Remainder Theorem

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 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.