p^{6} - p = (m^{2} + m + 6)(p - 1)

p( p^{5} - 1) = (m^{2} + m + 6)(p - 1)

p ( p^{4} + p^{3} + p^{2} + p + 1 )(p - 1) = (m^{2} + m + 6)(p - 1)

p ( p^{4} + p^{3} + p^{2} + p + 1 ) = (m^{2}+ m + 6)

If p is a prime greater than 2 then LHS is an odd number , but RHS is even for all integer value of m

So no solution for p> 2.

When p = 2 then m^{2} + m - 56 = 0 so m= -8, 7 Only one solution m=7 ; p=2

x + 1/x = - √3

Cubing both sides we get

x³ + 1/x³ + 3 ( x + 1/x) = - 3√3

x³ + 1/x³ - 3 √3 = -3√3

x³ + 1/x³ = 0

x^6 + 1 = 0. x³

x^6 = -1

x^42 + x^48 + x^54 + x ^60 + x^66 + x ^72

( -1)^7+ (-1)^8 + ( -1)^9 + ( -1)^10 + ( -1)^11 + ( -1)^12

0 .

got it (y)

x + 1/x = - √3

x³ + 1/x³ = 0

x^42 + x^48 + x^54 + x ^60 + x^66 + x ^72

= x^45(x³ + 1/x³) + x^57(x³ + 1/x³) + x^69(x³ + 1/x³)

= 0

Hello sir , Kindly share the solution to question 16 and 17

16)

D = b^2- 4ac

Any square is of the form 4k or 4k+1

So D can't be of the form 4k+3 hence option 1

17. For each element listed, there is exactly one other element such that the two elements sum to 11. Thus, we can list all the 10 numbers above as 5 pairs of numbers, such that each pair sums to 11. The problem then can be solved as follows:

in any given subset with no two elements summing to 11, at most one element from each pair can be present. Thus, there are 3 ways in which each pair can contribute to a given subset (no element, the first element in the pair, or the second element in the pair). Since there are 5 pairs, the total number of ways to construct a subset with no two elements summing to 11 is 3^{5} = 243.

Both the questions, please!!

98

97th answer is 5292 ?

Yes, answer for 97 is correct. Can you explain your method?

Alternative Approach:

Sir,Is this solution correct

Volume of pyramid : 1/3 of area of base x height

So 1/3 x 2562 x 40 - 1/3 x 144 x 30

In order to divisible by 6.

sum of digits should be a multiple of 3 and last digit should be even.

so, when 0 is not included

2346 => 4! numbers subtract cases when 3 is at units place

4!-3! = 18 numbers

When 2 is not included : 3460 => sum 13 , not divisible by 3

When 3 is not included : 2460=> sum 12

Total 4! numbers subtract cases when 0 is at thousands place : 24 - 6 = 18

When 4 is not included : 2306 => sum 11 , not divisible by 3

When 6 is not included : 0234 => sum 9

14 numbers

Total : 14+ 18 + 18 =50 numbers

What is the value of 2 power log2 power log2 power log2 and so on?

Sir , what should be the answer of this question?

Solution with explanation please

solution please

solution for questions 10 & 12

Solution to the Q. 10

a and b are roots

So , a + b = a² + b²

and ab = a²b²

=> ab(ab - 1)

i) a = 0

=> b = 0 or 1

ii) b = 0

=> a = 0 or 1

iii) ab = 1

a + 1/a = a² + 1/a²

=> a⁴ - a³- a + 1 = 0

=> (a³ - 1)(a - 1) = 0

=> a = 1 or w or w²

=> b = 1 or w² or w

So, equation are x² = 0

x² - x= 0

x² - 2x + 1 = 0 and x² + x + 1 = 0

4 such equations

Solution to the Q. 12.

12

Alternative Approach:

put x = 0 in the expression x² - 2ax + a² + a - 3

it reduces to

a² + a - 3 = 0

a = -1/2 + (√13)/2 and -1/2 + √13/2

Here roots are less than 2 hence option (B) and (C) got eliminated.

Hence , Option (A ) is the correct answer .

solution please

16 ,b , c are in A.P

so b=(c+16)/2

c=2b-16......................(1)

16 , b+3, c+15 are in G.p

(b+3)^2= 16(15+c)

putting value of c from (1)

b^2+9+6b =16(2b-1)

b^2 -26b +25 =0

b=25 or 1

putting value of b in (1)

c=49 or 1

so smallest value is 1 hence option B