FORMULAE


Remainder System


Find the remainder of a large numbers divided by a numbers:

Step 1: The remainder of multiplication of a numbers by a value is same as multiplication of remainder of individual terms.

1421 * 1423 * 1425 / 12 same as R (1421/12) * R(1423/12) * R(1425/12) where R(p/q) is remainder of p/q.
Therefore we get 5 * 7 * 9 / 12 = 35 * 9 / 12 = 99 / 12 = 3.



Method II --- Find the remainder of a large numbers divided by a numbers:

Step 1: The remainder of multiplication of a numbers by a value is same as multiplication of remainder of individual terms.

11 * 10 * 9 / 12 same as NR (11/12) * NR(10/12) * NR(9/12) where NR(p/q) is negative remainder of p/q.

So if 35/12 then remainder shall be 11 and negative remainder shall 11 - 12 = -1

Step 2: -1 * -2 * -3 / 12 = - 6 so remainder is 6.

Calculate remainder when dealing with large powers:

Step 1: Can question be converted to following formulae.
(ax + 1)n / a = Remainder is 1 and (ax - 1)n / a = Remainder is -1 i.e. a-1.



E.g:
37124556 / 9 = (9*4 + 1)124556 / 9 = Remainder is 1
35124556 / 9 = (9*4 - 1)124556 / 9 = Remainder is -1 i.e 9-1 = 8

Find the last two digits of a large multiplication:

Step 1: The last two digits of multiplication of a numbers by a value is same as finding the remainder of multiplication terms when divided by 100.

1421 * 1423 * 1425 / 100 same as R (1421/100) * R(1423/100) * R(1425/100) where R(p/q) is remainder of p/q.
Therefore we get 1 * 3 * 5 / 100 = 15

Thus the last two digits are 15.

Find the last digits of a large multiplication:

Step 1: The last digit of multiplication of a numbers by a value is same as finding the remainder of multiplication terms when divided by 10.

1421 * 1423 * 1425 / 10 same as R (1421/10) * R(1423/10) * R(1425/10) where R(p/q) is remainder of p/q.
Therefore we get 1 * 3 * 5 / 10 = 15 / 10 = 5

Thus the last digit is 5.



Practice Exercise: Remainder theory
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Progressions


Arithmetic Progressions


The sequence of numbers like 1,2,3,4... are said to be in arithmetic progression with common difference d = 1; Generalizing this the arithmetic series is of the form: a, a+d, a+2d, a+3d... a+(n-1)d.


Common difference d is Tn - T(n-1) i.e. next term minus previous term. This is uniform throughout.


Properties:




  1. Corresponding terms of A.P = first and last, second and second last etc. So if an AP is 1,2,3,4,5,6 then corresponding terms are (1,6) , (2,5) , (3,4).
  2. Average of the A.P = mean of the corresponding terms e.g: 1+6/2 or 2+5/2 etc
  3. Sum of A.P = ( Average of A.P. ) * ( Number of terms )
  4. Finding the Common difference given two terms in an A.P =If Tx and Ty are the terms in an AP at position 'x' and 'y'. The common difference = > ( Ty - Tx ) = Common difference * (y-x). E.g: So if 3rd term is 8 and 8th term is 28 the common difference is (28 - 8) = d * (8-3) = > 20 = 5*d and d = 4.



Geometric Progression


The series is in geometric progression if the numbers increase or decrease by a common ratio. So the series is a, ar, ar2, ar3, ar4.... arn-1


If r > 1 then Sum = a ( rn - 1) / ( r - 1)

If r < 1 then Sum = a ( 1 - rn) / ( 1 - r )


If an infinite geometric progression series is Sum = a / ( 1 - r )





Practice Exercise: Progressions
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