Basic Concepts

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Basic concepts, definitions and identities

Number System

Test of divisibility:

1. A number is divisible by ‘2’ if it ends in zero or in a digit which is a multiple of

‘2’i.e. 2,4, 6, 8.

2. A number is divisible by ‘3’, if the sum of the digits is divisible by ‘3’.

3. A number is divisible by ‘4’ if the number formed by the last two digits, i.e. tens

and units are divisible by 4.

4. A number is divisible by ‘5’ if it ends in zero or 5

5. A number is divisible by ‘6’ if it divisible by ‘2’ as well as by ‘3’.

6. A number is divisible by ‘8’ if the number formed by the last three digits, i.e,

hundreds tens and units is divisible by ‘8’.

7. A number is divisible by ‘9’ if the sum of its digit is divisible by ‘9’

8. A number is divisible by ‘10’ if it ends in zero.

9. A number is divisible by ‘11’ if the difference between the sums of the digits in

the even and odd places is zero or a multiple of ‘11’.

LCM: LCM of a given set of numbers is the least number which is exactly divisible by every number of the given set. HCF: HCF of a given set of numbers is the highest number which divides exactly every number of the given set. LCM, HCF: 1. Product of two numbers = HCF × LCM

2. HCF of fractions =

3. LCM of fractions =

4. One number =

5. LCM of two numbers=

6.



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Examples to Follow:

1. The square of an odd number is always odd.

2. A number is said to be a prime number if it is divisible only by itself and unity.

Ex. 1, 2, 3, 5,7,11,13 etc.

3. The sum of two odd number is always even.

4. The difference of two odd numbers is always even.

5. The sum or difference of two even numbers is always even.

6. The product of two odd numbers is always odd.

7. The product of two even numbers is always even.

Problems:

1. If a number when divided by 296 gives a remainder 75, find the remainder when

37 divides the same number.

Method:

Let the number be ‘x’, say

∴x = 296k + 75, where ‘k’ is quotient when ‘x’ is divided by ‘296’

= 37 × 8k + 37 × 2 + 1

= 37(8k + 2) + 1

Hence, the remainder is ‘1’ when the number ‘x’ is divided by 37.

2. If 232+1 is divisible by 641, find another number which is also divisible by ‘641’.

Method:

Consider 296+1 = (232)3 + 13

= (232 +1)(264-232 +1)

From the above equation, we find that 296+1 is also exactly divisible by 641,

since it is already given that 232+1 is exactly divisible by ‘641’.

3. If m and n are two whole numbers and if mn = 25. Find nm, given that n 1

Method:



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mn = 25 = 52

∴m = 5, n = 2

∴nm = 25 = 32

4. Find the number of prime factors of 610 × 717 × 5527

610 × 717 × 5527 = 210×310×717×527×1127

∴The number of prime factors = the sum of all the indices viz., 10 + 10 + 17 +

27 + 27 = 91

5. A number when successively divided by 9, 11 and 13 leaves remainders 8, 9 and

8 respectively.

Method:

The least number that satisfies the condition= 8 + (9×9) + (8×9×11) = 8 + 81 +

792 = 881

6. A number when divided by 19, gives the quotient 19 and remainder 9. Find the

number.

Let the number be ‘x’ say.

x = 19 × 19 + 9

= 361 + 9 = 370

7. Four prime numbers are given in ascending order of their magnitudes, the

product of the first three is 385 and that of the last three is 1001. Find the

largest of the given prime numbers.

The product of the first three prime numbers = 385

The product of the last three prime numbers = 1001

In the above products, the second and the third prime numbers occur in

common. ∴ The product of the second and third prime numbers = HCF of the

given products.

HCF of 385 and 1001 = 77

∴Largest of the given primes = = 13



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Square root, Cube root, Surds and Indices Characteristics of square numbers

1. A square cannot end with an odd number of zeros

2. A square cannot end with an odd number 2, 3, 7 or 8

3. The square of an odd number is odd

4. The square of an even number is even.

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

Ex.

4 × 4 = 16 = 5 × 3 + 1

5 × 5 = 25 = 8 × 3 + 1

7 × 7 = 49 = 16 × 3 + 1

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

Ex.

