subhashghosh wrote:
Could someone please explain in Question 10/20 how 211 is the second smallest prime factor of 11 * 211, what is the greatest prime factor then ? Not sure what am I missing.
Q:
If n is a positive integer greater than 1, then p(n) represents the product of all the prime numbers less than or equal to n. The second smallest prime factor of p(12)+11 is:
A. 2
B. 11
C. 13
D. 17
E. 211
Sol:
p(12) = 2*3*5*7*11 [Product of all prime numbers less than or equal to 12]
p(12)+11 = 2*3*5*7*11+11 = 11(2*3*5*7+1) = 11(210+1)=11*211
Now, we need to find out the second smallest prime number in "11*211"
We know, 11 is a prime number. We keep it aside and check how many prime numbers does 211 have. But, before that let's check whether 211 is a prime number in itself.
211 does not have factors 2*3*5*7 because two consecutive numbers are always co-prime. 210 has all these factors 2*3*5*7 and thus 211 won't have any of these factors.
211 is a number between 225(perfect square of 15) and 196(perfect square of 14). Thus, if 211 doesn't get divided by any prime number below 14 and more than 1, then 211 will be a prime number.
We already know that 211 won't divide by 2, 3, 5 or 7. Thus, we need to check the divisibility of 211 by a prime number between 7 and 14.
11,13 fit in this range.
211 doesn't divide by 11 or 13 and thus is a prime itself.
Finally, in the expression "11*211"; 11 become the smallest prime and 211 second smallest prime number.
Ans: "E"
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~fluke
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