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:
- 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.
- If last digit of a is 1, then last digit of its complete square root is either 1 or 9.
- If last digit of a is 4, then last digit of its complete square root is either 2 or 8.
- If last digit of a is 5 or 0, then last digit of its complete square root is either 5 or 0.
- If last digit of a is 6, then last digit of its complete square root is either 4 or 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:
- Find the approx square root of given.
- Divide the given number by the prime numbers less than approx square root of number.
- 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:
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 =
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 |
Some Properties of Divisibility:
- 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).
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.
Comments
write a comment