My choice is E since in either case we cannot decide if we have a fixed answer.
Please provide the OA.
Thank you
Integers and inequalities
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this_time_i_will
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imo B.
if k = 13!+2, then k = 2(1*3*4*5........*13+1) , and hence K has 2 as a factor.
if k = 13!+3, then k = 3(1*2*4*5....*13+1), and hence K has 3 as a factor.
similarly, till k = 13!+13, we will have atleast one factor for k.
if k = 13!+2, then k = 2(1*3*4*5........*13+1) , and hence K has 2 as a factor.
if k = 13!+3, then k = 3(1*2*4*5....*13+1), and hence K has 3 as a factor.
similarly, till k = 13!+13, we will have atleast one factor for k.
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hi.itz.mani
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IMO B
STATEMENT 1 : k > 4! ==> k > 24 Now we can have K = 29 which is a prime and hence won't have any factor P which is greater than 1
Statement 2: 13! + 2 <= K <= 13! + 13
so with this K can have values from 13! + 2 , 13! + 3 ......... 13! + 13
for each value above we can get a common factor
For example 13! + 2 = 2 ( 1.3.4.5.......13 + 1 )
13! + 3 = 3 ( 1.2.4.5 ........13 + 1 )
Hence Statement 2 is sufficient
STATEMENT 1 : k > 4! ==> k > 24 Now we can have K = 29 which is a prime and hence won't have any factor P which is greater than 1
Statement 2: 13! + 2 <= K <= 13! + 13
so with this K can have values from 13! + 2 , 13! + 3 ......... 13! + 13
for each value above we can get a common factor
For example 13! + 2 = 2 ( 1.3.4.5.......13 + 1 )
13! + 3 = 3 ( 1.2.4.5 ........13 + 1 )
Hence Statement 2 is sufficient
