The value of a* is the product of all positive, odd integers less the

Question

The value of  a^*  is the product of all positive, odd integers less then  a  but greater than 0. If  e = 16  and  o = 14  what is the largest prime factor of  e^* + 2o^* ?

(A) 23
(B) 19
(C) 17
(D) 13
(E) 11

stne · Answer

a^* =1 *3*5*7*9*11*13*15 ... (i)

(2o)^* = (2*14 )^* = 28^*

Hence   (2o)^*=28^* = (e^*) * 17 *19 *21*23*25*27 ... (ii)

Required highest prime from :  (a^*) + (2o)^*

=(a^*) + (a^*) * 17 *19 *21*23*25*27  ... Substituting the value of  (2o)^*  from (ii)

=(a^*)*(1 + 17 *19 *21*23*25*27) ... Taking common factor  (a^*)   out.

=(1*3*5*7*9*11*13*15)*(1 + 17 *19 *21*23*25*27 )  ...Substituing the value of  (a^*)  from (i)

Notice the expression:  ( 1 + 17 *19 *21*23*25*27 ) =   (odd + odd ) an even integer greater than  2

Hence this expr :  ( 1 + 17 *19 *21*23*25*27 )   cannot be a prime

Edit : We can still factor out a prime from the exp. Turns out I misinterpreted the question

Please refer to ensuing post by Ian sir for the correct soluton. Thanks to Ian sir.

Our highest prime has to be from the expression   (1 *3*5*7*9*11*13*15 )

Highest prime  = 13

Ans D

Hope it's clear.

IanStewart · Answer

From the notation, 16* is equal to the product of the odd numbers from 1 through 15, and 14* is equal to the product of the odd numbers from 1 through 13. Notice that 16* = (15)(13)(11)...(3)(1) = (15)(14*). So 16* + 2(14*) = 15(14*) + 2(14*) = 17(14*), and since 14* isn't divisible by any prime greater than 13, the answer is 17.

stne -- I think in your solution, you have solved as if the question asked about  e^* + (2o)^* , but here the star only applies to what its immediately next to, just as a factorial symbol would; 2n! is equal to 2 times n!, and not to (2n) factorial.

IanStewart · Answer

And be careful here: you're right that this huge even number is not prime. But it might still have a prime divisor larger than 13, which is what we care about here. And it turns out it has a huge prime factor, 907,811 (along with 29) though I had to use a computer program to figure that out -- there'd be no practical way to do that by hand.

stne · Answer

Right Sir, we can still factor out this huge even expression and still get a prime , completely forgot that an even number greater than  2  , can still have prime numbers as divisors. Thank you for letting me know.

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem ↧↧↧

https://www.youtube.com/watch?v=tX7rn_J7bQ4

e^* + 2o^*

e^*  =  16^*  is the product of all positive, odd integers less then  16  but greater than 0
 =>  e^*  = 1 * 3 * ... * 13 * 15

Similarly,  o^*  = 1 * 3 * ... * 13

=>   e^* + 2o^*   = 1 * 3 * ... * 13 * 15 + 2 * 1 * 3 * ... * 13 = 1 * 3 * ... * 13 * (15 + 2) = 1 * 3 * ... * 13 * 17
And 17 is a prime number

=> Largest Prime factor of  e^* + 2o^*  = 17

So,  Answer will be C 
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

https://www.youtube.com/watch?v=b7JoMX0gk8M

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