Division and its short-cut methods to find
A number (dividend) is said to be divisible by another number (called divisor) when quotient is a natural number and remainder is zero. In other words, we can say that when a number (dividend) is divisible by another number (divisor), the dividend can be expressed as multiple of divisor. The relation so obtained can be given by,
Dividend = Divisor × Quotient
Short-cut techniques to check divisibility of a number
a) Divisibility by 2:
A number is divisible by 2, when its unit's digit is either even or zero.
For example, the number 4, 6, 10, 24, 36, 388, 8964 etc. are divisible by 2.
b) Divisibility by 3:
A number is divisible by 3 when the sum of its digit is divisible by 3.
Example: 426: 4 + 2 + 6 = 12; which is divisible by 3. Hence, 426 too is divisible by 3.
8349 : 8 + 3 + 4 + 9 = 24; which is divisible by 3. Hence, 8349 is divisible by 3.
c) Divisibility by 4:
A number is divisible by 4, when the number formed by its two extreme right digits is either divisible by 4 or both these digits are zero.
Example, 324, 5632, 4500, 4720 are divisible by 4 as they satisfy the above mentioned conditions.
d) Divisibility by 5:
A number is divisible by 5, when its unit's digit is either 5 or zero.
For example, 2145, 890, 4500, are divisible by 5 they have either 5 or 0 at their unit's place.
e) Divisibility by 6:
A number is divisible by 6, when it is divisible by 2 as well as 3.
For example, 840 is divisible by 6 as it is divisible by 3 and as well as 2.
f) Divisibility by 7:
A number is divisible by 7, if unit digit of the given number is doubled and then it is subtracted from the number obtained after omitting the unit digit. If the remainder is divisible by 7, then the given number is also divisible by 7. If number is large enough, then the process is repeated till such number is obtained.
Example,
343 ====> doubling the unit digit 3 of 343, we get 6, then 34 - 6 = 28. Since, 28 is divisible by 7 then 343 is also divisible by 7.
5499838 ===> On doubling 8 and subtracting from the number which is obtained after omitting the unit digit, we get
549983 - 16 = 549967
54996 - 14 = 54982
5498 - 4 = 5494
549 - 8 = 541
54 - 2 = 52
Since 52 is not divisible by 7 then number is not divisible by 7.
g) Divisibility by 8:
A number is divisible by 8 if last three digits are divisible by 8. 5248 is divisible by 8 as 248 are divisible by 8.
h) Divisibility by 9:
A number is divisible by 9 when the sum of its digit is divisible by 9.
Example: 576: 5 + 7 + 6 = 18; this is divisible by 9. Hence, 576 too is divisible by 9.
i) Divisibility by 10:
A number is divisible by10 if it ends in zero. e.g. 100, 2310, 3450 etc.
j) Divisibility by 11:
A number is divisible by 11 if the difference of the sum of the digits in the odd places and sum of the digits in even place is zero or divisible by 11. e.g. 98518343; sum of digits of odd place (9 + 5 + 8 + 4) - sum of digit at even places (8 + 1 + 3 + 3) = 26 - 15 = 11. So given number is divisible by 11.
k) Divisibility by 12:
A number is divisible by 12, if it is divisible by both 3 and 4. This is because 4 and 3 are the two factors or sub-multiples of 12.
Example: 9612, 3400 etc. are divisible by 12.
Must know facts about divisibility:
If in a number the digits repeats thrice then the number is divisible by 3 and 37.
Example:
222, 777, 131313.
If in a number the digits repeats six times then the number is divisible by 3, 7, 11, 13, and 10101.
Example:
222222, 121212121212.
The highest power of a prime number p, which divides n! Exactly is given by
$$\frac{{\text{n}}}{{\text{p}}} + \frac{{\text{n}}}{{{{\text{p}}^2}}} + \frac{{\text{n}}}{{{{\text{p}}^3}}} + \,.\,.\,.\,.\,.\,.\,.\,.\,.$$
Where [x] denotes the greatest integer less than or equal to x.
For example:
i) Find the highest power of 5 which is present in the value of 127!
ii) When 127! Is divided by 5n the result is an integer. Find the highest possible value for n.
iii) Find the number of zeroes in 127!
Solution:
In each of above case, answer will be:
$$\eqalign{
& = \frac{{127}}{5} + \frac{{127}}{{{5^2}}} + \frac{{127}}{{{5^3}}} \cr
& = 25 + 5 + 1 \cr} $$
[Note that we have taken integral value only not the fractional. For example $$\frac{{127}}{5}$$ = 25.4 but we have taken 25 only and left the fraction and so on].
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