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Clearly 2,4,8,16……..1024 form a GP. With a = 2 and r = 4/2 = 2. 

Let the number of terms be n . 

Then 2 x 2^n-1 =1024 or 2^n-1 =512 

2^n-1= 2⁹. 

n - 1 = 9 or n = 10.

i) 5793405 x 9999 

= 5793405(10000-1)

= 57934050000 - 5793405

= 57928256595

ii) 839478 x 625 

= 839478 x 5⁴

= 8394780000/16

=  524673750

On dividing the given number by 342, let k be the quotient and 47 as remainder. 

Then, number – 342 k + 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9. 

The given number when divided by 19, gives (18k + 2) as quotient and 9 as remainder.

Required unit's digit = unit's digit in (4)¹⁰² + (4)¹⁰³. 

Now, 4² gives unit digit 6. 

(4)¹⁰² gives unit digit 6. 

(4)¹⁰³ gives unit digit of the product (6 x 4) i.e., 4. 

Hence, unit's digit in (264)m + (264)103 = unit's digit in (6 + 4) = 0.

Clearly, unit's digit in the given product = unit's digit in 7¹⁵³ x 1⁷² 

Now, 74 gives unit digit 1. 

7¹⁵² gives unit digit 1, 

7¹⁵³ gives unit digit (1 x 7) = 7. 

Also, 1⁷² gives unit digit 1. 

Hence, unit's digit in the product = (7 x 1) = 7.

In Hindu Arabic system, we use ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 called digits to represent any number. 

All natural numbers, 0 and negatives of counting numbers i.e., {…, - 3 , - 2 , - 1 , 0, 1, 2, 3,…..} together form the set of integers.

(i) Positive Integers : {1, 2, 3, 4, …..} is the set of all positive integers. 

(ii) Negative Integers : {- 1, - 2, - 3,…..} is the set of all negative integers. 

(iii) Non-Positive and Non-Negative Integers : 0 is neither positive nor negative. So, {0, 1, 2, 3,….} represents the set of non-negative integers, while {0, - 1 , - 2 , - 3 , …..} represents the set of non-positive integers

Counting numbers 1, 2, 3, 4, 5,..... are called natural numbers

All counting numbers together with zero form the set of whole numbers. Thus, 

(i) 0 is the only whole number which is not a natural number. 

 (ii) Every natural number is a whole number.

(i) Let x - 1936248 = 1635773.

Then, x = 1635773 + 1936248 =3572021. 

(ii) Let 8597 - x = 7429 - 4358. 

Then, x = (8597 + 4358) - 7429 

= 12955 - 7429 

= 5526. 

Given exp 

= (896)² - (204)² 

= (896 + 204) (896 - 204) 

= 1100 x 692 

= 761200.

Given exp 

= (387)² + (114)² + (2 x 387 x 114) 

= a² + b² + 2ab, where a = 387,b = 114 

= (a+b)² 

= (387 + 114 )² 

= (501)² 

= 251001.

2¹⁰ = 1024. Unit digit of 2¹⁰ x 2¹⁰ x 2¹⁰ is 4 [as 4 x 4 x 4 gives unit digit 4]. 

Unit digit of 231 is 8. 

Now, 8 when divided by 5, gives 3 as remainder. 

Hence, 231 when divided by 5, gives 3 as remainder.

Given exp 

= (81)² + (68)² – 2 x 81 x 68 

= a² + b² – 2ab, Where a = 81,b = 68 

= (a-b)² 

= (81 –68)² 

= (13)² 

= 169.

(i) Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3. Hence, 541326 is divisible by 3. 

(ii) Sum of digits in 5967013 =(5 + 9 + 6 + 7 + 0 + 1 + 3) = 31, which is not divisible by 3. Hence, 5967013 is not divisible by 3.

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 .

Hence, the digit in place of * must be 2.

The given numbers are 1, 3, 5, 7, ..., 99. 

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

Let it contain n terms. Then,  

1 + (n - 1) x 2 = 99 or n = 50. 

Required sum = n/2 (first term + last term) 

= 50/2 (1 + 99) 

= 2500.

(i) The number formed by the last two digits in the given number is 94, which is not divisible by 4. Hence, 67920594 is not divisible by 4.

(ii) The number formed by the last two digits in the given number is 72, which is divisible by 4. Hence, 618703572 is divisible by 4.

The required numbers are 14, 21, 28, 35, .... 77, 84. 

This is an A.P. with a = 14 and d = (21 - 14) = 7. 

Let it contain n terms. 

Then, T_n = 84 => a + (n - 1) d = 84 

 => 14 + (n - 1) x 7 = 84 or n = 11. 

Required number of terms = 11.

Since the given number is divisible by 5, so 0 or 5 must come in place of $. But, a number ending with 5 is never divisible by 8. So, 0 will replace $. 

Now, the number formed by the last three digits is 4*0, which becomes divisible by 8, if * is replaced by 4. 

Hence, digits in place of * and $ are 4 and 0 respectively.

All 2 digit numbers divisible by 3 are : 12, 51, 18, 21, ..., 99. 

This is an A.P. with a = 12 and d = 3. 

Let it contain n terms. 

Then, 12 + (n - 1) x 3 = 99 or n = 30. 

Required sum = 30/2 x (12+99) = 1665.

(Sum of digits at odd places) - (Sum of digits at even places) 

 = (8 + 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11

Hence, 4832718 is divisible by 11.

Smallest number of 6 digits is 100000. 

On dividing 100000 by 111, we get 100 as remainder. 

Number to be added = (111 - 100) - 11. 

Hence, required number = 100011.

On dividing 3105 by 21, we get 18 as remainder. 

Number to be added to 3105 = (21 - 18) - 3. 

Hence, required number = 3105 + 3 = 3108.

24 = 3 x 8, where 3 and 8 are co-primes. 

The sum of the digits in the given number is 36, which is divisible by 3. So, the given number is divisible by 3. 

The number formed by the last 3 digits of the given number is 744, which is divisible by 8. So, the given number is divisible by 8. 

Thus, the given number is divisible by both 3 and 8, where 3 and 8 are co-primes. So, it is divisible by 3 x 8, i.e., 24.

On dividing 3000 by 19, we get 17 as remainder. 

Number to be added = (19 - 17) = 2.

Everyone knows that Human and Horses have one head and human have two legs and horses have four legs.

Let number of human = x and number of horses = y

Now given that Total number of heads = 74 and Total number of legs = 196

So, x + y = 74 ---------------------(1) and 

2x + 4y = 196 ---------------------(2)

Solving the equations we get

x = 50

y = 24

Hence, the number of humans are 50 and the number of horses are 24