p6 - p = (m2 + m + 6)(p - 1)
p( p5 - 1) = (m2 + m + 6)(p - 1)
p ( p4 + p3 + p2 + p + 1 )(p - 1) = (m2 + m + 6)(p - 1)
p ( p4 + p3 + p2 + p + 1 ) = (m2+ 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 m2 + 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 35 = 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