MHB How many Numbers can appear as product?

[Marcelo Arevalo]

We increase by 1 each of three prime numbers, not necessarily distinct. Then we
form the product of these three sums. How many numbers between 1999 to 2021
can appear as such a product?

 

[Marcelo Arevalo]

Modified Trial & Error Solution :
Assume the three prime numbers as p < q < r, we consider the different cases for p,
q and r.

For the case of 2016= 2^5 x 3^2 x 7 , we now express 2016 as the product of three even
numbers in which some are greater than or equal to 3:
2016 = 3 x 4 x 168, where 3 = 2 +1, 4 = 3+1, 168 =167 + 1. But 2, 3 and 167 are prime numbers, so it meets
the condition of the problem. Hence, 2016 is the solution.
For the case of 2019 = 3 x 673, in the three factors except the prime number 3 the
other two numbers are both neither even number greater than 3, so has not met the
condition.
Based on all the cases above, we conclude that between 1999 to 2021, there is one
that meets the condition of the problem and that is 2016.

answers on the book, I don't quite understand his explanation.
been squeezing my head to come up with explanation still I din't get how they did it. can anyone here help me to understand it further?? thank you.

 

[kaliprasad]

Because all 3 numbers are even so we get product as multiple of 8
now the possible candidates are 2000, 2008 and 2016.
each of these to be checked.
I do not have an elegant way to analyse
2000 = 2^4 * 5^3

now 5 *2 is not a factor meeting criteria as 9 is not prime. 5*2^2 is (19+1) but there is not enough 2 (as 2^2 has to go) to give 3 even numbers

2008 = 8 * 501= 8 * 3 * 167 and as 8 is $2^3$ and it does not have 3 odd factors it is out

you have found for 2016.

 

[Marcelo Arevalo]

Sorry It took me a while to fully understood this number theory.