(Puzzles) Challenging Mathematical Puzzles (32)

Puzzles : Challenging Mathematical Puzzles

107. When Socrates was imprisoned for being a disturbing influence, he was held in high esteem by his guards. All four of them hoped that something would occur that would facilitate his escape. One evening, the guard who was on duty intentionally left the cell door open so that Socrates could leave for distant parts.

Socrates did not attempt to escape, as it was his philosophy that if you accept society's rules, you must also accept it's punishments. However, the open door was considered by the authorities to be a serious matter. It was not clear which guard was on that evening. The four guards make the following statements in their defense:

Aaron:
A) I did not leave the door open.
B) Clement was the one who did it.

Bob:
A) I was not the one who was on duty that evening.
B) Aaron was on duty.

Clement:
A) Bob was the one on duty that evening.
B) I hoped Socrates would escape.

David:
A) I did not leave the door open.
B) I was not surprised that Socrates did not attempt to escape.

Considering that, in total, three statements are true, and five statements are false, which guard is guilty

Answer : - David is the guilty.

Note that "All four of them hoped that something would occur that would facilitate his escape". It makes Clement's statement B True and David's statement B False.

Now consider each of them as a guilty, one at a time.

 

Aaron

Bob

Clement

David

True
Stmts

 

A

B

A

B

A

B

A

B

If Aaron is guilty

False

False

True

True

False

True

True

False

4

If Bob is guilty

True

False

False

False

True

True

True

False

4

If Clement is guilty

True

True

True

False

False

True

True

False

5

If David is guilty

True

False

True

False

False

True

False

False

3


Since in total, three statements are true and five statements are false. It is clear from the above table that David is?

 

108. Given any whole number take the sum of the digits, and the product of the digits, and multiply these together to get a new whole number.
For example, starting with 6712, the sum of the digits is (6+7+1+2) = 16, and the product of the digits is (6*7*1*2) = 84. The answer in this case is then 84 x 16 = 1344.
If we do this again starting from 1344, we get (1+3+4+4) * (1*3*4*4) = 576
And yet again (5+7+6) * (5*7*6) = 3780

At this stage we know what the next answer will be (without working it out) because, as one digit is 0, the product of the digits will be 0, and hence the answer will also be 0.
Can you find any numbers to which when we apply the above mentioned rule repeatedly, we never end up at 0? 

 

Given any whole number take the sum of the digits, and the product of the digits, and multiply these together to get a new whole number.

For example, starting with 6712, the sum of the digits is (6+7+1+2) = 16, and the product of the digits is (6*7*1*2) = 84. The answer in this case is then 84 x 16 = 1344.

If we do this again starting from 1344, we get (1+3+4+4) * (1*3*4*4) = 576

And yet again (5+7+6) * (5*7*6) = 3780

At this stage we know what the next answer will be (without working it out) because, as one digit is 0, the product of the digits will be 0, and hence the answer will also be 0.

Can you find any numbers to which when we apply the above mentioned rule repeatedly, we never end up at 0?




109. Brain Teaser No : 00474
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.

 

110. 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

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