# (PUZZLES) Collection of Brain Teasers - PART TWO

**(PUZZLES)
Collection of Brain Teasers - PART TWO
**

**Brain Teaser No : 00021
**A rich man died. In his will, he has divided his gold coins among his 5 sons, 5 daughters and a manager.

According to his will: First give one coin to manager. 1/5th of the remaining to the elder son. Now give one coin to the manager and

1/5th of the remaining to second son and so on..... After giving coins to 5th son, divided the remaining coins among five daughters

equally.

All should get full coins. Find the minimum number of coins he has?

**Answer**

We tried to find out some simple mathematical method and finally we wrote small C program to find out the answer. The

answer is 3121 coins.

Here is the breakup:

First son == 624 coins

Second son == 499 coins

Third son == 399 coins

Forth son == 319 coins

Fifth son == 255 coins

Daughters == 204 each

Manager == 5 coins

**Brain Teaser No : 00022**

There were N stations on a railroad. After adding X stations 46 additional tickets have to be printed.

Find N and X.

**Answer**

Let before adding X stations, total number of tickets

t == N(N-1)

After adding X stations total number of tickets are

t + 46 == (N+X)(N+X-1)

Subtracting 1st from 2nd

46 == (N+X)(N+X-1) - N(N-1)

46 == N2 + NX - N + NX + X2 - X - N2 + N

46 == 2NX + X2 - X

46 == (2N - 1)X + X2

X2 + (2N - 1)X - 46 == 0

Now there are only two possible factors of 46. They are (46,1) and (23,2)

Case I: (46,1)

2N - 1 == 45

2N == 46

N == 23

And X == 1

Case II: (23,2)

2N - 1 == 21

2N == 22

N == 11

And X == 2

Hence, there are 2 possible answers.

**Brain Teaser No : 00023**

There is a grid of 20 squares by 10 squares. How many different rectangles are possible?

Note that square is a rectangle.

**Answer**

11550

The Generic solution to this is:

Total number of rectangles == (Summation of row numbers) * (Summation of column numbers)

Here there are 20 rows and 10 columns or vice versa. Hence, total possible rectangles

== ( 20 + 19 + 18 + 17 + 16 + .... + 3 + 2 + 1 ) * ( 10 + 9 +8 + 7 + .... + 3 + 2 + 1)

== ( 210 ) * (55)

== 11550

Hence, total 11,550 different rectangles are possible.

If you don't believe it, try formula on some smaller grids like 4x2, 3x2, 3x3 etc...

**Brain Teaser No : 00024**

A person wanted to withdraw X rupees and Y paise from the bank. But cashier made a mistake and gave him Y rupees and X

paise. Neither the person nor the cashier noticed that.

After spending 20 paise, the person counts the money. And to his surprise, he has double the amount he wanted to withdraw.

Find X and Y. (1 Rupee == 100 Paise)

**Answer**

As given, the person wanted to withdraw 100X + Y paise.

But he got 100Y + X paise.

After spending 20 paise, he has double the amount he wanted to withdraw. Hence, the equation is

2 * (100X + Y) == 100Y + X - 20

200X + 2Y == 100Y +X - 20

199X - 98Y == -20

98Y - 199X == 20

Now, we got one equation; but there are 2 variables. We have to apply little bit of logic over here. We know that if we

interchange X & Y, amount gets double. So Y should be twice of X or one more than twice of X i.e. Y == 2X or Y == 2X+1

Case I : Y==2X

Solving two equations simultaneously

98Y - 199X == 20

Y - 2X == 0

We get X == - 20/3 & Y == - 40/2

Case II : Y==2X+1

Solving two equations simultaneously

98Y - 199X == 20

Y - 2X == 1

We get X == 26 & Y == 53

Now, its obvious that he wanted to withdraw Rs. 26.53

**Brain Teaser No : 00025**

What is the remainder left after dividing 1! + 2! + 3! + ? + 100! By 7?

Think carefully !!!

**Answer**

A tricky one.

7! onwards all terms are divisible by 7 as 7 is one of the factor. So there is no remainder left for those terms i.e. remainder left

after dividing 7! + 8! + 9! + ... + 100! is 0.

The only part to be consider is

== 1! + 2! + 3! + 4! + 5! + 6!

== 1 + 2 + 6 + 24 + 120 + 720

== 873

The remainder left after dividing 873 by 7 is 5

Hence, the remainder is 5.

**Brain Teaser No : 00026**

Find the last digit of summation of the series:

199 + 299 + 399 + 499 + ??? + 9899 + 9999

**Answer**

The last digit of the series is 0.

