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Difficulty:
75%
(hard)
Question Stats:
52% (01:59) correct
48%
(02:23)
wrong
based on 162
sessions
History
Date
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A house has \(x\) cakes and \(y\) people, with \(x\geq{2}\) and \(y\geq{1}\). How many values of \(y\) are there, such that all the cakes can be distributed among the people, with each receiving an equal number and none left over?
(1) \(x = a^2\)\(b^3\), where a and b are distinct primes.
(2) \(b = a+1\), where a and b are distinct primes.
(1) \(x = a^2\)\(b^3\), where a and b are distinct primes.
(2) \(b = a+1\), where a and b are distinct primes.
01 May 2020, 22:42
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A house has \(x\) cakes and \(y\) people, with \(x\geq{2}\) and \(y\geq{1}\). How many values of \(y\) are there, such that all the cakes can be distributed among the people, with each receiving an equal number and none left over?
(1) \(x = a^2\)\(b^3\)
(2) \(b = a+1\)
(1) \(x = a^2\)\(b^3\)
(2) \(b = a+1\)
HI Cellchat
Are you sure that you have posted a complete question with correct OA?
Cause the question has not defined what are a and b and in absence of this information, the question can NOT have answer Option A
The question must mention that a and b are prime numbers for x to have unit number of factors (i.e. possible values of y)
Question: How many values of y exist such that y is a divisor of x i.e. y is a factor of x?
After combining the statements
e.g. If x = 2^2*3^3 then the total factors of x i.e. total possible values of y will be (2+1)*(3+1) = 12
BUT
e.g. If x = 4^2*5^3 = 2^4*5^3 then the total factors of x i.e. total possible values of y will be (4+1)*(3+1) = 20
So with the given input the answer of question should be Option E
02 May 2020, 11:40
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A house has \(x\) cakes and \(y\) people, with \(x\geq{2}\) and \(y\geq{1}\). How many values of \(y\) are there, such that all the cakes can be distributed among the people, with each receiving an equal number and none left over?
(1) \(x = a^2\)\(b^3\), where a and b are distinct primes.
(2) \(b = a+1\), where a and b are distinct primes.
(1) \(x = a^2\)\(b^3\), where a and b are distinct primes.
(2) \(b = a+1\), where a and b are distinct primes.
Given: A house has \(x\) cakes and \(y\) people, with \(x\geq{2}\) and \(y\geq{1}\).
Asked: How many values of \(y\) are there, such that all the cakes can be distributed among the people, with each receiving an equal number and none left over?
y must be a factor of x.
(1) \(x = a^2\)\(b^3\), where a and b are distinct primes.
Number of factors of x = 3*4 = 12
Number of values of y such that all the cakes can be distributed among the people, with each receiving an equal number and none left over = 12
SUFFICIENT
(2) \(b = a+1\), where a and b are distinct primes.
Since there is no relation provided between x & y
NOT SUFFICIENT
IMO A
