Find the sum of square of first 35 natural numbers.

Sum of first n natural numbers = n(n+1)/2
= 25 ( 25 + 1 ) / 2
= ( 25 x 26 ) / 2
= 325

For n =1
7⁶ⁿ - 6⁶ⁿ = 7⁶ - 6⁶
? ( 7³ )² - ( 6³ )²
? ( 7³ - 6³ )( 7³ + 6³ )
= ( 343 - 216 )( 343 + 216 )
= 127 x 559

? it is clearly divisible by 127.

Solving the question by taking two odd numbers greater than 1 i.e. 3 and 5,
for n =3, n( n² -1 ) = 3( 9 - 1) = 24
for n = 5 n( n² -1 ) = 5( 25 - 1) = 120

Using option we find that both the number are divisible by 24.

Factors of composite number = (P₁ +1)(P₂+1)(P₃+1)......(P_n+1).
Where P₁,P₂,P₃... P_n are the powers of respective prime factors.

So,
120= 2³ x 3¹ x 5¹
Factors=(3+1)(1+1)(1+1)=16.

the number of pieces of chocolate left with manju =1- [(1/4) + (1/3) + (1/6)]
= 1 - [ (3 + 4 + 2)/12)
= 1 - (9/12)
= 3/12
Hence number of piece of chocolate left with in manju is 3

Sum of cubes of first n natural numbers = [ n(n+1) ]²
Given n = 15

? Required sum = [ ( 15 x 16 ) / 2 ] ²
= (15 x 8 )²
= 120²
= 14400

Sum of first n even numbers = n( n +1 )
Given n = 84
? Required sum = 84 ( 84 + 1 )
= 84 x 85
= 7140

Sum of first n odd number = n²
Given, n = 37
? Required sum = 37²
= 37 x 37
= 1369

First 15 multiple of 8 are 8, 16, 24, ......, 120
Sum = 8 (1 + 2 + 3 + 4 ......+ 15)
= 8 [ n(n + 1) / 2]
= 8 x 15 x 8
960

Required unique digit = Unique digit in (7)⁷⁵⁴
= Unique digit in { (7⁴)¹⁸⁸ x 7² }
= Unique digit in ( 1 x 49 ) = 9