
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. a1.
E.g:
37^{124556} / 9 = (9*4 + 1)^{124556} / 9 =
Remainder is 1
35^{124556} / 9 = (9*4  1)^{124556} / 9 =
Remainder is 1 i.e 91 = 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.
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+(n1)d.
Common difference d is T_{n}  T_{(n1) }i.e.
next term minus previous term. This is uniform throughout.
_{
}
Properties:
 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).
 Average of the A.P = mean of the
corresponding terms e.g: 1+6/2 or 2+5/2 etc
 Sum of A.P = ( Average of A.P. ) * (
Number of terms )
 Finding the Common difference given two terms in
an A.P =If T_{x} and T_{y} are
the terms in an AP at position 'x' and 'y'. The common
difference = > (_{ }T_{y}  T_{x}
) = Common difference * (yx). E.g: So
if 3rd term is 8 and 8th term is 28 the common difference
is (28  8) = d * (83) = > 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, ar^{2}, ar^{3}, ar^{4}.... ar^{n1}
^{
}
If r > 1 then Sum = a ( r^{n}  1) / ( r  1)
If r < 1 then Sum = a ( 1  r^{n}) / ( 1  r )^{
}
^{
}
If an infinite geometric progression series is Sum = a / (
1  r )^{
}
^{
}