# Solved Example

Example-1

**Evaluate 313**Ã—

**313 + 287**Ã—

**287**

Solution

aÂ² + bÂ² = 1/2 ((a + b)Â² + (a - b)Â²)

= Â½(313 + 287)Â² + (313 - 287)Â² = Â½(600Â² + 26Â²)

= Â½(360000 + 676) = 180338

= Â½(313 + 287)Â² + (313 - 287)Â² = Â½(600Â² + 26Â²)

= Â½(360000 + 676) = 180338

Example-2

**Which of the following 241, 337, 391 is a prime number?**

Solution

__241__

16>âˆš241. Hence take the value of Z = 16.

241 is not divisible by any of these. Hence we can conclude that 241 is a prime number.

Prime numbers less than 16 are 2, 3, 5, 7, 11, 13 and 17.

337 is not divisible by any of these. Hence we can conclude that 337 is a prime number.

Prime numbers less than 16 are 2, 3, 5, 7, 11, 13, 17 and 19

391 is divisible by 17. Hence we can conclude that 391 is not a prime number.

Prime numbers less than 16 are 2, 3, 5, 7, 11 and 13.

241 is not divisible by any of these. Hence we can conclude that 241 is a prime number.

__337__

Prime numbers less than 16 are 2, 3, 5, 7, 11, 13 and 17.

337 is not divisible by any of these. Hence we can conclude that 337 is a prime number.

__391__

Prime numbers less than 16 are 2, 3, 5, 7, 11, 13, 17 and 19

391 is divisible by 17. Hence we can conclude that 391 is not a prime number.

Example-3

**On dividing 15968 by a certain number the quotient is 89 and the remainder is 37. Find the divisor?**

Solution

Divisor = (Dividend - Remainder)/Quotient

= (15968 - 37)/89 â†’ 179

= (15968 - 37)/89 â†’ 179

Example-4

**Which of the following numbers is divisible by 3?**

**5****41326****5967013**

Solution

- Sum of digits in 541326 = 5 + 4 + 1 + 3 + 2 + 6 = 21 divisible by 3.
- Sum of digits in 5967013 = 5 + 9 + 6 + 7 + 0 + 1 + 3 = 31 not divisible by 3.

Example-5

**What least value must be assigned to * so that the number 197*5462 is divisible by 9?**

Solution

Let the missing digit be x

Sum of digits = (1 + 9 + 7 + x + 5 + 4 + 6 + 2) = 34 + x

For 34 + x to be divisible by 9, x must be replaced by 2

The digit in place of x must be 2.

Sum of digits = (1 + 9 + 7 + x + 5 + 4 + 6 + 2) = 34 + x

For 34 + x to be divisible by 9, x must be replaced by 2

The digit in place of x must be 2.

Example-6

**What least number must be added to 3000 to obtain a number exactly divisible by 19?**

Solution

On dividing 3000 by 19 we get 17 as remainder Therefore number to be added = 19 - 17 = 2

Example-7

**Find the smallest number of 6 digits which is exactly divisible by 111?**

Solution

Smallest number of 6 digits is 100000

On dividing 10000 by 111 we get 100 as remainder

Number to be added = 111 - 100 = 11.

Hence, required number = 10011.

On dividing 10000 by 111 we get 100 as remainder

Number to be added = 111 - 100 = 11.

Hence, required number = 10011.

Example-8

**A number when divided by 342 gives a remainder 47. When the same number is divided by 19 what would be the remainder?**

Solution

Number = 342 K + 47 = 19 Ã— 18 K + 19 Ã— 2 + 9 = 19 (18K + 2) + 9.

The given number when divided by 19 gives 18 K + 2 as quotient and 9 as remainder.

The given number when divided by 19 gives 18 K + 2 as quotient and 9 as remainder.

Example-9

**Find the remainder when 2**

^{31}is divided by 5?Solution

2

Unit digit of 2

Unit digit of 2

Unit digit of 2

Unit digit of 2

Unit digit of 2

â€¦. And so on

So unit digit of 2

Now 8 when divided by 5 gives 3 as remainder.

^{10}= 1024.Unit digit of 2

^{1}= 2Unit digit of 2

^{2}= 4Unit digit of 2

^{3}= 8Unit digit of 2

^{4}= 6Unit digit of 2

^{5}= 2â€¦. And so on

So unit digit of 2

^{31}= 8Now 8 when divided by 5 gives 3 as remainder.

Example-10

**How many numbers between 11 and 90 are divisible by 7?**

Solution

The required numbers are 14, 21, 28, . . . . . . . . . . . , 84

This is an A. P with a = 14, d = 7.

Let it contain n terms then by using the formula of last term of A.P.

T = a + (n - 1) d

Here T = 84

84 = 14 + (n - 1) 7

84 = 7 + 7n

> 7n = 77

> n = 11.

This is an A. P with a = 14, d = 7.

Let it contain n terms then by using the formula of last term of A.P.

T = a + (n - 1) d

Here T = 84

84 = 14 + (n - 1) 7

84 = 7 + 7n

> 7n = 77

> n = 11.

