**Properties of Numbers**

**1.**The numbers, which give a perfect square on adding as well as subtracting its reverse, are rare and hence termed as Rare Numbers.

If X is a positive integer and X1 is the integer obtained from X by writing its decimal digits in reverse order, then X + X1 and X - X1 both are perfect square then X is termed as Rare Number. For example:

For X=65, X1=56

X+X1 = 65+56 = 121 = 11^2

X-X1 = 65 - 56 = 9 = 3^2

So 65 is a Rare Number.

**2.**When n is odd, n( n2 – 1 ) will be divisible by 2 For e.g. n = 9 then n(n2-1) = 9(92 – 1) = 720 is divisible by 24

**3.**If n is odd, 2n + 1 will be divisible by 3, e.g. n=5, 25+1 =33, which is divisible by 3

And if n is even, 2n – 1 is divisible by 3, e.g. n=6, 26-1 =63, which is divisible by 3

**4.**If n is prime, then n (n4-1) will be divisible by 30, e.g. n=3, 3(34-1) = 240, which is divisible by 30

**5.**If n is odd, 22n + 1 is divisible by 5, e.g. n=5, 22*5+1 =1025, which is divisible by 5

And if n is even, 22n – 1 is divisible by 5, e.g. n=6, 22*6-1 = 4095, which is divisible by 5

**6.**If n is odd, 52n + 1 is divisible by 13, e.g. n=3, 52*3+1 =15626, which is divisible by 13

And if n is even, 52n – 1 is divisible by 13, e.g. n=4, 52*4-1 =390624, which is divisible by 13

**7.**The number of divisors of a given number N ( including 1 and the number itself ) where N = ambncp where a,b,c are prime numbers , are ( 1 + m ) ( 1 + n ) ( 1 + p ).

**8.**xn + yn = ( x + y ) ( xn-1 – xn-2 y + …. + yn-1 ), xn + yn is divisible by x + y when n is odd.

**9.**xn - yn = ( x + y ) ( xn-1 – xn-2 y + …. - yn-1 ) when n is even, so xn - yn is divisible by (x+y)

**10.**xn - yn = ( x - y ) ( xn-1 + xn-2 y + …. + yn-1 ) when n is either odd or even, so xn - yn is divisible by (x-y)

Example: The remainder, when ( 1523 + 2323 ) is divided by 19, is : (CAT 2004)

(a) 4 (b) 15

(c) 0 (d) 18

Thus 1523 + 2323 is always divisible by 15 + 23 = 38. As 38 is a multiple of 19, 1523 + 2323 is divisible by 19. So we get a remainder of 0.

**Square of a number: I**f a number is multiplied by itself, it is called the square of that number. Example 4 × 4 = 16 is square of 4.

**Important Properties:**

1. A square cannot end in odd number of zeroes.

2. The square of an odd number is odd and that of an even number is even.

3. Every square number is a multiple of 3 or exceeds a multiple of 3 by unity.

4. Every square number is a multiple of 4 or exceeds a multiple of 4 by unity.

5. If a square number ends in 9, the preceding digit is even.

Square root: The square root of a number is the number, whose square is the given number.

Example is 4, as 4 × 4 = 16

**Methods for finding square roots:**

1. Factorization: Resolve the number into prime factors and deduce if there are numbers which are repeating themselves (square of numbers).

Example: Find

Here 2601 = 32 × 172 = = 3 × 17 = 68

**2.**

**Approximation:**The approximation method is the simplest method to find the square root of a number, but as the name suggests it is an approximate method.

This method is best explained with an example. Suppose you want to find the square root of , you know the square root of 100 is 10 and 121 is 11, now 104 lies between 100 and 121. Difference is 21, and number is 4 more than the lower number which is 100. Therefore we can say the square root is

10 (of 100) + 4/21 = 10.19

**Cube of a number:**when a number is multiplied three times with the same number, it is called the cube of a number.

**Example:**4 × 4 × 4 = 64

**Cube root:**The cube root of a number is the number, which if multiplied three times by same number gives the given number.

Example: 64. It is represented by or with the power of 1/3, example (64)1/3 = (4^3)1/3 = 4.