We group the sum as follow:

(199 + 1199 + ... + 9199) + (299 + 2299 + ... 9299) + ...... + (999 + 1999 + ... + 9999) + (1099 + 2099 + 3099 + ... 9099)

All the terms in a single group have the same last digit (i.e. last digits of 199 + 1199 + ... + 9199 are same, is 1, & similarly for the

other groups).

Also, there are 10 terms in each group except for the last one. Therefore the last digit of the sum of terms in first 9 groups is 0.

(as whatever be the last digit, we have to multiply it by 10) And the last digit of the sum of the terms in the group 10 is obviously

0.

Hence, the last digit of the series is 0.

Brain Teaser No : 00027

Brain Teaser No : 00027

Find the smallest number such that if its rightmost digit is placed at its left end, the new number so formed is precisely 50% larger

than the original number.

**Answer**

The answer is 285714.

If its rightmost digit is placed at its left end, then new number is 428571 which is 50% larger than the original number 285714.

The simplest way is to write a small program. And the other way is trial and error !!!

**Brain Teaser No : 00028**

There are 10 boxes containing 10 balls each. 9 boxes contain 10 balls of 10 kg each and one box contains 10 balls of 9 kg each.

Tool is available for proper weighing. How can you find out the box containing 9 kg balls?

You are allowed to weigh only once. You can remove balls from the boxes. All balls are of same size and color.

**Answer**

1. Mark the boxes with numbers 1, 2, 3, 4, ... upto 10

2. Take 1 ball from box 1, take 2 balls from box 2, take 3 balls from box 3, take 4 balls from box 4 and so on

3. Put all of them on the scale at once and take the measurement.

4. Now, subtract the measurement from 550 ( 1*10 + 2*10 + 3*10 + 4*10 + 5*10 + 6*10 + 7*10 + 8*10 + 9*10 + 10*10)

5. The result will give you the box number which has a ball of 9 Kg

**Brain Teaser No : 00029**

A fly is flying between two trains, each travelling towards each other on the same track at 60 km/h. The fly reaches one engine,

reverses itself immediately, and flies back to the other engine, repeating the process each time.

The fly is flying at 90 km/h. If the fly flies 180 km before the trains meet, how far apart were the trains initially?

**Answer**

Initially, the trains were 240 km apart.

The fly is flying at the speed of 90 km/h and covers 180 km. Hence, the fly flies for 2 hours after trains started.

It's obvious that trains met 2 hours after they started travelling towards each other. Also, trains were travelling at the speed of

60 km/h. So, each train traveled 120 km before they met.

Hence, the trains were 240 km apart initially.

**Brain Teaser No : 00030**

A certain street has 1000 buildings. A sign-maker is contracted to number the houses from 1 to 1000. How many zeroes will he

need?

**Answer**

The sign-maker will need 192 zeroes.

Divide 1000 building numbers into groups of 100 each as follow:

(1..100), (101..200), (201..300), ....... (901..1000)

For the first group, sign-maker will need 11 zeroes.

For group numbers 2 to 9, he will require 20 zeroes each.

And for group number 10, he will require 21 zeroes.

The total numbers of zeroes required are

== 11 + 8*20 + 21

== 11 + 160 + 21

== 192

**Brain Teaser No : 00031**

Find sum of digits of D.

Let

A== 19991999

B == sum of digits of A

C == sum of digits of B

D == sum of digits of C

(HINT : A == B == C == D (mod 9))

**Answer**

The sum of the digits od D is 1.

Let E == sum of digits of D.

It follows from the hint that A == E (mod 9)

Consider,

A == 19991999

< 20002000

== 22000 * 10002000

== 1024200 * 106000

< 10800 * 106000

== 106800

i.e. A < 106800

i.e. B < 6800 * 9 == 61200

i.e. C < 5 * 9 == 45

i.e. D < 2 * 9 == 18

i.e. E <== 9

i.e. E is a single digit number.

Also,

1999 == 1 (mod 9)

so 19991999 == 1 (mod 9)

Therefore we conclude that E==1.

**Brain Teaser No : 00032**

Find the smallest number N which has the following properties:

1. its decimal representation has 6 as the last digit.

2. If the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as great as the

original number N.

**Answer**

The smallest such number is 153846.

Assume that the number N is

N == BnBn-1Bn-2 ... B3B26

as its given that 6 is the last digit.

