Determine the largest possible integer n such that 942! is divisible by 15n.
The answer to the question as posted is n=942!/(15), which is a large number approximately equal to: 1.45*10^2393.
I suspect the factor 15n should read 15n.
If that's the case, I offer the following.
942! is the product 942*941*940*....3*2*1
If n=1, then 942!/(15) should give a whole number, i.e. 942! contains at least the factor 5 and 3.
If n=2, then 942!/(15²) should give a whole number, i.e. 942! contains at least the factor 5² and 3².
If n=3, then 942!/(15³) should give a whole number, i.e. 942! contains at least the factor 5³ and 3³.
... and so on.
So the problem reduces to counting the number of factors of 5 in 942! and number of factors of 3 in 942!.
The number of factors of 3 is likely more numerous in 942! since 3 is a smaller number, so it would appear more frequently. We therefore concentrate on the power of 5 which divides 942!.
942/5= 188.4,
so there are 188 factors of 5.
942/25=37.68, so there are 37 more factors of 5.
942/125=7.5, there are 7 more factors of 5.
942/625=1.5, there is 1 more factor of 5.
In total, there are (188+37+7+1)=233 factors of 5, so
maximum n=233.
As a check, dividing 942!/15^233 gives ...0944 as the last digits, i.e. it does not divide by 5 (nor 15) any more.
Well, to find the largest possible integer n, we need to figure out the prime factors of 15n.
First, let's break down 15 into its prime factors: 15 = 3 x 5. Since n can be any integer, we can assume it to be a multiple of both 3 and 5.
Since 942! is already divisible by 3 and 5 (since it contains numbers that are multiples of both), we just need to figure out how many factors of 3 and 5 we have in the factorial.
To determine the number of factors of 3, we can divide 942 by 3, then divide the result by 3 again, and so on, until we can no longer divide evenly. By doing this, we find that there are 314 multiples of 3 in 942!.
Similarly, to find the number of factors of 5, we divide 942 by 5, then divide the result by 5 again, and so on. By doing this, we find that there are 235 multiples of 5 in 942!.
Now, we need to find the smaller count out of 314 and 235, which is 235.
Therefore, the largest possible integer n such that 942! is divisible by 15n is 235. And since we're talking about clowns here, let's call it the "Biggest Clown Factor"!
To determine the largest possible integer n such that 942! is divisible by 15n, we need to find the highest power of 5 that divides 942! and multiply it by 3.
First, let's determine the highest power of 5 that divides 942!. We use the formula for finding the highest power of a prime factor in a factorial.
Step 1: Divide 942 by 5 and count the number of multiples of 5 in the result
942 ÷ 5 = 188 (There are 188 multiples of 5 in the range 1 to 942)
Step 2: Divide 942 by 25 and count the number of multiples of 25 in the result
942 ÷ 25 = 37 (There are 37 multiples of 25 in the range 1 to 942)
Step 3: Divide 942 by 125 and count the number of multiples of 125 in the result
942 ÷ 125 = 7 (There are 7 multiples of 125 in the range 1 to 942)
Step 4: Divide 942 by 625 and count the number of multiples of 625 in the result
942 ÷ 625 = 1 (There is 1 multiple of 625 in the range 1 to 942)
Step 5: Add up the multiples from each step
188 + 37 + 7 + 1 = 233
So, the highest power of 5 that divides 942! is 233.
Next, we multiply the highest power of 5 (233) by 3 to determine the largest possible integer n.
n = 3 * 233
n = 699
Therefore, the largest possible integer n such that 942! is divisible by 15n is 699.
In order to determine the largest possible integer n such that 942! is divisible by 15n, we need to consider the prime factors of 15.
The prime factorization of 15 is 3 * 5.
Now, let's break down the problem step by step:
Step 1: Determine the largest power of 3 that divides 942!
To find the largest power of 3, we divide 942 by 3, then divide the result by 3 again, and so on, until we obtain a quotient less than 3.
942 ÷ 3 = 314
314 ÷ 3 = 104
104 ÷ 3 = 34
34 ÷ 3 = 11
11 ÷ 3 = 3
The quotient is now less than 3, so we stop dividing. We have found that 3^314 is the largest power of 3 that divides 942!.
Step 2: Determine the largest power of 5 that divides 942!
To find the largest power of 5, we divide 942 by 5, then divide the result by 5 again, and so on, until we obtain a quotient less than 5.
942 ÷ 5 = 188
188 ÷ 5 = 37
The quotient is now less than 5, so we stop dividing. We have found that 5^37 is the largest power of 5 that divides 942!.
Step 3: Determine the largest common power of 3 and 5 that divides 942!
To determine the largest common power, we take the minimum of the exponents we found in steps 1 and 2.
The minimum of 314 and 37 is 37. So, the largest common power of 3 and 5 that divides 942! is 5^37.
Step 4: Determine the largest integer n such that 942! is divisible by 15n.
Since 15 can be expressed as 3 * 5, and we have already determined the largest power of 3 and 5 that divides 942!, we need to find the largest integer n such that n is divisible by 5^37.
In this case, n can be expressed as 5^37 multiplied by any positive integer. Therefore, the largest possible integer n that satisfies the condition is n = 5^37.
So, the largest possible integer n such that 942! is divisible by 15n is n = 5^37.
942!/(15n)
= 942x941x940x939x938!/(15n)
= 942(941)(188)(313)(938!)/n
I divided by 5 and by 3 leaving the division by n