5 × 5 = 25 = 6 × 4 + 1

7 × 7 = 49 = 12 × 4 + 1

7. If a square numbers ends in ‘9’, the preceding digit is even.

Ex.

7 × 7 = 49 ‘4’ is the preceding even numbers

27 × 27 = 729 ‘2’ is the preceding even numbers.

Characteristics of square roots of numbers

1. If a square number ends in ‘9’, its square root is a number ending in’3’ or ‘7’.


3. If a square number ends in ‘5’, its square root is a number ending in’5’



6. If a square number ends in ‘0’, its square root is a number ending in ‘0’.

Ex.



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(i)

(ii)

(iii)

(iv)

(v)

(vi)



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THEORY OF INDICES

1. am × an = am + n 2. (am)n = amn

3.

4. (ab)m = ambm 5. a0 = 1

6. = qth root of ap =

7. = pth root of a

8.

9. a∞ = ∞ 10. a-∞ = 0

1.Find the square root of 6561 (Factor method)

6561 = (3×3)×(3×3)×(3×3)×(3×3) =(9×9)×(9×9) = 81×81

= 81 2. Find the least number with which you multiply 882, so that the product may be a perfect square. First find the factors of 882. 882 = 2 × 3 × 3 × 7 × 7 Now, 882 has factors as shown above, ‘3’ repeated twice, ‘7’ repeated twice and ‘2’ only once. So when one more factor ‘2’ is used, then it becomes a perfect square. 882 × 2 = (2 × 2) × (3 × 3) × (7 × 7) The least number required is ‘2’



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3. Find the cube root of 2985984 (Factor method)

2985984 = 23 × 23 × 23 × 23 × 33 × 33

= 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144

4. Find the value of

5.

6. Simplify



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= = = =

6.Find the least number with which you multiply 882. So that the product may be a perfect square.

First find the factors of 882.

Now, 882 has factors as shown above.

‘3’ repeated twice, ‘7’ repeated twice and ‘2’ only once. So, when one more factor ‘2’ is used, then it becomes a perfect square.

∴882×2=(2×2)×(3×3)×(7×7) ∴The least number required is ‘2’.

7. Find the cube root of 2985984 (Factor method)

∴2985984 = 23×23×23×23×33×33



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∴ = = = 144

8.Find the value of

= (approx)

9. = = = = 30

10. Simplify -

- = = =

11. Find the least number with which 1728 may be added so that the resulting number is a perfect square.

Note:

Take the square root of 1728 by long division method. It comes to 41.+ something. As shown, if 128 is made 164, we get the square root as an integer. The difference between 164 and 128 i.e., 36 must be added to 1728, so that 1764 is a perfect

square =42

Theory of Indices

Problems:

1. A certain number of persons agree to subscribe as many rupees each as there

are subscribers. The whole subscription is Rs.2582449. Find the number of

subscriber?



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Let the number of subscribers be x, say since each subscriber agrees to

subscribe x rupees.

The total subscription = no. of persons × subscription per person

= x × x = x2

given x2 = 2582449

∴x = 1607

2. Simplify:

Use the 2 formulas

3. Simplify

Sol.

4. Find the number whose square is equal to the difference between the squares of

75.12 and 60.12

Sol.

Let ‘x’ be the number required

∴x2 = (75.12)2 – (60.12)2

= (75.12 + 60.12) (75.12 – 60.12)



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= 135.24 ×15 = 2028.60

∴ 45.0399

5. Reduce to an equivalent fraction write rational denominator

Sol.


Sol.

approx.

7. An army general trying to draw his 16,160 men in rows so that there are as

many men as true are rows, found that he had 31 men over. Find the number of

men in the front row.

Let ‘a’ be the number of men in the front row.

a2 + 31 = 161610

No. of men in the front row = 127

a2 = 161610 – 31 = 16129

∴a = 127

8. A man plants his orchid with 5625 trees and arranges them so that there are as

many rows as there are trees in a row. How many rows are there?

Sol.

Let ‘x’ be the number of rows and let the number of trees in a row be ‘x’ say

x2 = 5625

∴x = 75

∴There are 75 trees in a row and 75 rows are arranged.

Basic Concepts

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