If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?

Question

Also... 10^d is a factor of f and d > 6

Answer 1

f = 30 x 25 x 20 x 15 x 10 x 5 x the other 24 integers.

So f contains 7 factors of 5 (25 supplies 2 of them) and more than 7 factors of 2.

(The 15 even numbers supply a total of more than 15 such factors as there are three powers of 2 and 3 of the other even numbers (12, 20 & 28) are also divisble by 4 and 24 is divisible by 8)

But 7 factors of 2 is all we need for our purposes and even if we had a multitude of surplus 2s, they cannot give us another 10

i.e. f contains 7 factors of 10. And ONLY 7 such factors as each 10 can only be made up of a 5 and a 2

so d = 7 is the only possible answer, given that we are told that d >6

GENERALISING:

50, 75, 100, also supply two 5s to a factorial they are part of,

so for f = 55! and d> 12, d must = 13
so for f = 80! and d> 18, d must = 19
so for f = 105! and d> 24, d must = 25

however

for f = 130! and d> 30, d is not 31 but 32
CAN YOU SEE WHY?

Answer 2

d = 7.

The product of the first n integers in maths is referred to as n factorial which is written as n! ie n followed by the exclamation mark.

f is 30! (30 factorial, ie 1 x 2 x 3 x ... x 28 x 29 x 30).

The question basically boils down to how many 0's will be on the end of 30!.

10^d is a power of 10, ie 10, 100, 1000 etc. So for 10^d to be a factor of 30 !, d can be no larger than the number of zeroes on 30!
29! has 6 zeroes, the last 9 digits being 616000000. If you mutiply that by 30 you get 1848000000.

As d must be > 6 and no more than 7, therefore d = 7.

Answer 3

f = 30! (30 factorial) = 30*29*28*...*3*2*1

If 10^d is a factor of f and d > 6, then that means 10^d = 10^7 or 10^8 or 10^9 etc.

But for any 10^d for d=7, 8, 9, 10, etc. that was a factor of 30!, then so would 10^6, 10^5, 10^4, 10^3, 10^2 and 10 be a factor.

So I'm not sure your question makes sense, really, since for any d>6 such that 10^d was a factor of f, then ANY d <= 6 would ALSO be a factor. Perhaps your question is just worded a little oddly, and you mean find the LARGEST d such that 10^d is a factor of f?

30! = 265,252,859,812,191,000,000,000,000,000,000

If your question is to find the LARGEST d such that 10^d divides evenly into 30!, then you would choose d=18, since 30! = 10^18 * 265,252,859,812,191

NRT

Answer 4

Let's find the largest power of 10 that divides 30!.
First 15 of the numbers between 1 and 30 are
even and some are divisible by 4, so there are
more than 15 2's in the product.
Now count the number of 5's.
The numbers divisible by 5 are
30, 25, 20, 15,10 and 5.
5 occurs twice in 25 and once in each of the other
numbers, so there are 7 5's in the product.
Combining these 7 5's with
7 2's we see that 7 is the maximum power
of 10 dividing 30!, so d = 7 is the solution
of the problem.

Answer 5

f=30!=2.65.....x 10^32
one factor would be 10^32 of "f" or 10^d
so, d=32 and d>6