(3 Δ 47) + 2 = 3*5*7*...*47+2 = odd + 2 = odd.

Now, 3*5*7*...*47 and 3*5*7*...*47 +2 are consecutive odd numbers. Consecutive odd numbers are co-prime, which means that they do not share any common factor but 1. For example, 25 and 27 are consecutive odd numbers and they do not share any common factor but 1.

Naturally every odd prime between 3 and 47, inclusive is a factor of 3*5*7*...*47, thus none of them is a factor of 3*5*7*...*47 +2. Since 3*5*7*...*47+2 = odd, then 2 is also not a factor of it, which means that the smallest prime factor of 3*5*7*...*47 +2 is greater than 50.

Answer: A.

Similar question from GMAT Prep: for-every-positive-even-integer-n-the-function-h-n-is-126691.html

Hope it helps.
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Since each prime number from 3 upto 47 is a factor of (3 Δ 47) , none of them can be a factor of (3 Δ 47) + 2 . Also 48, 49 and 50 are not prime factors. And y cannot be 2 because (3 Δ 47) +2 is odd. Therefore y>50.
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Hi Bunuel,

Thanks for the solution above. Is their a significance of the term " Every Odd prime between 3 & 47"..It can very well be every prime between 3 and 47.
Please confirm.

Thanks
Mridul
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Yes, that's correct.
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Hi Bunuel,

I just wanted to undertsand in what case 2 can be a smallest prime factor. For Eg if the Q. said that the smallest prime in (3 Δ 47) + 1.Then, the no (3 Δ 47) + 1 will be odd+1=even. Can we say 2 will be the smallest prime in this case.

Also, 2 consecutive integers will also be co-prime and therefore none of the factors in (3 Δ 47) will be factors of (3 Δ 47) + 1.

Thanks for your reply to my queries earlier.

Mridul
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That is correct. The smallest prime of (3 Δ 47) + 1 is naturally 2, since (3 Δ 47) + 1 = even, and the smallest prime of any positive even integer is 2 (notice that 2 is the smallest prime).

Similar question to practice: for-every-positive-even-integer-n-the-function-h-n-is-126691.html

Hope it helps.
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(3*5*7*...*47)/y + 2/y = Integer

2/y is a fraction hence (3*5*7*...*47)/y also has to be a fraction. Only possibility is y>50

HEY,
(3 Δ 47) AND (3 Δ 47) +1 should be consecutive numbers right? so they dont share the any common factors other than 1 .Hence the smallest prime factor should be more than 47 and hence E.
Please can some one explain whether my understanding is right or wrong ? if wrong please say where i made the mistake?

Since (3 Δ 47) is odd and so (3 Δ 47) +1 is even. All even numbers have smallest factor as 2.
Hence E is the answer.

Please give a kudo if my post is useful.

HI Bunnel,

Please explain what the smallest prime factor will not be greater than 49? as 48,49,50 &51..... none are prime factors. then why we are taking cut of 50 but not from 49 or 48.

Thanks in advance,

47 is a prime. The next prime is 53. y (prime number) must be more than 47, so 53 or larger. But it does not matter whether we say that it's more than 47, more than 48, ... or more than 52, it still must be 53 or larger.
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(3 Δ 47) + 2 - is 100% an odd number, so E is out right away.
(3 Δ 47) + 2 is not divisible by ANY of the prime factors between 3 and 47
since the next prime factor after 47 is 53, y must be greater than 50.

A

The function (3 Δ 47) equals the product (3)(5)(7)…(43)(45)(47). This product is a very large odd number, as it is the product of only odd numbers and thus does not have 2 as a factor. Therefore, (3 Δ 47) + 2 = Odd + Even = Odd, and (3 Δ 47) + 2 does not have 2 as a factor either.

Every odd prime number between 3 and 47 inclusive is a factor of (3 Δ 47), since each of these primes is a component of the product. For example, (3 Δ 47) is divisible by 3, since dividing by 3 yields an integer — the product (5)(7)(9)…(43)(45)(47). Now, consider the sum (3 Δ 47) + k, where k is an integer. The sum will only be divisible by 3 if k is
also divisible by 3. In other words, when we divide (3 Δ 47) + k by 3, we are evaluating (3 Δ 47)/3 + k/3. Because (3 Δ 47)/3 is an integer, k/3 must also be an integer to yield an integer sum.

In this problem, k = 2, which is not divisible by any of the odd primes between 3 and 47. Since (3 Δ 47) IS divisible, but 2 is NOT divisible, we conclude that the sum (3 Δ 47) + 2 is NOT divisible by any of the odd primes between 3 and 47. So, (3 Δ 47) + 2 is not divisible by any prime number less than or equal to 47. The smallest prime
factor of (3 Δ 47) + 2 must be greater than 47. Thus, the minimum possible prime factor is 53, since that is the smallest prime greater than 47.

The correct answer is A.
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