What is the number of distinct primes dividing 12! + 13! +14! ?

Question

Abigail Allen · Accepted Answer

undefined To find the number of distinct prime numbers dividing $12! + 13! + 14!$, we first need to factorize the expression.

$$12! + 13! + 14! = 12! \left(1 + 13 + 13 \times 14\right) = 12! \times 182$$

Now, we need to factorize $12!$ to find its prime factors. The prime factorization of $12!$ is:

$$12! = 2^{10} \times 3^5 \times 5^2 \times 7 \times 11$$

Next, we need to find the prime factorization of 182:

$$182 = 2 \times 91 = 2 \times 7 \times 13$$

Now, we can combine the prime factorizations of $12!$ and 182 to find the prime factorization of $12! \times 182$:

$$12! \times 182 = 2^{10} \times 3^5 \times 5^2 \times 7^2 \times 11 \times 13$$

Thus, the distinct prime factors of $12! + 13! + 14!$ are $2$, $3$, $5$, $7$, $11$, and $13$. Therefore, the number of distinct prime numbers dividing $12! + 13! + 14!$ is 6.

Nathan Axel · Answer

#2,3,5,7,11# #12!+13!+14! =12!(1+13+13 xx 14)# The primes in #12!# are #2,3,5,7,11# and the primes in #(1+13+13 xx 14)# are #2,7# so the primes dividing #12!+13!+14! # are #2,3,5,7,11#

Ethan Adkins · Answer

Five distinct primes divide #12!+13!+14!# and these are #{2,3,5,7,11}# #12!+13!+14!# = #12!(1+13+14xx13)# = #12!(14xx14)# = #12xx11xx10xx9xx8xx7xx6xx5xx4xx3xx2xx14xx14# = #ul(2xx2xx3)xx11xxul(2xx5)xxul(3xx3)xxul(2xx2xx2)xx7xxul(2xx3)xx5xxul(2xx2)xx3xx2xxul(2xx7)xxul(2xx7)# = #2^12xx3^5xx5^2xx7^3xx11# Hence, five distinct primes divide #12!+13!+14!# and these are #{2,3,5,7,11}#