If m is a positive integer such that m! is divisible by 3^{10} but not by 3^{11}, and (m+5… — GMAT

If $m$ is a positive integer such that $m!$ is divisible by $3^{10}$ but not by $3^{11}$ , and $(m+5)!$ is divisible by $3^{13}$ but not by $3^{14}$ , what is the total number of distinct positive factors of $m$ ?

A
6
8Answer
C
10
D
12
E
14

Answer

The total number of distinct positive factors of

m

is 8.

Using Legendre's formula, the highest power of 3 dividing

m!

is 10 when

m

is 24, 25, or 26. Evaluating

(m+5)!

29!

(from

m=24

) has

E_3(29!) = 9 + 3 + 1 = 13

, whereas

30!

(from

m=25

) has

E_3(30!) = 10 + 3 + 1 = 14

. Therefore,

m

must equal 24. Since

24 = 2^3 \times 3^1

, its number of distinct positive factors is

(3+1)(1+1) = 8

Step-by-Step Solution

Apply Legendre's formula to find the highest power of 3 dividing

m!

The exponent of 3 in

m!

is given by

E_3(m!) = \lfloor \frac{m}{3} \rfloor + \lfloor \frac{m}{9} \rfloor + \lfloor \frac{m}{27} \rfloor + \dots = 10

Legendre's formula determines the exact power of a prime

p

n!

Test integer values for

m

to satisfy

E_3(m!) = 10

For

m=24

E_3(24!) = \lfloor \frac{24}{3} \rfloor + \lfloor \frac{24}{9} \rfloor = 8 + 2 = 10

. Thus,

m \in \{24, 25, 26\}

Integers 24, 25, and 26 each contain 10 factors of 3 in their factorial products.

Apply the second condition

E_3((m+5)!) = 13

to identify the unique value of

m

m=24

m+5=29 \implies E_3(29!) = \lfloor \frac{29}{3} \rfloor + \lfloor \frac{29}{9} \rfloor + \lfloor \frac{29}{27} \rfloor = 9 + 3 + 1 = 13

. For

m=25

m=26

m+5 \ge 30

, giving

E_3(30!) = 10+3+1 = 14

. Hence,

m = 24

Only

m=24

satisfies both power constraints simultaneously.

Calculate the number of positive factors of

m = 24

24 = 2^3 \times 3^1

. Total positive factors

= (3+1)(1+1) = 4 \times 2 = 8

The number of factors is found by adding 1 to each prime exponent and multiplying.

Key Concept

Legendre's Formula and Prime Factorization Divisor Counting

Estimated Time:2m 0s

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