NUMBERS AND THEIR CLASSIFICATION
A Number p may be,
(i). a natural number (N)
(ii). a whole number (w)
(iii) an integer (Z)
(iv) a rational number (Q)
(v) a real number (R)
(vi) an irrational number
For Example,
Set of natural number is {1, 2, 3,…..}
Set of whole number is {0, 1, 2, 3,….}
Set of integers is {… -3, -2, -1, 0, 1, 2, 3, …}
Set of rational numbers is {…. }
Besides the above cited number,
we often come across numbers like -8, -7, -5, -1, 3 + -7 and so on.Thes are undefined numbers, called complex numbers.
A positive integer, except 1, is a prime number, if its factors are 1 and the number itself 2, 3, 5, 7, … are prime numbers.
A Number p may be,
(i). a natural number (N)
(ii). a whole number (w)
(iii) an integer (Z)
(iv) a rational number (Q)
(v) a real number (R)
(vi) an irrational number
For Example,
Set of natural number is {1, 2, 3,…..}
Set of whole number is {0, 1, 2, 3,….}
Set of integers is {… -3, -2, -1, 0, 1, 2, 3, …}
Set of rational numbers is {…. }
Besides the above cited number,
we often come across numbers like -8, -7, -5, -1, 3 + -7 and so on.Thes are undefined numbers, called complex numbers.
A positive integer, except 1, is a prime number, if its factors are 1 and the number itself 2, 3, 5, 7, … are prime numbers.
TEST OF DIVISIBILITY OF NUMBERS
There are certain tests for divisibility of numbers by any of the numbers 2, 3, 4, 5, 6, 8, 9, 10 and 11 such that by simply examining the digits in the given number, one can easily determine whether or not a given number is divisible by any of these numbers. Such tests are detailed as follows: 1. Divisibility by 2 If the last digit is an even number or it has zero (0) at the end. Example: 74, 148, 1210 are all divisible by 2. 2. Divisibility by 3 If the sum of the digits of the given number is divisible by 3. Example: The sum of the digits of number 3705 is 3 + 7 + 0 + 5 = 15 Since 15 (the sum of digits) is divisible by 3. the number 3705 is also divisible by 3. 3. Divisibility by 4 If the number formed by the last two digits of the given number is divisible by 4, or if the last two digits are ‘00’. Example: 216560 is a number whose last two digit are 60. Since 60 is divisible by 4, the given number 216560 is also divisible by 4. 4. Divisibility by 5 If the last digit of the given number is 0 or 5. Example: 865, 1705, 4270 are all divisible by 5. 5. Divisibility by 6 If the given number is divisible by both 2 and 3. Example: Let us consider the number 629130. It has 0 as the last digit, so it is divisible by 2. Sum of the digits = 6 + 2 + 9 + 1 + 3 + 0 = 21 This sum 21 is divisible by 3, so the number is divisible by 3. Since, 620130 is divisible by both 2 and 3, the number is also divisible by 6. 6. Divisibility by 8 If the number formed by the last three digits of the given number is divisible by 8 or if the last three digits are ‘000’. Example: The number 81976 has 976 as the last three digits. Since 976 is divisible by 8, 81796 is also divisible by 8. The number 6145000 ends with ‘000’ and so, it is divisible by 8. 7. Divisibility by 9 If the sum of the digits of the given number is divisible by 9. Example : 870111 is a number the sum of whose digits = 8 + 7 + 0 + 1 + 1 + 1 = 18 Since 18 (sum of digits) is divisible by 9, the number 870111 is also divisible by 9. 8. Divisibility by 10 If the last digit of the number is zero (0). Example : 730 has 0 at the end, so it is divisible by 10. 9. Divisibility by 11 If the difference of the sum of its digits in odd places (i.e. first, third, fifth …) and the sum of its digits in even places (i.e. second, fourth, sixth …) is either zero (0) or a multiple of 11. Example : Let us consider the number 647053. Sum of digits at odd places = 6 + 7 + 5 = 18 Sum of digits at even places = 4 + 0 + 3 = 7 Difference of the sums = 18 − 7 = 11 Since the difference (= 11) is a multiple of 11, 647053 is also divisible by 11 Let us consider another number 9610260. Sum of digits at odd places = 9 + 1 + 2 + 0 = 12 Sum of digits at even places = 6 + 0 + 6 = 12 Difference of the sums = 12 − 12 = 0 Since the difference is 0, 9610260 is divisible by 11. |
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General Properties of Divisibility
There are some general properties of divisibility that help in determining the divisibility of a natural number by other natural numbers (other than detailed in 1.2)
Property 1
If a number x is divisible by another number y, then any number divisible by x, will also be divisible byy and by all the factors of y.