To find the cube root of a number you have to find prime factors of the numbers, and deduce if in those numbers if a number is repeated thrice.

Example: 3 × 17 = 51

**Complex numbers**

Since i = , i2 = –1, i3 = –1 × i = –i

and i4 = (i2)2 = (–1)2 = 1

Just like surds, to rationalize complex numbers, the rationalizing factor or conjugates are used like (a + ib) and (a – ib) are relative conjugates.

HCF AND LCM

**HCF:**HCF is the Highest Common Factor or Greatest Common Divisor (GCD). Actually GCD explains it well, that is the greatest division that divides given set of numbers. Example: HCF of 10, 15 and 30 = 5 and HCF of 15, 30 and 45 is 15. It is obvious to see in each case 5 and 15 are the highest numbers which can divide the three numbers.

To find the HCF of given numbers, resolve the numbers into their prime factors and then pick the common term from them and multiplying them will give you the HCF. The HCF is 1 when no common prime factors are there, as 1 is the only number which divides the two and is the highest.

Example : Find the HCF of 24, 48, 102

Prime factors 2 × 2 × 2 × 3, 2 × 2 × 2 × 2 × 3, 17 × 3 × 2

Common numbers = 2 × 3 = 6, therefore 6 is the HCF

**LCM:**LCM is Least common multiple. It represents the smallest number which is divisible by all of the given numbers.

Example: LCM of 3, 4 and 5 = 60, as it is smallest number divisible by them.

To find the LCM, resolve all the numbers into their prime factors, take the ones which are common and the ones which are left (uncommon) and multiplying them will give you the LCM of the number.

Example : Find the LCM of 24, 48, 102

Prime factors - 2 × 2 × 2 × 3, 2 × 2 × 2 × 2 × 3, 17 × 3 × 2

Common numbers = 2 × 3

Numbers left = 2 × 2 × 2 × 2 × 2 × 17

LCM - common numbers x numbers left

= 2 × 3 × 2 × 2 × 2 × 2 × 2 × 17 = 3264

**Important Note:**

**1.**For two numbers, LCM × HCF = product of two numbers

Example: LCM of 4, 5 = 20 and HCF is 1, 20 × 1 = 4 × 5, 20 = 20

**2.**HCF of fractions = (HCF of Numerators)/(LCM of Denominators)

Example: Find HCF of 3/5 and 4/10 .

Here HCF of 3, 4 is 1

And LCM of 5, 10 is 10

HCF of 3/5and 4/10 = 1/10

**3.**LCM of fractions = (LCM of Numerators)/(HCF of Denominators)

Example: Find LCM of 3/5 and 4/10.

Here LCM of 3, 4 is 12

And HCF of 5, 10 is 5

LCM of 3/5 and 4/10 = 12/5

**4.**HCF and LCM of decimals: To calculate the HCF and LCM of decimals, remove the decimals and convert them into non-decimals, by multiplying with 10 or 100 or…. Post that calculate the HCF and LCM in regular fashion and once you have the regular HCF and LCM, convert that number into decimal by dividing it with power of 10 with which you multiplied earlier.

Example: Find the HCF and LCM of 0.6, 0.9, 1.5

Convert the numbers by multiplying them with 10, therefore numbers are 6, 9, 15

HCF of 6, 9, and 15 is 3; dividing by 10 it is 0.3

LCM of 6, 9, and 15 is 90; dividing by 10 it is 9

For large numbers the way to find the HCF is using Euclid’s Algorithm, but it works for two numbers only. From the larger number, subtract the biggest multiple of the smaller number without getting a negative answer. Replace the larger number with the answer and repeat this until the last number is zero, and the HCF is the next-to-last number computed.

347236 – 1 × 297228 = 50008

297228 – 5 × 50008 = 47188

50008 – 1 × 47188 = 2820

47188 – 16 × 2820 = 2068

2820 – 1 × 2068 = 752

2068 – 2 × 752 = 564

752 – 1 × 564 = 188

564 – 3 × 188 = 0

The HCF here is 188. Once we have the HCF, the LCM is the product of the two numbers divided by the HCF as per the properties mention on LCM and HCF.

