The idea of factorial is very often
tested in competitive exams. Have given below two interesting questions on this
concept
Questions
1.
How many values can natural number n take, if n! is a multiple of 76 but not 79?
2.
How many values can natural number n take, if n! is a multiple of 220 but not 320?
Correct
Answers
1.
14 values
2.
21 values
Explanatory
Answers:
1.
How many values can natural number n take, if n! is a multiple of 76 but not 79?
The smallest
factorial that will be a multiple of 7 is 7!
14! will be
a multiple of 72
Extending
this logic, 42! will be a multiple of 76
However,
49! will be a multiple of 78 as 49 (7 * 7) will contribute two 7s to
the factorial. (This is a standard question whenever factorials are discussed).
Extending beyond this, 56! will be a multiple of 79
In general
for any natural number n,
n! will be
a multiple of [n/7] + [n/49] + [n/343] + …..
where [x]
is the greatest integer less than or equal to x. A more detailed discussion of
this is available on this link
So, we see
than 42! is a multiple of 76. We also see that 56! is the smallest factorial
that is a multiple of 79. So, n can take values { 42, 43, 44, 45, ……55}
There are
14 values that n can take.
2.
How many values can natural number n take, if n! is a multiple of 220
but not 320?
The highest
power of 2 that will divide n! = [n/2] + [n/4] + [n/8] + [n/16]….. and so on.
So, let us try to find the smallest n such that n! is a multiple of 220,
If n = 10,
the highest power of 2 that will divide n! = [10/2] + [10/4] + [10/8] = 5 + 2 +
1 = 8
If n = 20,
the highest power of 2 that will divide n! = [20/2] + [20/4] + [20/8] + [20/16] = 10 + 5 + 2 + 1 = 18
If n = 24,
the highest power of 2 that will divide n! = [24/2] + [24/4] + [24/8] + [24/16] = 12 + 6 + 3 + 1 = 22
{Here we
can also see that each successive number is just the quotient of dividing the
previous number by 2. As in, [12/2] = 6, [6/2] = 3, [3/2] = 1. This is a
further shortcut one can use.}
So the
lowest number of n such that n! is a multiple of 2^20 is 24
Now, moving
on to finding n! that is a multiple of 3. The highest power of 3 that will
divide n! = [n/3] + [n/9] + [n/27] + [n/81] and so on,
When n =
20, the highest power of 3 that can divide 20! = [20/3] + {6/3] = 6 + 2 = 8
When n = 35,
the highest power of 3 that can divide 35! = [35/3] + {11/3] + [3/3] = 11 + 3 +1
=
15
When n = 45,
the highest power of 3 that can divide 45! = [45/3] + [15/3] + [5/3] = 15 + 5 +1
=
21
The lowest
number n such that n! is a multiple of 3^20 is 45.
When n
takes values from 24 to 44, n! will be a multiple of 2^20 and not 3^20. n can
take 21 values totally.