Example : The number 84 is divisible by 6. Thus any number that is divisible by 84, will also be divisible by 6 and also by the factors of 6, i.e. By 2 and by 3.
Property 2
If a number x is divisible by two or more than two co-prime numbers than x is also divisible by the product of those numbers.
Example : The number 2520 is divisible by 5, 4, 13 that are prime to each other (i.e. co-prime), so, 2520 will also be divisible by 20 (= 5 × 4), 65 (= 5 × 13), 52 (= 4 × 13).
Property 3
If two numbers x and y are divisible ‘p’, then their sum x + y is also divisible by p.
Example : The numbers 225 and 375 are both divisible by 5. Thus their sum 225 + 375 = 6000 will also be divisible by 5.
Note : It is also true for more than two numbers.
Property 4
If two numbers x and y are divisible by a number ‘p’, then their difference x ~ y is also divisible by p.
Example : The numbers 126 and 507 are both divisible by 3. Thus their difference 507 − 126 = 381 will also be divisible by 3.
There are some general properties of divisibility that help in determining the divisibility of a natural number by other natural numbers (other than detailed in 1.2)
Property 1
If a number x is divisible by another number y, then any number divisible by x, will also be divisible byy and by all the factors of y.
Example : The number 84 is divisible by 6. Thus any number that is divisible by 84, will also be divisible by 6 and also by the factors of 6, i.e. By 2 and by 3.
Property 2
If a number x is divisible by two or more than two co-prime numbers than x is also divisible by the product of those numbers.
Example : The number 2520 is divisible by 5, 4, 13 that are prime to each other (i.e. co-prime), so, 2520 will also be divisible by 20 (= 5 × 4), 65 (= 5 × 13), 52 (= 4 × 13).
Property 3
If two numbers x and y are divisible ‘p’, then their sum x + y is also divisible by p.
Example : The numbers 225 and 375 are both divisible by 5. Thus their sum 225 + 375 = 6000 will also be divisible by 5.
Note : It is also true for more than two numbers.
Property 4
If two numbers x and y are divisible by a number ‘p’, then their difference x ~ y is also divisible by p.
Example : The numbers 126 and 507 are both divisible by 3. Thus their difference 507 − 126 = 381 will also be divisible by 3.
TEST OF A PRIME NUMBER
A prime number is only divisible by 1 and by the number itself. The first prime number is 2. Every prime number other than 2 is odd, but every odd number is not necessarily a prime number. Again any even number(other than 2) can’t be a prime number. To test whether any given number(if odd) is prime number or not, following steps are to be considered:
Step1 find an integer (x) which is greater than the approximate square root of the given number.
Step2 Test the divisibility of the given number by every prime number less than x.
Step3 (i) If the given number is divisible by any of them in step 2, then the given number is NOT a prime number.
(ii) If the given number is not divisible by any of them in step 2, then the given number is a prime number.
Example: Consider a number 203. test is it is a prime number or not.
Step1 The approximate square root of 203 is 14 plus. Take x = 15.
Step 2 Check the divisibility of 203 by the prime number less than 15 i.e. by 2, 5, 7, 11, 13.
Step 3 203 is divisible 7. Thus, it is not a prime number.
Division and Remainder
When a given number is not exactly divisible by any number, then there is a remainder number at the end of such division.
Suppose we divide 25 by 7 as,
7) 25 (3
21/4
A prime number is only divisible by 1 and by the number itself. The first prime number is 2. Every prime number other than 2 is odd, but every odd number is not necessarily a prime number. Again any even number(other than 2) can’t be a prime number. To test whether any given number(if odd) is prime number or not, following steps are to be considered:
Step1 find an integer (x) which is greater than the approximate square root of the given number.
