Study Notes on Number Systems

By Yash Bansal|Updated : May 6th, 2019

Method to find the multiplication of 2-digit numbers:

Type - 1: AB × CD = AC / AD + BC / BD

For Example:

35 × 47 = 12 / 21 + 20 / 35 = 12 / 41 / 35 = 1645

Type - 2: AB × AC = A2/ A (B + C) / BC

For Example:

74 × 76 = 72 / 7(4 + 6) / 4 × 6

= 49 / 70 / 24 = 5624

Type - 3: AB × CC = AC / (A + B)C / BC

For Example:

35 × 44 = 3 × 4 / (3 + 5) × 4 / 5 × 4

= 12 / 32 / 20 = 12 / 32 / 20 = 1540

Method to find the multiplication of 3-digit numbers:

Type - 1: ABC × DEF = AD / AE + BD / AF + BE + CD / BF + CE / CF

For Example:

456 × 234 = 4 × 2 / 4 × 3 + 5 × 2 / 4 × 4 + 5 × 3 + 6 × 2 / 5 × 4 + 6 × 3 / 6 × 4

= 8 / 12 + 10 / 16 + 15 + 12 / 20 + 18 / 24 = 106704

To find the sum of series containing all repeating numbers, we start from the left multiply 7 by 1, 2, 3, 4, 5 & 6.

For Example:

777777 + 77777 + 7777 + 777 + 77 + 7 = ?

= 7 × 1 / 7 × 2 / 7 × 3 / 7 × 4 / 7 × 5 / 7 × 6

= 7 / 14 / 21 / 28 / 35 / 42 = 864192

Some properties of square and square root:

  1. Complete square of a is possible if its last digit is 0, 1, 4, 5, 6 & 9. If last digit of a no. is 2, 3, 7, 8 then complete square root of this no. is not possible.
  2. If last digit of a is 1, then last digit of its complete square root is either 1 or 9.
  3. If last digit of a is 4, then last digit of its complete square root is either 2 or 8.
  4. If last digit of a is 5 or 0, then last digit of its complete square root is either 5 or 0.
  5. If last digit of a is 6, then last digit of its complete square root is either 4 or 6.
  6. If last digit of a is 9, then last digit of its complete square root is either 3 or 7.

To Check whether the given number is Prime or not:

  1. Find the approx square root of given.
  2. Divide the given number by the prime numbers less than approx square root of number.
  3. If given number is not divisible by any of these prime number, then the given number is prime otherwise not.

For example: To check 359 is a prime number or not.

Approx sq. root = 19

Prime no. < 19 are 2, 3, 5, 7, 11, 13, 17

359 is not divisible by any of these prime nos. So 359 is a prime number.

  • There are 15 prime from 1 to 50.
  • There are 25 prime from 1 to 100.
  • There are 168 prime from 1 to 1000.

If a number is in the form of xn + an, then it is divisible by (x + a); if 'n' is odd.

If xn ÷ (x – 1), then remainder is always 1

If xn ÷ (x + 1) then

  • If n is even, then remainder is 1.
  • If n is odd, then remainder is x.

To Find the value of:

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To find Number of Divisors:

CASE 1:

If N is any number and is represented as N = an × bm× cp ×...... where a, b, c are prime no's, then the

Number of divisors of N = (n + 1) (m + 1) (p + 1) ....

For Example:

Find the number of divisors of 90000.

N = 90000 = 22 × 32 × 52 × 102 = 24 × 32 × 54

So, the number of divisors of N are = (4 + 1) (2 + 1) (4 + 1) = 75

CASE 2:

If N = an × bm × cp, where a, b, c are prime

Then set of co-prime factors of N = [(n + 1) (m + 1) (p + 1) – 1 + nm + mp + pn + 3mnp]

CASE 3:

If N = an × bm× cp..., where a, b & c are prime, then sum of the divisors =

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To find the last digit or digit at the unit's place of an:

Type - 1: 

If the last digit or digit at the unit’s place of a is 1, 5 or 6, whatever be the value of n, it will have the same digit at unit’s place, i.e.,

(.....1)n = (........1)

(.....5)n = (........5)

(.....6)n = (........6)

Type - 2:

