# FORMULAE: Remainder System

### 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.

Step 2: 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

Eg:

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 21 * 23 * 25 / 100 = 525 * 23 / 100 ; 25 * 23 / 100 = 575 / 100

Thus the last two digits are 75.

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

### 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 = 4.4

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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