There are four consecutive prime numbers written in

The prime numbers less than 50 are...

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41...etc.

First three numbers must be around approx 20 ish for the product to be around 4000.

13*17*19 = 4,199
17*19*23 = 7,429

So largest of the four ==> 23
Smallest of the four ==> 13

I used a little intuition and brute force

a*b*c = 4199 , b*c*d= 7429 ---> (b*c) * ( d-a) = 3230 = 323*10
Since b*c can't be a multiple of 10 ---> d-a must be a multiple of 10
since 323 has no factors of 2,3,4,5 ,6, 7 (this we can judge by using divisibility tricks) ---> The best d-a is 10 (coz if d-a = 8 or larger ----> there are A LOT other primes in between ----> violate the "consecutive" characteristics of this sequence)

we have 20*20*20= 8000>> 4199 ----> the least number must be < 20
1) a=2 ---> d=12 , not a prime --> out
2) a=3 ---> d= 13, there're 3 primes in between ---> out
3) a=5 --> d=15 , not a prime ---> out
4) a=7 -->d=17 ---> 7,11,13,17 ...but 7*11*13 has unit digit of 1 instead of 9 ---> out
5) a=11 -->d=21 , not a prime --> out
6) a= 13 ---> d=23 ---> 13,17,19,23 : WOW!
7) a= 17 ---> d=27, not a prime --> out
8) a= 19 ---> d=29 ---> there's only 1 prime in between ---> out

So the best sequence is 13,17,19,23.

It looks redundant but in fact, the reasoning process is very short and quick.

WOW, easy and nice way to solve this problem.

Did this an old timer way. Got first 13 and the rest just came.

13, 17, 19, 23