MHB How many Numbers can appear as product?

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The discussion centers on determining how many numbers between 1999 and 2021 can be expressed as the product of three sums derived from increasing three prime numbers by one. The analysis reveals that only 2016 meets the criteria, as it can be expressed as the product of three even numbers, specifically 3, 4, and 168, where the primes are 2, 3, and 167. Other candidates like 2000 and 2008 do not satisfy the conditions due to their factor compositions. The conclusion emphasizes that 2016 is the sole valid product in the specified range. Understanding the reasoning behind these calculations involves complex number theory concepts.
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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?
 
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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.
 
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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.
 
Sorry It took me a while to fully understood this number theory.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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