CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

Aarav

This is an easy question but interesting.

Two budding MBAs who are also mathematicians, Srikar and arbit_rageur, play a game. The computer selects some secret positive integer N < 60 (both Srikar and arbit_rageur know that , but that they don't know what the value of N is). The computer tells Srikar the unit digit of N, and it tells arbit_rageur the number of divisors of N. Then, Srikar and arbit_rageur have the following dialogue:

Srikar: I don't know what N is, and I'm sure that you don't know either. However, I know that N is divisible by at least two different primes.

arbit_rageur: Oh, then I know what the value of N is.

Srikar: Now I also know what N is.

Assuming that both Srikar and arbit_rageur speak truthfully and to the best of their knowledge, how many possible values of N are there?

(a)0 (b) 1 (c) 2 (d) 3 (e) none of these

srikar2097

removed double posting of question

sabsebadapaagal

Quote:

Originally Posted by Aarav

This is an easy question but interesting.

Two budding MBAs who are also mathematicians, Srikar and arbit_rageur, play a game. The computer selects some secret positive integer N < 60 (both Srikar and arbit_rageur know that , but that they don't know what the value of N is). The computer tells Srikar the unit digit of N, and it tells arbit_rageur the number of divisors of N. Then, Srikar and arbit_rageur have the following dialogue:

Srikar: I don't know what N is, and I'm sure that you don't know either. However, I know that N is divisible by at least two different primes.

arbit_rageur: Oh, then I know what the value of N is.

Srikar: Now I also know what N is.

Assuming that both Srikar and arbit_rageur speak truthfully and to the best of their knowledge, how many possible values of N are there?

(a)0 (b) 1 (c) 2 (d) 3 (e) none of these

I tried analysing the cases for various last digits but when we consider the last digit as 2 and 3, we get two unique solution . The two nos being 22 and 33

Thus, the answer would be option c

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Please let me know if I am wrong...

selebratinglife

Quote:

Originally Posted by Aarav

This is an easy question but interesting.

Two budding MBAs who are also mathematicians, Srikar and arbit_rageur, play a game. The computer selects some secret positive integer N < 60 (both Srikar and arbit_rageur know that , but that they don't know what the value of N is). The computer tells Srikar the unit digit of N, and it tells arbit_rageur the number of divisors of N. Then, Srikar and arbit_rageur have the following dialogue:

Srikar: I don't know what N is, and I'm sure that you don't know either. However, I know that N is divisible by at least two different primes.

arbit_rageur: Oh, then I know what the value of N is.

Srikar: Now I also know what N is.

Assuming that both Srikar and arbit_rageur speak truthfully and to the best of their knowledge, how many possible values of N are there?

(a)0 (b) 1 (c) 2 (d) 3 (e) none of these

I'll go with option (c)..
If we take the unit digit to be either 7 or 3 then we have a unique value..
For 17,27,37,47,57 only 57 satisfies the condition. Others have only one prime factor..
For 13,23,33,43,53 only 33 satisfies the condition ..Other numbers have a lone prime factor..

srikar2097

Since srikar is able to tell that there are atleast 2 primes divisors of N by just knowing the units digit, then N must end in either 6 or 0.
So we have the N in the for or X6 or X0.
Also arbit_rageur knows the no. of divisors of N. and this info is enough to find the number.

Checking for various cases N=X6-
N=6, then divisors = 4
N=16, not possible
N=26 then Divisors = 4
N=36, divisors=9
N=46, divisors=4
N=56, divisors=8

For N=X0 -
N=10, divisors=4
N=20, divisors=6
N=30, divisors=8
N=40, divisors=8
N=50, divisors=6

Out of all these since N=36 has unique divisors amongst them both of them are able to figure out the value of N. So although the values of N are multiple in only one case will it yield unique result. So N=36 and Choice (B)

srikar2097

Quote:

Originally Posted by selebratinglife

I'll go with option (c)..
If we take the unit digit to be either 7 or 3 then we have a unique value..
For 17,27,37,47,57 only 57 satisfies the condition. Others have only one prime factor..
For 13,23,33,43,53 only 33 satisfies the condition ..Other numbers have a lone prime factor..

