8888 + 888 + 88 + 8 = 9872

Let x - 1936248 = 1635773. Then, x = 1635773 + 1936248 = 3572021

We may analyse the given equation as shown : Clearly, 2 + P + R + Q = ll. So, the maximum value of Q can be (11 - 2) i.e., 9 (when P = 0, R = 0);

 
    1 2
 + 5 P 9
 + 3 R 7
 + 2 Q 8
---------
 = 1 11 4    

5793405 x 9999 = 5793405(10000-1) = 57934050000-5793405 = 57928256595.b

986 x 137 + 986 x 863 = 986 x (137 + 863) = 986 x 1000 = 986000.

1605 x 1605 = \((1605)^2\)= \((1600 + 5)^2 \)= \((1600)^2 + (5)^2\) + 2 x 1600 x 5 = 2560000 + 25 + 16000 = 2576025

\((a^2 + b^2)\) = \(\sqrt{(a+b)^2+(a-b)^2}\)....formula
\((313)^2 + (287)^2\)= \(\sqrt{(313 + 287)^2 + (313 - 287)^2}\) = \(\sqrt{(600)2 + (26)2}\) = \(\sqrt{(360000 + 676)}\) = 180338.

Clearly, 16 > Ö241. Prime numbers less than 16 are 2, 3, 5, 7, 11, 13. 241 is not divisible by any one of them. 241 is a prime number

Clearly, unit's digit in the given product = unit's digit in \(7^{163} * 1^{72}.\)
Now, 74 gives unit digit 1. \(7^{162}\) gives unit digit 1, =\(7^{163}\) gives unit digit (l x 7) = 7. Also, \(1^{72}\)gives unit digit 1. Hence, unit's digit in the product = (7 x 1) = 7.

Required unit's digit = unit's digit in \((4)^{102} + (4)^{103}\). Now,\( 4^2\) gives unit digit 6 =\((4)^{102}\) gives unjt 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.

\((4)^{11} * (7)^5 * (11)^2\)= \((2*2)^{11} *(7)^5 *(11)^2\) = \(2^{11} * 2^{11} * 7^5 * 11^2\) = \(2^{22} * 7^5 * 11^2\) Total number of prime factors = (22 + 5 + 2) = 29.

Given exp = \((896)^2 - (204)^2\) = (896 + 204) (896 - 204) = 1100 * 692 = 761200.

Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3. Hence, 541326 is 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 number formed by the last two digits in the given number is 72, which is divisible by 4. Hence, 618703572 is divisible by 4.

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.

(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.

24 = 3 x 8, where 3 and 8 are co-primes. The sum of the digits in the 52563744 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.

On dividing 2000 by 17, we get 11 as remainder. =Required number to be subtracted = 11.

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.

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.

Divisor = (Dividend - Remainder)/ Quotient= (15968-37)/89 =179

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

\(2^{10}\) = 1024. Unit digit of \(2^{10}\) x \(2^{10}\) x \(2^{10}\) 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.

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, Tn = 84 => a + (n - 1) d = 84 => 14 + (n - 1) x 7 = 84 or n = 11. Required number of terms = 11.

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) * 2 = 99 or n = 50. Required sum = \(\frac{n *(first term + last term)}{2}\) = \(\frac {50 *(1 + 99)}{2}\) = 2500.

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 = \(\frac {30 * (12+99)}{2}\) = 1665.

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 = \(2^9\).

Let 8597 - x = 7429 - 4358. Then, x = (8597 + 4358) - 7429 = 12955 - 7429 = 5526.

839478 x 625 = 839478 x \(5^4\) = \(\frac {8394780000}{16}\)= 524673750.

983 x 207 - 983 x 107 = 983 x (207 - 107) = 983 x 100 = 98300.

1398 x 1398 - (1398)2 = (1400 - 2)2= (1400)2 + (2)2 - 2 x 1400 x 2 =1960000 + 4 - 5600 = 1954404.

Given exp = \((387)^2\)+ \((114)^2\)+ (2 x 387x 114) = \(a^2\) + \(b^2\) + 2ab, where a = 387,b=114 = \((a+b)^2\) = \((387 + 114 )^2\) = \((501)^2\) = 251001.

Given exp = \((81)^2\) + \((68)^2\) – 2x 81 x 68 = \(a^2\) + \(b^2\) – 2ab,Where a =81,b=68 = \((a-b)^2\)= \((81 –68)^2\) = \((13)^2\) = 169.