Now after erasing 6 and putting it in front of the remaining digits, we get

Nnew == 6BnBn-1Bn-2 ... B3B2

Also given that Nnew is 4 times the N. Also note that the last digit Nnew is second last digit of N and so on. The required result is

BnBn-1Bn-2 ... B3B26

X 4

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

6BnBn-1Bn-2 ... B3B2

So start multiplying and put nth digit of Nnew to (n + 1)th digit of N and you will get result as

1 5 3 8 4 6

X 4

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

6 1 5 3 8 4

Hence, the number is 153846

**Brain Teaser No : 00033**

There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd

coin?

**Answer**

It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.

1. Take 8 coins and weigh 4 against 4.

o If both are not equal, goto step 2

o If both are equal, goto step 3

2. One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly,

name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter.

Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.

o If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.

? If both are equal, L4 is the odd coin and is lighter.

? If L2 is light, L2 is the odd coin and is lighter.

? If L3 is light, L3 is the odd coin and is lighter.

o If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2

? If both are equal, there is some error.

? If H1 is heavy, H1 is the odd coin and is heavier.

? If H2 is heavy, H2 is the odd coin and is heavier.

o If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4

? If both are equal, L1 is the odd coin and is lighter.

? If H3 is heavy, H3 is the odd coin and is heavier.

? If H4 is heavy, H4 is the odd coin and is heavier.

3. The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.

o If both are equal, there is some error.

o If X is heavy, X is the odd coin and is heavier.

o If X is light, X is the odd coin and is lighter.

**Brain Teaser No : 00034**

My friend collects antique stamps. She purchased two, but found that she needed to raise money urgently. So she sold them for

Rs. 8000 each. On one she made 20% and on the other she lost 20%.

How much did she gain or lose in the entire transaction?

**Answer**

She lost Rs 666.67

Consider the first stamp. She mades 20% on it after selling it for Rs 8000.

So the original price of first stamp is

== (8000 * 100) / 80

== Rs 6666.67

Similarly, consider second stamp. She lost 20% on it after selling it for Rs 8000

So the original price of second stamp is

== (8000 * 100) / 80

== Rs 10000

Total buying price of two stamps

== Rs 6666.67 + Rs 10000

== Rs 16666.67

Total selling price of two stamps

== Rs 8000 + Rs 8000

== Rs 16000

Hence, she lost Rs 666.67

**Brain Teaser No : 00035**

In a sports contest there were m medals awarded on n successive days (n > 1).

1. On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded.

2. On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.

3. On the nth and last day, the remaining n medals were awarded.

How many days did the contest last, and how many medals were awarded altogether?

**Answer**

Total 36 medals were awarded and the contest was for 6 days.

On day 1: Medals awarded == (1 + 35/7) == 6 : Remaining 30 medals

On day 2: Medals awarded == (2 + 28/7) == 6 : Remaining 24 medals

On day 3: Medals awarded == (3 + 21/7) == 6 : Remaining 18 medals

On day 4: Medals awarded == (4 + 14/7) == 6 : Remaining 12 medals

On day 5: Medals awarded == (5 +7/7) == 6 : Remaining 6 medals

On day 6: Medals awarded 6

I got this answer by writing small program. If anyone know any other simpler method, do submit it.

**Brain Teaser No : 00036**

A number of 9 digits has the following properties:

? The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3,

the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the

first n digits is divisible by n, 2<==n<==9.

? Each digit in the number is different i.e. no digits are repeated.

? The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order.

Find the number.

**Answer**

The answer is 381654729

One way to solve it is Trial-&-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even

positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur.

The other way to solve this problem is by writing a computer program that systematically tries all possibilities.

**Brain Teaser No : 00038**

The population of an island consists of two and only two types of people : the knights, who invariably tell the truth and the knaves

who always lie.

? three of the inhabitants called X, Y and Z were standing together. A newcomer to the island asked, "Are you a knight or a

knave?" X mumbled his answer rather indistinctly, so the stranger could not quite make out what he had said. The

stranger than asked Y, "What did X say?" Y replied, "X said that he was a knave." Whereupon Z said, "Don't believe Y,

he's lying." What are Y and Z?

? Suppose that the stranger asked X, instead, "How many knights among you?" Again X replies indistinctly. So the stranger

asks Y, "What did X say?" Y replies, "X said that there is one knight among us." Then Z says, "Don't believe Y, he is lying!"

Now what are Y and Z?

? There are only two inhabitants, X and Y. X says, "At least one of us is a knave." What are X and Y?

? Suppose X says, "Either I am a knave, or Y is a knight?" What are X and Y?

? Consider once more X, Y and Z each of who is either a knight or a knave. X says, "All of us are knaves." Y says, "Exactly

one of us is a knight." What are X, Y and Z?

**Answer**

Teaser 1 : A Simple one. The statement made by Y is false - "X said that he was a knave".

