f=30! = 10^7*q where q is not multiple of 10
a)10^d is factor of f. d can be 1,2,3,4,5,6,7. Insufficient
b)d>6 d can be 7,8,9,..... insufficient
a&b together) d>6 thus only possible value from a) is 7
Sufficient
IMO C

i fail to understand

see we know f is product of first 30 integers
or f=1*2*3......*30 = 30!
a) 10^d is factor of f.
factor means f is divisible by 10^d
Now as f=30! , if we write it in terms of power of 10, we get 10*20*30*(2*5)*(4*15)*(8*25)*q
(we get 10 by either multiple of 10 or by multiplying 2 with multiple of 5)
Till 30 multiples of 10 are 10,20,30 = 3
and till 30, multiples of 5 are 5,15,25. Also 25 is 5*5 thus can be used twice. Thus total 4 fives.
So we get 7(=3+4) tens
or f can be written as 10^7*q where q is not multiple of 10 (as we have already used all 10's)
thus 10^7 is factor of f. Also if 10^7 is factor, 10^1 is also a factor and similarly 10^2,10^3,10^4..10^6 all are factors.
Thus d can take any value from 1 to 7. Insufficient.
b)d>6 we just know that d>6 and its a positive integer. Thus it can take any value like 7,8,100,1000 .. Insufficient.

a&b) d can take any value from 1 to 7 and d>6
Thus only 7 is possible
Sufficient.

divya23 wrote:i fail to understand

Hi,
I think you did not understand the part of d=7
In 30! we have 6 multiples of 5 (5,10,15,20,25,30)
10,20,30 when multiplied gives integer times 10^3
each of 5,15 when multiplied with even numbers likes 2,4,6,... gives integer times 10.
But 25 = 5.5. This when multiplied by 2 evens or a multiple of 4 gives integer times 10^2
So, in total you have 3+2+2 = 7 multiples of 10. Those numbers when multiplied yields 10^7.

You can straight away use the following formula to find maximum value of 'k' if 10^k is a factor of n!
k = [n/5]+[n/5^2]+[n/5^3]+... where [a] is the integer part of a
Use this formula in this case
n=30
So, k = [30/5]+[30/5^2] =6+1 =7. You need not use the next etrms such as [30/5^3] because 5^3 > 30 and all the numbers from there will end up to 0.

IMO C

1) 10^d is a factor of f.

10 can be written as a product of 2 and 5

so any power of 10 , say 10^3, can be written as 2^3 * 5^3.

As f= 30! = 30*29*28*27..........5*4*3*2*1, we can break down each factor from 30 to 1 into its prime factors.
eg = 30 = 2*5*3 ---> this contributes 2 and 5 resulting in 10

similarly we can calculate the pair of 2 and 5 in order to get the highest power of 10.

Given : 10^d is a factor of f

As @cans pointed out that 30! can be written as 10^7 * (q) , here q is the product of all prime factors other than 2 and 5 present in 30! .

now 10 ^ 1 to 10 ^ 7 can be factor of 10^7 * (q). This means d can be 1 or 2 or 3 or 4 or 5 or 6 or 7

As we can have more than 1 value of d, the statement is insufficient.

2) d>6 , d can be any value such as 7,8,9...So insufficient.

Combining 1 and 2

d can be 1 or 2 or 3 or 4 or 5 or 6 or 7 AND d>6

Only value which satisfies both the statements is "7"

Since we have only one value ,its Sufficient

divya23 wrote:i fail to understand

A good small technique. What power of 2 is there in 5!

Step 1: Now to reach 5 we need to multiply 2 with atleast '2' Jot it down
Step 2: Now to reach 2 we need to multiply 2 atleast with '1'. Jot this as well
5
----
2 (step 1)
1 (step 2)

Step 3: Add step 1 and step 2. [3 is the power of 2 in 5!]

Now try the same for power of 10 in 30!. but be careful that my formula is only for prime factors and the least power for prime factors is valid.
(2,5) in here.
Powers of 2 in 30!: 15+7+3+1 = 26
powers of 5 in 30!: 6+1 = 7

Hope this helps...

f = 30! is given
d > 0 (integer), is given

d = ?

1) The integer constraint in the question stem signals that number/scenario testing is likely.

If 10^d = (2*5)^d is a factor of f, then the possibilities for d are constrained by the factors of 5 in the numbers from 1 to 30, inclusive, since there are many more factors of 2 in these numbers.

5 (1), 10 (1), 15 (1), 20 (1), 25 (2), 30 (1) for a total of 7 factors of 5.

Therefore, we know that d could take the values 1, 2, 3, 4, 5, 6, or 7. Insufficient.

2) There are infinite integers greater than 6. Insufficient.

Combined) If x = 1, 2, 3, 4, 5, 6, or 7 AND x > 6, then x = 7. Sufficient.

C