Therefore LCM of 347236 and 297228 = 347236 × 297228/188 = 549090936.

LCM × HCF = Product of two numbers

LCM = (32 × 40)/8 = 160Solution:- Since both the trucks visited today, next they will visit together on the LCM of 4 and 5, which is 20, so on the 20th day from now

Ans. General form of such a number = (LCM of 7 and 11) k + 4

= 77 k + 4 where k is any whole number

Now largest 3-digit number of the form (77k+4) is for k = 12

i.e. 77×12 + 4 = 928.

Ans. Since HCF = 8,

So, let the two numbers be ‘8a’ and ‘8b’, where a and b are co-prime to each other.

So, LCM of (8a and 8b) = 8 × a × b

8ab = 248

ab = 31

Since a and b are co-primes, the possible values for a and b can be 1 and 31.

So the numbers are 8 × 1 and 8 × 31 i.e. 8 and 248.

Remainder Theorem

We have to take care about the following things while finding remainder:1. 'x' when divided by 'x' the remainder will be 0.

2. 'x + a' when divided by 'x' the remainder will be a.

3. 'x - a' when divided by 'x' the remainder will be ‘-a' i.e. '-a + x'.

4. (mx + a)^n when divided by 'x' the remainder will be a^n.

5. (mx - a)^n when divided by 'x' the remainder will be (-a)^n.

Here we will always express numerator in terms of denominator.

For example

If we want to find remainder when 2^80 is divided by 5.Here, we will express such that the base of 2^80 get near to 5.

We can write 2^80 as 4^40

Now 4 can be written as (5-1)

So 2^80 = 4^40 = (5-1)^40

Now when (5-1)^40 is divided by 5, the remainder will be (-1)^40 = 1

Example: What is the remainder when 1044 × 1047 is divided by 33?

Ans. Here when 33 divide 1044, remainder is 21.

And when 33 divide 1047 is 24

Now the rule is the remainders multiplication has to be divided by divisor to get eh remainder

Therefore 21 × 24 = 504. Dividing by 33 remainder is 9, which is the answer.

Example: What is the reminder when 91 + 92 + 93 + ...... + 99 is divided by 6?

Ans. It is obvious that 6 is leaving remainder 3 with powers of 9, which is actually a property, but students can check by dividing first 2–3 numbers in the series, therefore total remainder is 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 27, when 27 is divided by 6, remainder is 3, which is the answer.

To find power of any number in the factorial of any number.

The power of 'a' in n! Can be obtained by adding the quotients obtained by dividing 'n' by a, a2,a3 … till we get zero as the quotients.

For example power of 2 in 36! Will be

36/(2^1) + 36/(2^2) + 36/(2^3) + …...

= 18 + 9 + 4 + 2 + 1 + 0

= 34

Example: What maximum power of 6 would perfectly divide 50!?

Ans. It will be the number of 6’s in 50!, as 6 = 2 × 3, it will be equal to number of these pairs in 50!.

No. of 2’s (leave decimals)

= 50/2 + 50/22 + 50/23 + 50/24 + 50/25

Onwards powers of 2 will exceed 50,

= 25 + 12 + 6 + 3 + 1 = 47

No. of 3’s (leave decimals)

= 50/3 + 50/32 + 50/33

Onwards powers of 3 will exceed 50,

= 16 + 5 + 1 = 22

Since 3’s are 22, there will be 22 pairs; maximum power of 6 is 22.

Example. How many zeros will be there in the value of 25!?

Ans. The number of zeros are number’s of (5 × 2)’s

No. of 5’s (leave decimals)

= 25/5 + 25/52 onwards powers of 5 will exceed 25,

= 5 + 1 = 6, as twos are plenty, but 5’s are 6 therefore number of zeros are 6

**Last Two Digits**

Where 'ab' is a 2-digit number and n is any number, we will first find out the unit's digit and then the ten's digit.

We will convert all the calculations in which we have unit digit of the base as 1.

**1.**So let’s start with those calculations which have 1 as unit digit in base

Now if the base has 1 as the unit digit then we can say that the unit digit of the result will also be 1.

Now tens digit will be the product of tens digit of base and unit digit of power.