Case 1 Case 2 Case 3 Case 4

X Knight Knight Knave Knave

Y Knight Knave Knight Knave

Analyse the above 4 cases. In all the cases statement made by Y is contradicory and therefore false. Hence, Y is Knave and Z

is Knight.

Teaser 2 : Again the statement made by Y is false - "X said that there is one knight among us". Analyse these statement with 4

possible cases as above. In all the cases statement made by Y is false. Hence, Y is Knave and Z is Knight.

Teaser 3 : X is Knight and Y is Knave.

Teaser 4 : Both are Knight.

Teaser 5 : X and Z are Knaves, Y is Knight.

**Brain Teaser No : 00039**

Find next number in the series :

3, 7, 31, 211, ?

**Answer**

1831

All the numbers in the series are Prime Numbers. So the next number will also be a prime number.

Two consecutive numbers in the

series

(A and B)

Total Prime numbers bewtween A

and B

(C)

Prime number just before

B

(D)

E == (C +

D)

3 and 7 1 5 6

7 and 31 6 29 35

31 and 211 35 199 234

? Number after 7 is the prime number on skipping 6 prime numbers after 7 i.e. 31

? Number after 31 is the prime number on skipping 35 prime numbers after 31 i.e. 211

? Hence, number after 211 is the prime number on skipping 234 prime numbers after 211 i.e. 1831

The other possible answer is 1891.

Subtract 1 from each number in the given series: 2, 6, 30, 210

First number == 2*1 == 2

Second number == 2*3 == 6

Third number == 6*5 == 30

Fourth number == 30*7 == 210

Fifth number == 210*9 == 1890

Sixth number == 1890*11 == 20790

Thus, the pattern is : multiply previous number by next odd number and add one to the multiplication. Thus, the series is 3, 7,

31, 211, 1891, 20791, ...

Thanks to N. Anand for this much more simpler answer.

**Brain Teaser No : 00040**

Assume for a moment that the earth is a perfectly uniform sphere of radius 6400 km. Suppose a thread equal to the length of the

circumference of the earth was placed along the equator, and drawn to a tight fit.

Now suppose that the length of the thread is increased by 12 cm, and that it is pulled away uniformly in all directions.

By how many cm. will the thread be separated from the earth's surface?

**Answer**

The cicumference of the earth is

== 2 * PI * r

== 2 * PI * 6400 km

== 2 * PI * 6400 * 1000 m

== 2 * PI * 6400 * 1000 * 100 cm

== 1280000000 * PI cm

where r == radius of the earth, PI == 3.141592654

Hence, the length of the thread is == 1280000000 * PI cm

Now length of the thread is increasd by 12 cm. So the new length is == (1280000000 * PI) + 12 cm

This thread will make one concentric circle with the earth which is slightly away from the earth. The circumfernce of that circle is

nothing but (1280000000 * PI) + 12 cm

Assume that radius of the outer circle is R cm

Therefore,

2 * PI * R == (1280000000 * PI) + 12 cm

Solving above equation, R == 640000001.908 cm

Radius of the earth is r == 640000000 cm

Hence, the thread will be separatedfrom the earth by

== R - r cm

== 640000001.908 - 640000000

== 1.908 cm

**Brain Teaser No : 00041**

A polygon has 1325 diagonals. How many vertices does it have? taken from www.johnsjm.blogspot.com

**Answer**

The formula to find number of diagonals (D) given total number of vertices or sides (N) is

N * (N - 3)

D == -----------

2

Using the formula, we get

1325 * 2 == N * (N - 3)

N2 - 3N - 2650 == 0

Solving the quadratic equation, we get N == 53 or -50

It is obvious that answer is 53 as number of vertices can not be negative.

Alternatively, you can derive the formula as triange has 0 diagonals, quadrangel has 2, pentagon has 5, hexagon has 9 and so

on......

Hence the series is 0, 0, 0, 2, 5, 9, 14, ........ (as diagram with 1,2 or 3 vertices will have 0 diagonals).

Using the series one can arrive to the formula given above.

Brain Teaser No : 00042

An emergency vehicle travels 10 miles at a speed of 50 miles per hour.

How fast must the vehicle travel on the return trip if the round-trip travel time is to be 20 minutes?

**Answer**

75 miles per hour

While going to the destination, the vehicle travels 10 mils at the speed of 50 miles per hour. So the time taken to travel 10 miles

is

== (60 * 10) / 50

== 12 minutes

Now it's given that round-trip travel time is 20 minutes. So the vehicle should complete its return trip of 10 miles in 8 minutes.

So the speed of the vehicle must

== (60 * 10) / 8

== 75 miles per hour