Finding the Maximum Power of 3 Dividing a Number

The question asks for the maximum possible integer value of 'm' such that the given number N is completely divisible by \(3^m\). The number N is defined as a product of several integers: N = 90 × 42 × 324 × 55.

To find the highest power of a prime number (in this case, 3) that divides a product of numbers, we need to find the prime factorization of each number in the product and then sum up the exponents of that prime number.

Step 1: Prime Factorization of Each Number

Let's find the prime factors for each component of N:

Step 2: Combining the Powers of 3

Now, we multiply these numbers together to get N. When multiplying numbers, we add the exponents of the same prime factors.

N = 90 × 42 × 324 × 55

N = \((3^2 \times 2 \times 5) \times (2 \times 3^1 \times 7) \times (2^2 \times 3^4) \times (5 \times 11)\)

To find the total power of 3 in N, we sum the exponents of 3 from each factor:

Total power of 3 = Exponent of 3 in 90 + Exponent of 3 in 42 + Exponent of 3 in 324 + Exponent of 3 in 55

Total power of 3 = \(2 + 1 + 4 + 0 = 7\)

So, the prime factorization of N includes \(3^7\) as a factor. This means N is divisible by \(3^7\).

Step 3: Determining the Maximum Value of m

The question states that N is divisible by \(3^m\). For N to be divisible by \(3^m\), the power of 3 in the prime factorization of N must be greater than or equal to m.

We found that N is divisible by \(3^7\). Therefore, the maximum value of m for which N is divisible by \(3^m\) is 7.

Any power of 3 greater than 7, like \(3^8\) or \(3^9\), would not divide N because N only has \(3^7\) as its highest power of 3.

Maximum Value of m for N Divisible by 3m

The maximum value of m is 7.

Revision Table: Key Concepts for Divisibility

Additional Information: Exponents and Divisibility Rules

Understanding exponents is crucial when dealing with divisibility questions involving powers of prime numbers. If a number A can be written as \(A = p^a \times q^b \times r^c \times \dots\) (where p, q, r are distinct prime factors and a, b, c are their positive integer exponents), then A is divisible by \(p^x\), \(q^y\), \(r^z\), etc., only if \(x \le a\), \(y \le b\), \(z \le c\), and so on.

In our problem, N has a prime factorization that includes \(3^7\). This means N is divisible by \(3^1\), \(3^2\), \(3^3\), \(3^4\), \(3^5\), \(3^6\), and \(3^7\). It is not divisible by \(3^8\), \(3^9\), or any higher power of 3, because it simply doesn't have enough factors of 3.

The maximum power of 3 that divides N is often referred to as the 'valuation' of N with respect to the prime 3, denoted as \(\nu_3(N)\). In this case, \(\nu_3(N) = 7\).