If a and b are odd integers, a Δ b represents the product of

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

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.

(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

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.

(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.

How is 53 a factor of (3 Δ 47) + 2 though....

Question is bad

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The question is fine. The point is that the smallest prime factor of 3*5*7*...*47 +2 is greater than 47. The next prime is 53, so y > 50. That does not mean that y = 53. Please review the question more carefully. 

Bunuel anairamitch1804
I am not able to figure out why in all the answers above, it has been assumed that (3*5*7*...*47) +2 is not a Prime Number and so, divisible by a prime number y>50.

It may so be the case that (3 Δ 47) + 2 is not divisible by any number other than 1 and itself.

Please clarify.

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No such assumption was made above. If (3 Δ 47) + 2 is not divisible by 1 and itself, then (3 Δ 47) + 2 is prime, so y = (3 Δ 47) + 2 > 50.