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12 May 2012, 03:49
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A
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Difficulty:
85%
(hard)
Question Stats:
43% (02:02) correct
57%
(02:18)
wrong
based on 516
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At a particular moment, a restaurant has x biscuits and y patron(s), with x>2 and y>1. How many values of y are there, such that all the biscuits can be distributed among the patrons, with each patron receiving an equal number of whole biscuits and with no biscuits left over?
(1) x=a^2*b^3, where a and b are different prime numbers
(2) b=a+1
(1) x=a^2*b^3, where a and b are different prime numbers
(2) b=a+1
12 May 2012, 04:06
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At a particular moment, a restaurant has x biscuits and y patron(s), with x>2 and y>1. How many values of y are there, such that all the biscuits can be distributed among the patrons, with each patron receiving an equal number of whole biscuits and with no biscuits left over?
Basically the questions asks: how many values of y are there, such that x/y=integer, which means that we are asked to find the # of factors of x.
(1) x=a^2*b^3, where a and b are different prime numbers --> the # of factors of x (including 1 and x itself) is (2+1)(3+1)=12, so y can take 12-1=11 values (y can take all the values except 1 since given that y>1)
(2) b=a+1. Not sufficient.
Answer: A.
Basically the questions asks: how many values of y are there, such that x/y=integer, which means that we are asked to find the # of factors of x.
(1) x=a^2*b^3, where a and b are different prime numbers --> the # of factors of x (including 1 and x itself) is (2+1)(3+1)=12, so y can take 12-1=11 values (y can take all the values except 1 since given that y>1)
(2) b=a+1. Not sufficient.
Answer: A.
General Discussion
14 May 2012, 01:46
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Bunuel
Dear Bunuel,
Could you pls elaborate on "the # of factors of x (including 1 and x itself) is (2+1)(3+1)=12, so y can take 12-1=11 values (y can take all the values except 1 since given that y>1)"
How you have zeroed on 2 and 3, of all prime numbers?
