For any integer P greater than 1, P! denotes the product of all the in

I think the answer should be 11.

45m = (3)^2 x 5 x m

48! can hold upto 22 3s, 10 5s and ___ Ms.

Thus the largest value that m can take will be 11, since 48! can also hold 4 11s.

Thus, 3^2 x 5 x 11 would be the factor of 48! where 11 is the largest value of m among the options.

There was a typo: 45m instead of 45^m. Edited.

Ans: 10; D
45=3^2+5^1
48/3+48/9+48/27= 16+5+1= 22; 3^2 occurs 11 times.
48/5+48/25=9+1= 10. 5 occurs once.
Limit is set by 5 and ans is 10.
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\(45=3^2*5\)

highest power of \(3^2\) that can divide 48! is \(\frac{1}{2}\) {\([\frac{48}{3}]+[\frac{48}{3^2}]+[\frac{48}{3^3}] \)} = (16+5+1) /2 = 11

highest power of 5 that can divide 48! is \([\frac{48}{5}]+[\frac{48}{5^2}]\)= 9+1=10

Greatest integral value that m can have is minimum of {11,10}
= 10

Answer is (D)

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Thank you Sir, was wondering about the same. Wonder why Kinshook did not look into this.

45^m = (3^2 x 5)^m
Highest power of 3 by which 48! can be divided is 16+5+1 = 22
Highest power of 9 by which 48! can be divided is 11

Highest power of 5 by which 48! can be divided is 9+1 = 10

Hence, greatest value of m can be 10
D is correct.

Can you elaborate this solution a little more. Why total factors is divided by 2.
More appreciable if you can explain in detail.

Hi, it is because the power of factor 3 in 45=3^2*5. Notice that for factor 5, I didn't divide the result by 2.

Hope it clarifies

Okay, i got confused if it's \(45^m\) or \(45m\)
So, had it been about finding highest power 27 then it would 7 (22/3 = 7.33 lower integer value)
Basically, what we do is breaking down to prime form and then calculate by dividing by power of that prime.

Given:

For any positive integer P, P! = P(P-1)(P-2) . . . * 3* 2 * 1
Since 48 is a positive integer, 48! = 48*47*46*45* . . . *3*2*1
To find: The greatest integer m for which 45m is a factor of 48!

Approach:

1. To determine the greatest possible value of m, we first need to understand when will 45m be a factor of 48!

45m = (32*5)m = 32m*5m
So, 45m is a factor of 48! only if:

32m is a factor of 48! AND
5m is a factor of 48!

We need to find the greatest value of m that satisfies this constraint

2. We’ll find the greatest value of m for which 32m is a factor of 48! Let this value be x
To find x, we’ll calculate the power of 3 in the prime-factorized expression of 48!

3. Next, we’ll find the greatest value of m for which 5m is a factor of 48! Let this value be y

To find y, we’ll calculate the power of 5 in the prime-factorized expression of 48!

4. The greatest value of m for which 32m and 5m are both factors of 48! will be equal to the lesser value between x and y

Working Out:

Finding the value of x
We first need to calculate the power of 3 in the prime-factorized expression for 48!


As inferred above, 48! = 48*47*46*45* . . . *3*2*1

The multiples of 3 in the above product are: {3, 6, 9, 12, 15, 18, 21, 24 . . . 48}
Upon counting, we see that there are 16 multiples of 3
Note: To ease the counting, here’s a simple trick: 3 = 3*1 and 48 = 3*16. This means, starting from 3 till 48, inclusive, there are 16 multiples of 3

So, each of these multiples of 3 contributes at least one 3 to the prime-factorized form of 48!

So, the number of 3s we’ve counted so far = 16

Now, note that some multiples of 3 in the list above are also multiples of 9 (= 32). These multiples therefore contain not one but two 3s. The number 16 calculated above considers only one 3 contributed by each multiple. So, we need to know how many multiples of 9 are there between 1 and 48 to know how many additional 3s they contribute.

There are 5 multiples of 9 between 1 and 48.

So, these 5 multiples contribute at least one more 3 each to the prime-factorized form of 48! (in addition to the one 3 already counted when counting the multiples of 3 in the above point)

So, the number of 3’s counted so far = 16 + 5 = 21

Note that one multiple of 9 in the list above is also a multiple of 27 = 33. This is the number 27 itself. So, this number contributes one more 3 to the prime-factorized form of 48!

So, the total number of 3s in the prime-factorized form of 48! = 21 + 1 = 22

Therefore, the power of 3 in the prime-factorized form of 48! Is 22


Hence, the maximum power of 3 which can divide 48! completely is 322

So, for 32m to divide 48!
The maximum value of 2m = 22
The maximum value of m = 11
That is, x = 11

Important Takeaway:

We can summarize the process we used above to find the power of 3 in 48!, in the following formula:


(Power of 3 in 48!)

= (Number of multiples of 3 in 48!) + (Number of multiples of 32 in 48!) + (Number of multiples of 33 in 48!)

Good so far?

Now comes the interesting point:

We can follow this same process to find the power of any prime number X in P! where P is any positive integer

The general formula to summarize the above process will be:

(Power of prime-factor X in P!)

= (Number of multiples of X in P!) + (Number of multiples of X2 in P!) + (Number of multiples of X3 in P!) + ...

Finding the value of y
We first need to find the power of 5 in the prime-factorized expression for 48! This time, we’ll use the formula to find this power.

Once again, we’ll start by writing 48! = 48*47*46*45* . . . *3*2*1

The multiples of 5 in the above product are: {5, 10, 15 . . . 45}
Upon counting, we see that there are 9 multiples of 5

There is only 1 multiple of 52 (that is, 25) in 48!. This multiple is number 25 itself.

So, the power of 5 in the prime-factorized form of 48! = 9 + 1 = 10

Therefore, the maximum power of 5 which can divide 48! completely is 510

So, for 5m to divide 48!
The maximum value of m = 10
That is, y = 10

Finding the greatest possible value of m
We’ve determined that x = 11 and y = 10
The lesser value out of 10 and 11 is 10
So, the greatest value of m for which 45m is a factor of 48! Is 10 (because if m = 11 then 32m = 322 will be a factor of 48! But 5m = 511 will not be a factor of 48!)
Looking at the answer choices, we see that the correct answer is Option D

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