Example-11

**Find the sum of all odd numbers up to 100?**

Solution

The given numbers are 1, 3, 5. . . . . . . . . 99.

This is an A. P with a = 1, d = 2.

Let it contain n terms then by using the formula of last term of A.P.

T = a + (n - 1) d

= >1 + (n - 1) 2 = 99

= >n = 50

Then required sum = n/2 (first term + last term) = (50/2) (1 + 99) = 2500.

This is an A. P with a = 1, d = 2.

Let it contain n terms then by using the formula of last term of A.P.

T = a + (n - 1) d

= >1 + (n - 1) 2 = 99

= >n = 50

Then required sum = n/2 (first term + last term) = (50/2) (1 + 99) = 2500.

Example-12

**The sum of three prime numbers is 100. If one of them exceeds another by 36 then one of the numbers is?**

Solution

x + (x + 36) + y = 100

2x + y = 64

Therefore y must be even prime which is 2

2x + 2 = 64 = >x = 31. Third prime number = x + 36 = 31 + 36 = 67.

2x + y = 64

Therefore y must be even prime which is 2

2x + 2 = 64 = >x = 31. Third prime number = x + 36 = 31 + 36 = 67.

Example-13

**The sum of all possible two digit numbers formed from three different one digit natural numbers when divided by the sum of the original three numbers is equal to?**

Solution

Let the one digit numbers x, y, z

Sum of all possible two digit numbers=

= (10x + y) + (10x + z) + (10y + x) + (10y + z) + (10z + x) + (10z + y) = 22 (x + y + z)

Therefore sum of all possible two digit numbers when divided by sum of one digit numbers gives 22.

Sum of all possible two digit numbers=

= (10x + y) + (10x + z) + (10y + x) + (10y + z) + (10z + x) + (10z + y) = 22 (x + y + z)

Therefore sum of all possible two digit numbers when divided by sum of one digit numbers gives 22.

Example-14

**A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as remainder. The number is?**

Solution

Number = (555 + 445) x (555 - 445) Ã— 2 + 30

= (555 + 445) Ã— 2 Ã— 110 + 30 = 220000 + 30 = 220030.

= (555 + 445) Ã— 2 Ã— 110 + 30 = 220000 + 30 = 220030.

Example-15

**The difference between two numbers s 1365. When the larger number is divided by the smaller one the quotient is 6 and the remainder is 15. The smaller number is?**

Solution

Let the smaller number be x, then larger number = 1365 + x

Therefore 1365 + x = 6x + 15

5x = 1350 = >x = 270

Thus, Required number is 270.

Therefore 1365 + x = 6x + 15

5x = 1350 = >x = 270

Thus, Required number is 270.

Example-16

**In doing a division of a question with zero remainder, a candidate took 12 as divisor instead of 21. The quotient obtained by him was 35. The correct quotient is?**

Solution

Dividend = 12 Ã— 35 = 420.

Now dividend = 420 and divisor = 21; Therefore correct quotient = 420/21 = 20.

Now dividend = 420 and divisor = 21; Therefore correct quotient = 420/21 = 20.

Example-17

**Find the L.C.M of 2, 4, 6, 8, 10.**

Solution

The numbers can be factorised as follows 2, 2 Ã— 2, 2 Ã— 3, 2 Ã— 2 Ã— 2, 2 Ã— 5

Take the highest powers of all the primes. i.e., 2

Take the highest powers of all the primes. i.e., 2

^{3}Ã— 3 Ã— 5 = 120Example-18

**Find the H.C.F. of 2, 4, 6, 8, 10.**

Solution

The numbers can be factored as:

2, 2 Ã— 2, 2 Ã— 3, 2 Ã— 4, 2 Ã— 5; The only common factors in all the number is 2. HCF = 2

2, 2 Ã— 2, 2 Ã— 3, 2 Ã— 4, 2 Ã— 5; The only common factors in all the number is 2. HCF = 2

Example-19

**What is the greatest number which exactly divides 110, 154 and 242?**

Solution

The required number is the HCF of 110, 154 & 242.

110 = 2 Ã— 5 Ã— 11

154 = 2 Ã— 7 Ã— 11

242 = 2 Ã— 11 Ã— 11

HCF = 2 Ã— 11 = 22

110 = 2 Ã— 5 Ã— 11

154 = 2 Ã— 7 Ã— 11

242 = 2 Ã— 11 Ã— 11

HCF = 2 Ã— 11 = 22

Example-20

**What is the greatest number, which when divides 3 consecutive odd numbers produces a remainder of 1.**

Solution

If x, y, z be 3 consecutive odd numbers, then the required number will be the HCF of

x - 1, y - 1 and z - 1.

Since x - 1, y - 1 & z - 1 are 3 consecutive even integers, their HCF will be 2. So answer is 2.

x - 1, y - 1 and z - 1.

Since x - 1, y - 1 & z - 1 are 3 consecutive even integers, their HCF will be 2. So answer is 2.