What are the next three numbers in the given series?
1 2 3 2 1 2 3 4 2 1 2 3 4 3 2 ? ? ?
Answer
The next three numbers are 3, 4 and 5.
The pattern is - the number of letters in the Roman numeral representation of the numbers i.e. number of letters in I, II, III, IV, V, VI, VII, VIII, IX, X, XI, XII, XIII, XIV, XV, .....
Hence, the next numbers in the given series are 3(XVI), 4(XVII), 5(XVIII), 3(XIX), 2(XX), 3(XXI), 4(XXII), 5(XXIII), 4(XXIV), 3(XXV), etc.
There are 4 mathematicians - Brahma, Sachin, Prashant and Nakul - having lunch in a hotel. Suddenly, Brahma thinks of 2 integer numbers greater than 1 and says, "The sum of the numbers is..." and he whispers the sum to Sachin. Then he says, "The product of the numbers is..." and he whispers the product to Prashant. After that following conversation takes place :
Sachin : Prashant, I don't think that we know the numbers. Prashant : Aha!, now I know the numbers. Sachin : Oh, now I also know the numbers. Nakul : Now, I also know the numbers. What are the numbers? Explain your answer.
Answer
The numbers are 4 and 13.
As Sachin is initially confident that they (i.e. he and Prashant) don't know the numbers, we can conclude that -
1) The sum must not be expressible as sum of two primes, otherwise Sachin could not have been sure in advance that Prashant did not know the numbers.
2) The product cannot be less than 12, otherwise there would only be one choice and Prashant would have figured that out also.
If the sum of two numbers is 11, Sachin will think that the numbers would be (2,9), (3,8), (4,7) or (5,6).
Sachin : "As 11 is not expressible as sum of two primes, Prashant can't know the numbers."
Here, the product would be 18(2*9), 24(3*8 ), 28(4*7) or 30(5*6). In all the cases except for product 30, Prashant would know the numbers.
- if product of two numbers is 18:
Prashant : "Since the product is 18, the sum could be either 11(2,9) or 9(3,6). But if the sum was 9, Sachin would have deduced that I might know the numbers as (2,7) is the possible prime numbers pair. Hence, the numbers must be 2 and 9." (OR in otherwords, 9 is not in the Possible Sum List)
- if product of two numbers is 24:
Prashant : "Since the product is 24, the sum could be either 14(2,12), 11(3,8 ) or 10(4,6). But 14 and 10 are not in the Possible Sum List. Hence, the numbers must be 3 and 8."
- if product of two numbers is 28:
Prashant : "Since the product is 28, the sum could be either 16(2,14) or 11(4,7). But 16 is not in the Possible Sum List. Hence, the numbers must be 4 and 7."
- if product of two numbers is 30:
Prashant : "Since the product is 30, the sum could be either 17(2,15), 13(3,10) or 11(5,6). But 13 is not in the Possible Sum List. Hence, the numbers must be either (2,15) or (5,6)." Here, Prashant won't be sure of the numbers.
Hence, Prashant will be sure of the numbers if product is either 18, 24 or 28.
Sachin : "Since Prashant knows the numbers, they must be either (3,8), (4,7) or (5,6)." But he won't be sure. Hence, the sum is not 11.
Here, Prashant will be sure of the numbers if the product is 52.
Sachin : "Since Prashant knows the numbers, they must be (4,13)."
For all other numbers in the Possible Sum List, Prashant might be sure of the numbers but Sachin won't. Here is the step by step explaination:
Sachin : "As the sum is 17, two numbers can be either (2,15), (3,14), (4,13), (5,12), (6,11), (7,10) or (8,9). Also, as none of them is a prime numbers pair, Prashant won't be knowing numbers either."
Prashant : "Since Sachin is sure that both of us don't know the numbers, the sum must be one of the Possible Sum List. Further, as the product is 52, two numbers can be either (2,26) or (4,13). But if they were (2,26), Sachin would not have been sure in advance that I don't know the numbers as 28 (2+26) is not in the Possible Sum List. Hence, two numbers are 4 and 13."
Sachin : "As Prashant now knows both the numbers, out of all possible products - 30(2,15), 42(3,14), 52(4,13), 60(5,12), 66(6,11), 70(7,10), 72(8,9) - there is one product for which list of all possible sum contains ONLY ONE sum from the Possible Sum List. And also, no such two lists exist. [see table above for 17] Hence, two numbers are 4 and 13."
Nakul figured out both the numbers just as we did by observing the conversation between Sachin and Prashant.
It is interesting to note that there are no other such two numbers. We checked all the possible sums till 500 !!!
A stick of length 1 is divided randomly into 3 parts.
What is the probability that a triangle can be made with those three parts?
Answer
The probability, that a triangle can be made by randomly dividing a stick of length 1 into 3 parts, is 25%
A triangle can be made, if and only if, sum of two sides is greater than the third side. Thus,
X1 < X2 + X3 X2 < X3 + X1 X3 < X1 + X2
Also, it is given that X1 + X2 + X3 = 1 From above equations: X1 < 1/2, X2 < 1/2, X3 < 1/2 Thus, a triangle can be formed, if all three sides are less than 1/2 and sum is 1.
Now, let's find the probability that one of X1, X2, X3 is greater than or equal to 1/2. Note that to divide stick randomly into 3 parts, we need to choose two numbers P and Q, both are between 0 & 1 and P
Now, X1 will be greater than or equal to 1/2, if and only if both the numbers, P & Q, are greater than or equal to 1/2. Thus, probability of X1 being greater than or equal to 1/2 is = (1/2) * (1/2) = 1/4
Similarly, X3 will be greater than or equal to 1/2, if and only if both the numbers, P & Q, are less than or equal to 1/2. Thus, probability of X3 being greater than or equal to 1/2 is = (1/2) * (1/2) = 1/4
Also, probability of X2 being greater than or equal to 1/2 is = (1/2) * (1/2) = 1/4 The probability that a triangle can not be made = (1/4) + (1/4) + (1/4) = (3/4) Thus, the probability that a triangle can be made
= 1 - (3/4) = (1/4) = 25 %
Thus, the probability that a triangle can be made by randomly dividing a stick of length 1 into 3 parts is 25%
Let's generalise the problem. What is the probability that a polygon with (N+1) sides can be made from (N+1) segments obtained by randomly dividing a stick of length l into (N+1) parts?
The probability is = 1 - (N+1)*(1/2)^N The probability tends to 1 as N grows. Thus, it is easier to make a N-sided polygon than it is to make a triangle!!!
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