Step2 Test the divisibility of the given number by every prime number less than x.
Step3 (i) If the given number is divisible by any of them in step 2, then the given number is NOT a prime number.
(ii) If the given number is not divisible by any of them in step 2, then the given number is a prime number.
Example: Consider a number 203. test is it is a prime number or not.
Step1 The approximate square root of 203 is 14 plus. Take x = 15.
Step 2 Check the divisibility of 203 by the prime number less than 15 i.e. by 2, 5, 7, 11, 13.
Step 3 203 is divisible 7. Thus, it is not a prime number.
Division and Remainder
When a given number is not exactly divisible by any number, then there is a remainder number at the end of such division.
Suppose we divide 25 by 7 as,
7) 25 (3
21/4
Then, divisor = 7, dividend = 25
Quotient = 3 and remainder = 4So, we can represent it as
divisor) dividend (quotient remainder
Thus dividend = (divisor × quotient) + remainderSo, if a number x is divided by k, leaving remainder ‘r’ and giving quotient ‘q’ then the number can be found by using (i)
x = kq + r
Hence, if the number x is exactly divisible by k, then remainder = r = 0
∴ x = kq
and so x = q, implying that x is divisible exactly by k and q is an integer.kMethods to Find a Number Completely Divisible by Another
Consider a given number x. When divided by d, it gives a quotient q and remainder ‘r’.
It implies that the given number ‘x’ is not exactly divisible by ‘d’
d) x_ (q
r
Quotient = 3 and remainder = 4So, we can represent it as
divisor) dividend (quotient remainder
Thus dividend = (divisor × quotient) + remainderSo, if a number x is divided by k, leaving remainder ‘r’ and giving quotient ‘q’ then the number can be found by using (i)
x = kq + r
Hence, if the number x is exactly divisible by k, then remainder = r = 0
∴ x = kq
and so x = q, implying that x is divisible exactly by k and q is an integer.kMethods to Find a Number Completely Divisible by Another
Consider a given number x. When divided by d, it gives a quotient q and remainder ‘r’.
It implies that the given number ‘x’ is not exactly divisible by ‘d’
d) x_ (q
r
Now, to find a number exactly divisible by ‘d’, we can use either of the following two methods to reduce the remainder to zero. (If a number exactly divisible, then remainder is zero).
Method 1
• By subtracting the remainder from the given number (dividend).
∴ the required number that is exactly divisible by ‘d’ = x − r
Hence ‘remainder’ is the least number that can be subtracted from any number to make it exactly divisible.
Method 2
• By adding the (divisor − remainder) to be given number.
∴ the required number that is exactly divisible by d = x + (divisor − remainder)
Therefore, (divisor − remainder) is the least number that can be added to any given number to make it exactly divisible.
Example : Find the least number, that must be
Solution: On dividing 5029 by 17, we find that
17) 5029 ( 295 34 162 153 99 85 14∴ remainder = 14
(a) The least number to be subtracted to make it exactly divisible = remainder = 14. (By method 1)
(b) The least number to be added to make it exactly divisible = divisor − remainder = 17 − 14 = 3. (By method 2)
Method 1
• By subtracting the remainder from the given number (dividend).
∴ the required number that is exactly divisible by ‘d’ = x − r
Hence ‘remainder’ is the least number that can be subtracted from any number to make it exactly divisible.
Method 2
• By adding the (divisor − remainder) to be given number.
∴ the required number that is exactly divisible by d = x + (divisor − remainder)
Therefore, (divisor − remainder) is the least number that can be added to any given number to make it exactly divisible.
Example : Find the least number, that must be
- subtracted from
Solution: On dividing 5029 by 17, we find that
17) 5029 ( 295 34 162 153 99 85 14∴ remainder = 14
(a) The least number to be subtracted to make it exactly divisible = remainder = 14. (By method 1)
(b) The least number to be added to make it exactly divisible = divisor − remainder = 17 − 14 = 3. (By method 2)