In four consecutive prime numbers, the product of the last three is 7429 and that of the first three is 4199. The largest of these prime number is :

Correct option is C

Given:

Product of the last three primes = 7429

Product of the first three primes = 4199

Concept Used:

Consecutive Primes: The primes follow each other without any non-prime in between (e.g., 5, 7, 11, 13).

Solution:

Four consecutive prime numbers: Let them be $$p_1, p_2, p_3, p_4$$​ where $$p_1 < p_2 < p_3 < p_4$$​​

First, factorize 4199:

Divide by small primes:

13:$$4199 \div 13 = 323$$​ → So, $$p_1 = 13$$​​

Now, factorize 323:

17: $$323 \div 17 = 19$$​ → So, $$p_2 = 17$$​ and $$p_3 = 19$$​​

Thus, the first three primes are 13, 17, 19

Given $$p_2 = 17 , p_3 = 19$$​, and the product of the last three primes is 7429:

​$$17 \times 19 \times p_4 = 7429$$​​

​$$323 \times p_4 = 7429$$​​

​$$p_4 = \frac{7429}{323} = 23$$​​

​$$p_4 = 23$$​​

The four consecutive primes are: 13, 17, 19, 23