What is the greatest value of x such that 9^x is a factor of 9! ?

let's rewrite it as:
1*2*3*4*5*6*7*8*9/9^x
x can only be 2.

answer is B.

incorrect tag for distance/rate prob

Edited. Thank you for noticing.

Can someone explain how there are two 9s in 9!?

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9! = 1*2*3*4*5*6*7*8*9 = 1*2*3*4*5*(2*3)*7*8*9

3 and 3 (in 6) give one more 9.

Hope it's clear.

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Hope this helps.

We need to determine the greatest values of x such that 9^x (or 3^2x) divides into 9!. In other words, we need to determine the number of factors of 9 in 9!. Since 9 = 3^2, let’s first determine the total number of factors of 3 in 9!.

To do so, we can use the following shortcut in which we divide 9 by 3, then divide the quotient (ignore any nonzero remainder) by 3 and continue this process until we no longer get a nonzero quotient.

9/3 = 3

3/3 =1

Since 1/3 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 3 within 9!.

Thus, there are 3 + 1 = 4 factors of 3 within 9!, and therefore there are 2 factors of 9 in 9!.

Alternate solution:

Recall that 9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1. We see that the first factor 9 is a factor of 9. We also see that 6 x 3 = 18 will contain another factor of 9. Thus there are only 2 factors of 9 in 9! since none of the other numbers contain or contribute a factor of 9 or 3. Thus, the largest value of x such that 9^x is a factor of 9! is 2.

Answer: B

It's an easy problem if you know the concept.

Prime factorization of 9 = 3^2

9/3 + 9/9 = 4

=> it appears twice

Ans B

ScottTargetTestPrep

I really struggle in these problems. There are two methods to solve these I believe.

1/ As this question requires, how many times 3 appears in 9!?
This can be answered in following way:
9/ 3 + 9/3^2

3+1 = 4 times

2/ Method which you described:
9/3 = 3
3/3 = 1

3+ 1 = 4 times

However, what is the difference between both them methods? And can I use method 1 to solve this?

Response:

As a matter of fact, the two methods are equivalent. I’ll only show that dividing n by k^2 and dividing n by k successively two times will produce the same quotient; however, the idea applies to any power of k.

Let n divided by k produce a quotient of s and a remainder of r. We have:

n = ks + r

Now, divide s by k. Let the quotient be q and the remainder be p. Then:

s = kq + p

Substitute s = kq + p in the first equality:

n = k(kq + p) + r = k^2 * q + kp + r

The above suggests that the quotient when n is divided by k^2 is q; however, we have to show that kp + r is less than k^2 before we can actually conclude that. Recall that the remainder is always strictly less than the divisor. So, we have r ≤ k - 1 from the first division and p ≤ k - 1 from the second division. Multiplying the second inequality by k, we obtain:

kp ≤ k^2 - k

Adding the above inequality and the inequality r ≤ k - 1 together, we obtain:

kp + r ≤ k^2 - 1

Notice that n = k^2 * q + kp + r and kp + r is less than k^2. Then, the quotient from the division of n by k^2 is q, which is the same quotient we obtain by successively dividing n by k two times. This shows that to determine the number of factors of k in n!, you can either divide n by k, k^2, k^3 etc. until you get a zero quotient or you can divide n by k and keep dividing the quotient by k, until you get a zero quotient. Both will give you the same answer.