*Consider*:(ab)^xyz

Now, if b=1 then we can say that unit digit of result will be 1,

and tens digit = unit digit of (a * z)

Example: (31)^142

Unit digit =1

Tens digit = 3 * 2 = 6

So, the last two digits = 61

Example: (81)^236

Unit digit =1

Tens digit = 8 * 6 => 8

So, the last two digits = 81

**2.**Now if we have 3, 7 or 9 as the unit digit in the base then we can square the base once or twice to obtain 1 as the unit digit in the base.

Example: (39)^132

(39)^132 = ((39)^2)^66

= (21)^66 (as square of 39 has 21 as last 2 digits)

= 21 (unit = 1 and tens = 2 * 6 = 2)

Example: (67)^148

(57)^144 = ((67)^2)^74

= (89)^74

= ((89)^2)^37

= (21)^37

= 41

3. Now for those numbers which have base as even numbers.

For this we have to find the last 2 digits of any power of 2.

Things to remember for this method:

**a.**Any odd power of 24 will have 24 as last 2 digits.

**b.**Any even power of 24 will have 76 as last 2 digits.

**c.**(2)^10 = 1024, or we can say 24 as last 2 digits.

Example: Now let us find last two digits of (2)^135

(2)^135 = ((2)^10)^13 * 2^5

= 24^13 * 2^5

= (24^odd) * (2^5)

= 24 * 32

= 68

Example: Find last two digits of 38 ^ 96

38 = 2 * 19

So (38)^96 = (2^96) * (19^96)

= ((2^10)^9) * (2^6) * ((19^2)^48)

= (24^9) * (2^6) * (61^48)

= 24 * 64 * 81

= 36 * 81

= 16

**Miscellaneous Examples**

**Example:**How many different positive integers exist between 106 and 107, the sum of whose digits is equal to 2?

**Ans.**Between 10 and 100, that is 101 and 102, we have 2 numbers, 11 and 20.

Similarly, between 100 and 1000, that is 102 and 103, we have 3 numbers, 101, 110 and 200. Extrapolating the trend, between 106 and 107, one will have 7 integers whose sum will be equal to 2.

Example: How many strokes on computer keyboard are needed to type numbers from 1 to 999?

Ans. 1 to 9 needs 9 strokes

10 to 99 needs 180 (90 × 2) strokes

100 to 999 needs 2700 (900 × 3) strokes

Total = 2892 strokes

Example: A certain number when successively divided by 8 and 11 leaves remainders of 3 and 7 respectively. What will be remainder when the number is divided by the product of 8 and 11?

Ans. Solution to this question comes by a rule although there is a longish method using the choices also. Here divisor 1 = 8, divisor 2 = 11, remainder 1 = 3, remainder 2 = 7, by rule it is equal to divisor 1 × remainder 2 + remainder 1 = 8 × 7 + 3 = 59.

Example: The number of positive integers not greater than 100, which are not divisible by 2, 3 or 5 is:

(CAT 1993)

(a) 26 (b) 18

(c) 31 (d) none of these

Ans. (a) There are 50 odd numbers less than 100 which are not divisible by 2.

Out of these 50 there are 17 numbers which are divisible by 3. [Total numbers divisible by 3 from 1 to 99 are 33, out of last number divisible is 99(33 × 3) is an odd number, so odd numbers divisible are 17] Out of remaining there are 7 numbers that are divisible by 5. [Total numbers divisible by 5 from 1 to 99 are 19, out of which 9 are even (10, 20, 90), and three odd numbers are divisible by 3(15, 45, 75), therefore 7 are left]

Hence numbers which are not divisible by 2, 3, 5 = (50 – 17 – 7) = 26

Example: Let x < 0.50, 0 < y <1, z > 1. Given a set of numbers; the middle number, when they are arranged in ascending order, is called the median. So the median of number x, y and z would be: (CAT 1993)

(a) Less than 1 (b) between 0 and 1

(c) Greater than 1 (d) cannot say

Ans. (b) The median is the middle number in ascending order, given x < 0.50, 0 < y < 1, z > 1, the numbers could be arranged as (x, y, z) or (y, x, z). Since x and y both are under 1, hence median will also lie between 0 and 1.

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