What is the exponent of 3 in the prime factorization of the integer S = 25! + 26! + 27!? — GMAT

What is the exponent of $3$ in the prime factorization of the integer $S = 25! + 26! + 27!$ ?

Answer: 16

The exponent of

3

in the prime factorization of

S

16

Factoring

25!

from the sum gives

S = 25!(1 + 26 + 26 \times 27) = 25!(729) = 25! \times 3^6

. Applying Legendre's formula to

25!

yields

\lfloor 25/3 \rfloor + \lfloor 25/9 \rfloor = 8 + 2 = 10

factors of

3

. Adding the

6

factors of

3

from

729 = 3^6

gives a total exponent of

10 + 6 = 16

Factor out

25!

from the sum

S = 25! + 26! + 27!

S = 25! \left(1 + 26 + 26 \times 27\right)

Factoring out the greatest common factorial term

25!

converts the sum into a product of

25!

and an integer factor.

Simplify the expression inside the parentheses

1 + 26 + 702 = 729 = 3^6

Using algebraic simplification,

1 + 26(1 + 27) = 1 + 26(28) = 729 = 27^2 = (3^3)^2 = 3^6

Compute the exponent of

3

25!

using Legendre's formula

E_3(25!) = \lfloor \frac{25}{3} \rfloor + \lfloor \frac{25}{9} \rfloor + \lfloor \frac{25}{27} \rfloor = 8 + 2 + 0 = 10

Legendre's formula counts the total prime factors of

3

contributed by all multiples of

3, 9, 27, \dots

up to

25

Combine the exponent of

3

from

25!

and the factor

729

Total exponent of

3 = 10 + 6 = 16

Since

S = 25! \times 3^6 = (3^{10} \times k) \times 3^6 = 3^{16} \times k

(where

3 \nmid k

), the total exponent of

3

10 + 6 = 16

Exponent of a prime in factorial expressions using Legendre's formula and algebraic factoring

Question