Math Problem Statement

N is a two digit number which when expressed in base 12 has 7 in its units place and 6 in its tens place. What are the number of trailing zeros when N! is expressed in the base

Solution

We are tasked with finding the number of trailing zeros when N!N! is expressed in base 12, where NN is a two-digit number that, when expressed in base 12, has 7 in its units place and 6 in its tens place.

Step 1: Understanding the Number NN

Given that NN is a two-digit number in base 12, it can be written as: N=6×12+7=72+7=79N = 6 \times 12 + 7 = 72 + 7 = 79 Thus, N=79N = 79 in base 10.

Step 2: Trailing Zeros in Factorial N!N!

To determine the number of trailing zeros in N!N! when expressed in base 12, we need to analyze the factors of 12. Since 12=22×312 = 2^2 \times 3, trailing zeros in base 12 are created by factors of 12=22×312 = 2^2 \times 3.

The number of trailing zeros in N!N! in base 12 depends on how many times N!N! contains the factor 222^2 (i.e., two factors of 2) and the factor 3.

Step 3: Counting Powers of 2 and 3 in N!N!

We need to count the powers of 2 and 3 in 79!79! (the factorial of 79).

Count of Powers of 2

The number of powers of 2 in 79!79! is given by the sum: 792+794+798+7916+7932+7964\left\lfloor \frac{79}{2} \right\rfloor + \left\lfloor \frac{79}{4} \right\rfloor + \left\lfloor \frac{79}{8} \right\rfloor + \left\lfloor \frac{79}{16} \right\rfloor + \left\lfloor \frac{79}{32} \right\rfloor + \left\lfloor \frac{79}{64} \right\rfloor Calculating these: 792=39,794=19,798=9,7916=4,7932=2,7964=1\left\lfloor \frac{79}{2} \right\rfloor = 39, \quad \left\lfloor \frac{79}{4} \right\rfloor = 19, \quad \left\lfloor \frac{79}{8} \right\rfloor = 9, \quad \left\lfloor \frac{79}{16} \right\rfloor = 4, \quad \left\lfloor \frac{79}{32} \right\rfloor = 2, \quad \left\lfloor \frac{79}{64} \right\rfloor = 1 So, the total number of factors of 2 in 79!79! is: 39+19+9+4+2+1=7439 + 19 + 9 + 4 + 2 + 1 = 74

Count of Powers of 3

Similarly, the number of powers of 3 in 79!79! is given by: 793+799+7927+7981\left\lfloor \frac{79}{3} \right\rfloor + \left\lfloor \frac{79}{9} \right\rfloor + \left\lfloor \frac{79}{27} \right\rfloor + \left\lfloor \frac{79}{81} \right\rfloor Calculating these: 793=26,799=8,7927=2,7981=0\left\lfloor \frac{79}{3} \right\rfloor = 26, \quad \left\lfloor \frac{79}{9} \right\rfloor = 8, \quad \left\lfloor \frac{79}{27} \right\rfloor = 2, \quad \left\lfloor \frac{79}{81} \right\rfloor = 0 So, the total number of factors of 3 in 79!79! is: 26+8+2=3626 + 8 + 2 = 36

Step 4: Trailing Zeros in Base 12

To form a factor of 12 in N!N!, we need two factors of 2 and one factor of 3. The number of trailing zeros is determined by the minimum of:

  • The number of pairs of factors of 2: 742=37\left\lfloor \frac{74}{2} \right\rfloor = 37
  • The number of factors of 3: 3636

Thus, the number of trailing zeros in 79!79! when expressed in base 12 is: min(37,36)=36\min(37, 36) = 36

Final Answer:

The number of trailing zeros in 79!79! when expressed in base 12 is 36.


Would you like any further details or have more questions about trailing zeros or factorials? Here are some follow-up questions for you:

  1. How does the number of trailing zeros change when the base is different, like base 10 or base 16?
  2. Can you calculate trailing zeros for factorials in other bases, like base 5 or base 2?
  3. How do you compute the number of trailing zeros for large factorials without direct computation of the factorial?
  4. What would happen if the number NN was not a two-digit number? How would that affect the calculation?
  5. How does the presence of higher powers of prime factors (like 535^3) in base 5k5^k impact trailing zeroes?

Tip: For large factorials, always break down the problem into prime factorizations of the base to avoid directly calculating the factorial, which would be computationally infeasible.

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Math Problem Analysis

Mathematical Concepts

Factorials
Trailing Zeros
Base Systems
Prime Factorization

Formulas

n! = n × (n-1) × ... × 1
Number of trailing zeros in base b = min(number of factors of b in n!, number of pairs of factors of prime divisors)

Theorems

Fundamental Theorem of Arithmetic
Counting factors in factorials

Suitable Grade Level

Grades 11-12