Divisibility
| Numbers | IF A Number | Examples |
|---|---|---|
| Divisible by 2 | End with 0,2,4,6,8 are divisible by 2 | 254,326,3546,4718 all are divisible by 2 |
| Divisible by 3 | Sum of its digits is divisible by 3 | 375,4251,78123 all are divisible by 3. [549=5+4+9][5+4+9=18]18 is divisible by 3 hence 549 is divisible by 3. |
| Divisible by 4 | Last two digit divisible by 4 | 5648 here last 2 digits are 48 which is divisible by 4 hence 5648 is also divisible by 4. |
| Divisible by 5 | Ends with 0 or 5 | 225 or 330 here last digit digit is 0 or 5 that mean both the numbers are divisible by 5. |
| Divisible by 6 | Divides by Both 2 & 3 | 4536 here last digit is 6 so it divisible by 2 & sum of its digit (like 4+5+3+6=18) is 18 which is divisible by 3.Hence 4536 is divisible by 6. |
| Divisible by 8 | Last 3 digit divide by 8 | 746848 here last 3 digit 848 is divisible by 8 hence 746848 is also divisible by 8. |
| Divisible by 10 | End with 0 | 220,450,1450,8450 all numbers has a last digit zero it means all are divisible by 10. |
| Divisible by 11 | [Sum of its digit in odd places-Sum of its digits in even places]= 0 or multiple of 11 |
Consider the number 39798847
(Sum of its digits at odd places)-(Sum of its digits at even places)(7+8+9+9)-(4+8+7+3)
(23-12)
23-12=11, which is divisible by 11. So 39798847 is divisible by 11. |
Division & Remainder Rules
Suppose we divide 45 by 6
hence ,represent it as:
dividend = ( divisor✘quotient ) + remainder
or
divisior= [(dividend)-(remainder] / quotient
could be write it as
x = kq + r where (x = dividend,k = divisor,q = quotient,r = remainder)
Example:
On dividing a certain number by 342, we get 47 as remainder. If the same number is divided by 18, what will be the remainder ?
Number = 342k + 47
( 18 ✘19k ) + ( 18 ✘2 ) + 11
18 ✘( 19k + 2 ) +11.
Remainder = 11
Sum Rules
(1+2+3+.........+n) = 1/2 n(n+1)
(12+22+32+.........+n2) = 1/6 n (n+1) (2n+1)
(13+23+33+.........+n3) = 1/4 n2 (n+1)2
Arithmetic Progression (A.P.)
a, a + d, a + 2d, a + 3d, ....are said to be in A.P. in which first term = a and common difference = d.
Let the nth term be tn and last term = l, then
b) Sum of n terms = n/2 [2a + (n-1)d]
c) Sum of n terms = n/2 (a+l) where l is the last term
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