**Numbers - Important Concept and Formulae**

**Finding the number of zeroes at the end of any factorial:**

**How to find the number of zeroes at the end of the product of numbers**

The
number of zeroes at the end of the product depends upon the number of 2‘s and
5’s in the expression.

The number of zeroes is the same as the number of pair of (5 x 2) units.

Since
the number of two’s will be more than the number of fives

So,
we have to count the number of five’s to get the number of zeroes at the end.

**Example : Find the number of zeroes in 10!?**

Solution
: 10!= 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1.

Since
the number of 2 is more than the number of 5’s, so the number of 5’s will be
counted to form the combination of 2 x 5.

We
have two 5’s, one digit 4 and another 5 is 10 ( since 10= 5 x2)

Thus,
there are only 2 zeroes at the end of 10!

**Example: Find the number of zeroes at the of the product of 2**

^{111}x 5^{222}
Solution
: In the given product, the number of 2’s are less than the number of 5’s,
hence we take the number of 2’s into consideration.

Thus
there will be only 111 pairs of(2 x5).

Therefore,
the number of zeroes at the end of the product of the given expression will be
111.

*Finding the largest power of a number contained in a factorial:***How to find the highest power of x that can exactly divide n! : W**e divide n by x, n by x

^{2}… and so on till we get $\frac{n}{x^{k}}$ equal to 1.

**Example : Find the largest power of 5 contained in 120!?**

Solution
: [$\frac{120}{5}$
] + [$ \frac{120}{5^{2}}$
] = 24 + 4 =28

We
need not further, because 120 is not divisible by 5

^{3 }(=125)**Example 2: Find the largest power of 7 that can exactly divide 800!?**

Solution
: $\frac{800}{7}$
+ $\frac{800}{7^{2}}$
+ $\frac{800}{7^{3}}$

= 114 + 16 +2 = 132
Thus
the highest power of 7 is 132 by which 800! can be divided.

**Concept of Remainder :**

**Definition of Remainder:**The part of the dividend that is not evenly divided by the divisor, is called the remainder.

**Rule 1 :**

When x1, x2, x3 … are divided by n individually, the respective
remainders obtained are R1, R2, and R3
etc .. when ( x1 + x2 + x3) is divided by n the remainders can be obtained by
(R1+ R2+R3+ ..) by n.

**Example : Find the remainder when 65+ 75+ 87 is divided by 7?**

Solution:
When we divided 65, 75 and 87, we obtain the respective remainders as 2, 5
and 3.

Now
we can get the required remainder by dividing the sum of the remainders by 7 =
$\frac{(2 + 5 + 3)}{7}$
=3

Instead
of dividing the sum of original numbers (65 +75+87 =227) by 7, which will also give the
same remainder 3.

**Example: Find the remainder when 719 x 121 x 237 x 725 x 727 is divided by 14 ?**

Solution
:

First, we find the remainders when each individual number is divided by 14.

First, we find the remainders when each individual number is divided by 14.

The
remainder when 719 is divided by 14 is 5

The
remainder when 121 is divided by 14 is 9

The
remainder when 237 is divided by 14 is 13

The
remainder when 725 is divided by 14 is 11

The
remainder when 727 is divided by 14 is 13

Now
we find the remainder of the remainders.

Hence, the remainder of $\frac{ 719 \,\times\,
121 \,\times\,
237 \,\times\,
725 \,\times\,
727}{14}$

=$\frac{5\,\times\, 9\,\times\, 13\,\times\, 11\,\times\, 13}{14}$

=$\frac{5\,\times\, 9\,\times\, 13\,\times\, 11\,\times\, 13}{14}$

Ã° $\frac{45
\,\times\,
13 \,\times\,
11 \,\times\,
13}{14}
$

Ã° $ \frac{3
\,\times\,
13 \,\times\,
11 \,\times\,
13}{14}$

Ã° $\frac{39
\,\times\,
143}{14}$

Ã° $\frac{11 \,\times\,
3}{4}$

Ã° $\frac{33}{4}$

Ã° 5

**Example: What is the remainder when 7**

^{9}is divided by 4?
Solution : $\frac{7^{9}}{4}$
= $ \frac{7\,\times\,
7 \,\times\,
7 \,\times\,
7 \,\times\,
7\,\times\,
7 \,\times\,
7 \,\times\,
7 \,\times\,
7}{4}
$

= $\frac{3 \,\times
3 \,\times\,
3 \,\times\,
3\,\times\,
3 \,\times\,
3 \,\times\,
3 \,\times\,
3 \,\times\,
3}{4}
$

= $ \frac{27 \,\times\,
27 \,\times\,
27}{4}$

= $\frac{3 \,\times\,
3 \,\times\,
3}{4}$

=4

Thus the required remainder is 4.

**Rule 2 :**

**A certain number when successively divided by two different numbers leaves some remainder. The same number, when divided by the product of the two divisors, will leave remainder equal to :**

**(First Divisor x Second Remainder ) + First Remainder.**

Example: A certain number when successively divided by 5 and 9 leaves the
remainders 2 and 7 respectively. What is the remainder when the same number is
divided by 45?

Solution
:

Remainder =( First Divisor x Second Remainder ) + First remainder

Remainder =( First Divisor x Second Remainder ) + First remainder

= (5 x 7) +2 =37

**Rule 3 :**

**If two numbers M and N when separately divided by x, leaving the remainders a and b respectively. Then M**

^{p }x N^{q}, when divided by x, will leave the remainder a^{m }x b^{n}**Example : Find the remainder when 80**

^{2}x 53^{3}is divided by 26.
Solution
:80 and 53, when divided by 26, leave remainders 4 and 1 respectively.

Therefore, 80

^{2}x 53^{3}when divided by 26 leaves remainder 4^{2}x^{ }1^{3}=16**Rule 4 :**

**Finding whether (a**

^{n}– k) is divisible by x?
Method
:

**1.**a

^{n}is divided into some parts in such a way that the value of each part is slightly more or less than the multiple of x

**2**. Take remainder when each part is divided by x

**3.**Multiply all the remainders obtained from each part

**4.**Subtract k from the product of remainders

**5**. If the result is a multiple of x or zero, then (a

^{n}– k) is exactly divisible by y.

**Example: Is 7**

^{8}-1 divisible by 24?
Solution
: 7

^{8}-1 = 7^{2 }x 7^{2 }x7^{2 }x7^{2 }-1= 49 x 49 x 49 x 49 – 1
The remainder we get when each part
49 is divided by 24 = 1 x 1 x 1 x 1 -1 =0

Therefore, 7

^{8}-1 divisible by 24**Concept of Negative Remainders:**

Remainders
are always non-negative by their definition.

When
14 is divided by 5, the usual remainder would be 4. But it can also be expressed
as -1. It is a negative remainder. The concept of negative remainders sometimes
helps the student in reducing calculations.

For
numbers like 12,19,26,33 etc., we can represent them as 7k+5 or 7k-5.

**Example:**

**Find the remainder when 262 x 68 is divided by 18?**

Thus
, $\frac{262 \,\times\,
68 }{18}$
will give remainder $\frac{-10 \,\times\,
-4 }{18}$ = $\frac{40
}{18}$ = 14

This
will give 14 as remainder.

So,
using the concept of negative remainders, we can reduce our calculations

**Example: What is the remainder when 11**

^{215}is divided by 12?
Solution:
The remainder we get when 11 divided by
12 is 11 = -1

The negative remainder is -1.

So 11

^{215}=(-1)^{215}= -1 =11
Using
the concept of the negative remainder, we can say the remainder is 11.