our next door neighbor."

The investigator went next door and came back and said: "Still not enough information". Jaclyn said: "Oh, I forgot to tell you that my oldest daughter is now in school".

The investigator found out the ages of her daughters immediately. Do you know their ages?

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Solution:

The possiblities are that makes the product 36 are

a) 1, 4 ,9, Sum=14
b) 1, 2, 18, Sum=21
c) 1, 1, 36,Sum=38
d) 1, 6, 6,Sum=13
e) 2, 2, 9,Sum=13
f) 2, 3, 6, Sum=11
g) 3, 3, 4, Sum=10

The investigator could not get the answer even after checking the street number of next door neighbor. This gives us the clue that there are more than one case where the sum of their ages is the street number of next door neighbor. From above possibilities, only sum 13 has two cases. The case 2, 2, 9 will be more likely the case because the oldest daughter(age 9) goes to school.

04 August 2007

Puzzle:

Use five 5 s to make 37.

Can you make a target number 37 by using five 5s? You can use any math operator as you want. There are at least two different ways.

5 5 5 5 5 = 37

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Solution:

Since the total of five 5s is 25, we have to choose an operator which can make a large number. Obviously, the multiplication operator is not good enough. We chose the factory operator!

((5!+5!)-55)/5 = (120+120-55)/5 = 185/5 = 37
5!/(5+5)+5*5 = 120/10 + 25 = 12 + 25 = 37
((5+5)/5)^5 + 5 = 37

03 August 2007

Puzzle:

Look at the figure below, which shows fifteen men. The figure is subdivided into three areas (upper left, upper right, and the bottom half).

By exchanging the upper two parts of the figure, one gets the figure below. This new figure however only shows fourteen men! (If you don't belief what happened here: please print it, cut it, and try it out yourself!)

Now the question is where did the missing man go?

Similar figures:

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Solution:

The trick of this puzzle is that every man has grown a little bit in the second figure. So you can say that the "missing" man is divided among the other men who became taller... But to really grasp it, take your time and study the figures, because it does look like magic!

Another example of the same kind of magic is depicted below:


For better understanding, refer to defective yeti

02 August 2007

Puzzle:

Look at the figure. How can exactly four sticks be moved to make three equal squares?

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Solution:

01 August 2007

Puzzle:

Which would you say is heavier, a pound of cotton or a pound of gold?

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Solution:

A pound of cotton is heavier than a pound of gold because cotton is weighted by the avoirdupois pound, which consists of 16 ounces, whereas gold, being a precious metal is weighted by the troy pound which consists 12 ounces.

Resources:
Wikipedia
sizes.com