FINDING THE REMAINDER IN DIVISIONS INVOLVING POWERS OF
NUMBERS
Q. Find the remainder when 2^58 is divided by 7?
Ans. Let us find the pattern that remainders follow when
successive powers of 2 are dividedby 7.
Remainder of 2^1 /7 is 2
Remainder of 2^2 /7 is 4
Remainder of 2^3 /7 is 1
Remainder of 2^4 /7 is 2
We can see that remainders are repeating after 3 steps, so
the cyclicity is 3
Divide the power i.e. 58 by cyclicity which is 3. Remainder
obtained is 1 when 58/3
So we can say that remainder when 2^58 is divided by 7 is
same as the remainder of 2^1/7 which is 2.
- Some Basic Formulae:
1. 1. Identities
- (a + b)(a - b) = (a2 - b2)
- (a + b)2 = (a2 + b2 + 2ab)
- (a - b)2 = (a2 + b2 - 2ab)
- (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
- (a3 + b3) = (a + b)(a2 - ab + b2)
- (a3 - b3) = (a - b)(a2 + ab + b2)
- (a3 + b3 + c3 - 3abc) = (a + b + c)(a2 + b2 + c2 - ab - bc - ac)
- When a + b + c = 0, then a3 + b3 + c3 = 3abc.
2. The
sum of first n natural numbers is = [n(n+1)]/2.
Example: The sum first 49 natural number is
= [49(49+1)]/2 = [49(50)]/2 = 49 * 25 = 1225
3. The sum of the squares of the first n natural numbers is = [n(n+1)(2n+1)]/6.
Example: The sum
of the squares of the first 10 natural numbers is = 10(11)(21)/6 = 385.
4. The
sum of the cubes of the first n natural numbers is = [n(n+1)/2]2
Example: The sum of the cubes of the first 10
natural numbers is = [10(10+1)/2]2
= [(10*11)/2]2 = 552 = 3025
5.
The sum of the first n odd natural numbers is = n2
Example: The sum of the first 24 odd natural numbers is = 242 = 576.
6.
The sum of the first n even natural numbers is = n(n+1)
Example: The sum of the first 10 even natural numbers is = 10(10 + 1) = 10 * 11 = 110
7.
The
p/q form of any recurring number = The non-recurring and recurring part written
once)-(the non-recurring part)/ As many 9’s as the number of digits in the
recurring part followed by as many 0’s as digits in the non-recurring part.
Example: : Express 0.5858585….. as fraction
= (585- 5/(990)
=58/99
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