If n is a positive integer, what is the maximum possible value of n

Question

If n is a positive integer, what is the maximum possible value of n such that 77!/720^n is an integer ?

(A) 35
(B) 18
(C) 17
(D) 36
(E) 38

gmatophobia · Answer

720 = 8 * 9 * 10 = 2^4 * 3^2 * 5

Maximum power of 2 in 77! = 73 
 Maximum power of  2^4  in 77! =  \frac{73 }{ 4}  = 18

Maximum power of 3 in 77! = 35 
 Maximum power of  3^2  in 77! =  \frac{35}{2}  = 17

Maximum power of 5 in 77! = 18

Min (18, 17, 18) = 17

Option C

Rabab36 · Answer

gmatophobia

SIR HOW TO CALCULATE MAX POWER OF 2 IN 77!

Bunuel · Answer

Finding the power of a  prime  number  p , in the  n!

The formula is:
 \frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k} , where  k  must be chosen such that  p^k\leq{n}

Example: 
What is the power of 2 in 25!?

\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22  (notice that we take only the quotient of the division, that its 25/2 = 12, 25/4 = 6, ..., 25/16 = 1).

So, the highest integer power of 2 in 77! is:

\frac{77}{2} + \frac{77}{2^2} + \frac{77}{2^3} + \frac{77}{2^4} + \frac{77}{2^5} + \frac{77}{2^6}=\frac{77}{2} + \frac{77}{4} + \frac{77}{8} + \frac{77}{16} + \frac{77}{32} + \frac{77}{64} = 38 + 19 + 9 + 4 + 2 + 1 =73 .

This means that 77! is divisible by 2^73 but not divisible by 2^74.

FYI:

77!=

=2^{73}*3^{35}*5^{18}*7^{12}*11^7*13^5*17^4*19^4*23^3*29^2*31^2*37^2*41*43*47*53*59*61*67*71*73

Check this:  Power of a Number in a Factorial Problems

Hope it helps.

Rabab36 · Answer

Bunuel

THANKYOU COACH

gmatophobia · Answer

Bunuel  has already shared the correct way to find the power, sharing a another way you can approach the problem (using successive division). Please note, both methods use the same underline principle.

Suppose you want to find the power of prime number p in n!. To so do, divide p (dividend) by n (divisor) successively and in the process keep a note of the quotient obtained. The division needs to be done till the quotient obtained is less than the divisor. Once the division is performed, add all the quotient to get the power

Let's see this with 77!.

We want to find the power of 2 in 77!.

Step 1  : Divide 77 by 2, and keep a track of the quotient.
 \frac{77}{2}  =  Q 38

Step 2  : Now divide the obtained quotient from step 1, with the divisor (2) in this case and keep a track of the quotient.

\frac{38}{2}  =  Q 19

Keep repeating the process, till the quotient is less than the divisor

\frac{19}{2}  =  Q 9

\frac{9}{2}  =  Q 4

\frac{4}{2}  =  Q 2

\frac{2}{2}  =  Q 1

As we have obtained a quotient, 1, which is less than the divisor we can stop the process.

Step 3  
Power = Sum of all the quotients obtained =  1 + 2 + 4 + 9 + 19 + 38  = 73

Kinshook · Answer

If n is a positive integer, what is the maximum possible value of n such that 77!/720^n is an integer ?

720 = 2^4*3^2*5

Maximum power of 2 in 77! = 38 + 19 + 9 + 4 + 2 + 1 = 73
Maximum power of 3 in 77! = 25 + 8+ 2 = 35
Maximum power of 5 in 77! = 15 + 3 = 18

77! = 2^{73}*3^{35}*5^{18}*k = (2^4)^{18}*2*(3^2)^{17}*3*5^{18}k = (2^4*3^2*5)^{17}*z = (720)^{17}*z

IMO C

If n is a positive integer, what is the maximum possible value of n

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GMAT Problem Solving (PS) Questions

If n is a positive integer, what is the maximum possible value of n

Prep Toolkit

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