If the last digit or digit at the units place of a is 2, 3, 5, 7 or 8, then the last digit of an depends upon the value of n and follows a repeating pattern in terms of 4 as given below :

n

last digit of (...2)n

last digit of (...3)n

last digit of (...7)n

last digit of (...8)n

4x+1

2

3

7

8

4x+2

4

9

9

4

4x+3

8

7

3

2

4x

6

1

1

6

Type - 3:

If the last digit or digit at the unit’s place of a is either 4 or 9, then the last digit of an depends upon the value of n and follows repeating pattern in terms of 2 as given 

n

last digit of (...4)n

last digit of (...9)n

2x

6

1

2x + 1

4

9

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Some Properties of Divisibility:

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  • A number of 3-digits which is formed by repeating a digit 3-times, then this number is divisible by 3 and 37.

For Example:

111, 222, 333, ....... all these numbers are divisible by 3 and 37 both.

  • A number of 6-digit which is formed by repeating a digit 6-times then this number is divisible by 3, 7, 11, 13 and 37.

For Example:

111111, 222222, 333333, 444444, .............all these numbers are divisible by 3, 7, 11, 13, 37 seperately

Special Tricks of Divisibility:

Divisible by 7 : We use osculator (– 2) for divisibility

For Example:

99995 : 9999 – 2 × 5 = 9989

9989 : 998 – 2 × 9 = 980

980 : 98 – 2 × 0 = 98

Now 98 is divisible by 7, so 99995 is also divisible by 7.

Divisible by 11 : In a number, if difference of sum of digit at even places and sum of digit at odd places is either 0 or multiple of 11, then is divisible by 11.

For example:

12342 ¸ 11

Sum of even place digit = 2 + 4 = 6 Sum of odd place digit = 1 + 3 + 2 = 6

Difference = 6 – 6 = 0

∴ 12342 is divisible by 11.

Divisible by 13 : We use osculator (+ 4) as

For Example:

876538 ÷ 13

876538: 8 × 4 + 3 = 35

5 × 4 + 3 + 5 = 28

8 × 4 + 2 + 6 = 40

0 × 4 + 4 + 7 = 11

1 × 4 + 1 + 8 = 13

13 is divisible by 13.

∴ 876538 is also divisible by 13.

Divisible by 17 : We use osculator (– 5) as

For Example:

294678: 29467 – 5 × 8 = 29427

27427: 2942 – 5 × 7 = 2907

2907: 290 – 5 × 7 = 255

255: 25 – 5 × 5 = 0

∴ 294678 is completely divisible by 17.

Divisible by 19 : We use osculator (+ 2)

For Example:

149264: 4 × 2 + 6 = 14

4 × 2 + 1 + 2 = 11

1 × 2 + 1 + 9 = 12

2 × 2 + 1 + 4 = 9

9 × 2 + 1 = 19

19 is divisible by 19

∴ 149264 is divisible by 19.

Highest Common Factor (HCF):

There are two methods to find the HCF–

(a) Factor method                 

(b) Division method

  • For two a and b if a < b, then HCF of a and b is always less than or equal to a.
  • The greatest number by which x, y and z completely divisible is the HCF of x, y and z.
  • The greatest number by which x, y, z divisible and gives the remainder a, b and c is the HCF of (x –a), (y–b) and (z–c).
  • The greatest number by which x, y and z divisible and gives same remainder in each case, that number is HCF of (x–y), (y–z) and (z–x).
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Least Common Multiple (LCM):

 There are two methods to find the LCM–

(a)  Factor method              

(b) Division method

  • For two numbers a and b if a < b, then C.M. of a and b is more than or equal to b.
  • If ratio between two numbers is a : b and their H.C.F. is x, then their L.C.M. = abx.
  • If ratio between two numbers is a : b and their C.M. is x, then their H.C.F. = x/ab
  • The smallest number which is divisible by x, y and z is L.C.M. of x, y and z.
  • The smallest number which is divided by x, y and z give remainder a, b and c, but (x – a) = (y – b) = (z – c) = k, then number is (L.C.M. of (x, y and z) – k).
  • The smallest number which is divided by x, y and z give remainder k in each case, then number is (L.C.M. of x, y and z) + k.
  • For two numbers a and b: LCM x HCF = a x b
  • If a is the H.C.F. of each pair from n numbers and L is L.C.M., then product of n numbers = an-1L.
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