Hello selebratinglife, I didn't get your approach.
Just by knowing the units digit as 7 or 3 (since they are prime) is there a way to tell for sure that the number is divisible by atleast 2 primes?

Also taking from your examples 17 is a prime and is divisible by one 1 prime i.e. 17

and 27 too. which is divisible by only 3.

This does not satisfy the condition that "N is divisible by at least two different primes."

Please elaborate your approach...

selebratinglife

Quote:

Originally Posted by srikar2097

Hello selebratinglife, I didn't get your approach.
Just by knowing the units digit as 7 or 3 (since they are prime) is there a way to tell for sure that the number is divisible by atleast 2 primes?

Also taking from your examples 17 is a prime and is divisible by one 1 prime i.e. 17

and 27 too. which is divisible by only 3.

This does not satisfy the condition that "N is divisible by at least two different primes."

Please elaborate your approach...

That was a wrong approach..I'm still trying to figure out the right method..I just got a bit confused with who knows what and who spoke what..Just ignore my above post..

neynetr

Quote:

Originally Posted by srikar2097

Since srikar is able to tell that there are atleast 2 primes divisors of N by just knowing the units digit, then N must end in either 6 or 0.
So we have the N in the for or X6 or X0.
Also arbit_rageur knows the no. of divisors of N. and this info is enough to find the number.

Checking for various cases N=X6-
N=6, then divisors = 4
N=16, not possible
N=26 then Divisors = 4
N=36, divisors=9
N=46, divisors=4
N=56, divisors=8

For N=X0 -
N=10, divisors=4
N=20, divisors=6
N=30, divisors=8
N=40, divisors=8
N=50, divisors=6

Out of all these since N=36 has unique divisors amongst them both of them are able to figure out the value of N. So although the values of N are multiple in only one case will it yield unique result. So N=36 and Choice (B)

As 16 is not possible so i guess we can rule out 6 as units digit because just by looking at the last digit srikar is able to say that it has at least two prime factors...

so we'r left only with 10,20,30,40 and 50... out if which only 10 has uniqe no of factors.... so answer should be only one value of N... (B).... (

still the same)

However i think therz a catch in the question...

Thru srikar, arbit comes to know that the number has two or more prime factors.. He still doesnt know of the 10 to 50 thing as srikar has not told him that he has come to know of two prime factors by looking at the last digit of the number (this information is limited to srikar only..... right..???)
so now arbit has a list of all the numbers <60 to choose from for which the no of prime factors is two or more than two....

Is my observation correct....???

sabsebadapaagal

Quote:

Originally Posted by srikar2097

Since srikar is able to tell that there are atleast 2 primes divisors of N by just knowing the units digit, then N must end in either 6 or 0.
(B)

didnt get your approach .... lets consider the case when N= 33 and N=22 the both above no have factors which are prime nos.... more over when srikar knows that last digit is 3 the only possible soln it 33... similarly in case of 22..

pls let me know if I have missed something or made some mistake ....

srikar2097

Quote:

Originally Posted by neynetr

As 16 is not possible so i guess we can rule out 6 as units digit because just by looking at the last digit srikar is able to say that it has at least two prime factors...

so we'r left only with 10,20,30,40 and 50... out if which only 10 has uniqe no of factors.... so answer should be only one value of N... (B).... (

still the same)

However i think therz a catch in the question...

Thru srikar, arbit comes to know that the number has two or more prime factors.. He still doesnt know of the 10 to 50 thing as srikar has not told him that he has come to know of two prime factors by looking at the last digit of the number (this information is limited to srikar only..... right..???)
so now arbit has a list of all the numbers <60 to choose from for which the no of prime factors is two or more than two....

Is my observation correct....???

How are you ruling out 6, just because 16 is not satisfying. There are other possible nos. with 6 as units digit for N<60.