Grade 6

Playing with Numbers

 A factor of a number is an exact divisor of that number.

For example: 3 is an exact divisor of 15 (15 = 3 × 5). So, 3 is a factor of 15. 2 is not
an exact factor of 17 (since any whole number multiplied with 2 does not give 17).
So, 2 is not a factor of 17.
 1 is a factor of every number. E.g., 5 = 1 × 5, 19 = 1 × 19
 Every number is a factor of itself. For example: 20 = 20 × 1. This shows that 20 is a
factor of 20.
 Every factor of a number is less than or equal to the number. For example, factors of
12 are 1, 2, 3, 4, 6 and 12. Here, 1, 2, 3, 4, 6 and 12 are less than or equal to 12.
 Though we may find difficulties in factorising bigger numbers. But the number of
factors of a number is finite.
For example: The factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. Here, we may
observe that there are only 8 factors of 30. Similarly, we can find a fixed number of
factors of bigger numbers like 999, 1001, etc., although it is difficult; but we may
observe that the number of factors of a number is finite.
 We can find the multiples of a given number by multiplying 1, 2, 3 …, to the
number. For example, to find the multiples of 12, we need to calculate as follows:
12 × 1 = 12, 12 × 2 = 24, 12 × 3 = 36, 12 × 4 = 48 …, and so on
So, the numbers 12, 24, 36, 48, etc., are the multiples of 12.
 A number is a multiple of each of its factors. For example, the factors of 15 are 1, 3,
5 and 15. So, we can say that 15 is a multiple of 1, 3, 5 and 15.
 Every multiple of a number is greater than or equal to that number.
For example, the multiple of 6 is 6, 12, 18 …
Here we may observe that the multiples of 6 are greater than or equal to 6.
 Every number is a multiple of itself. For example, 18 is a multiple of itself, that is,
18. Therefore, every number is a factor as well as a multiple of itself.
 The number of multiples of a given number is infinite. For example, the multiples of
11 are 11, 22, 33, 44 …. So, the number of multiples of 11 is infinite.
 Prime numbers are numbers having exactly two factors: 1 and the number itself. For
example, the factors of 23 are 1 and 23 only. So, 23 is a prime number.
 Composite numbers are numbers having more than two factors. For example, the
factors of 42 are 1, 2, 3, 6, 7, 14, 21, and 42. Since there 8 factors of 42, it is a
composite number.
 1 is neither prime nor composite as it has exactly one factor.
 The smallest prime number is 2.
 The smallest even prime number is 2 and the smallest odd prime number is 3.
 All even numbers except 2 are composite.
 The pairs of prime numbers whose difference is 2 are known as twin primes. For
example, 11 and 13 are twin primes.
Example:
Express 44 as the sum of four different primes.
Solution:
 44 3 11 13 17
44 5 7 13 19
44 3 5 17 19
44 3 7 11 23
In this way we can express 44 as the sum of four different primes.
 A number is divisible by 10, if the digit in one’s place is zero. For example, the
numbers 9520, 67120, 830, 1200, etc., are divisible by 10.
 A number is divisible by 5, if the digit in one’s place is either 0 or 5. For example,
3615, 92185, 370 are divisible by 5.
 A number is divisible by 2, if the digit in one’s place is either 0, 2, 4, 6, or 8. For
example, the numbers 9218, 6054, 932 are divisible by 2.
 A number is divisible by 3, if the sum of the digits in the number is a multiple of 3.
For example, the sum of the digits of 9528 is 9 + 5 + 2 + 8 = 24, which is a
multiple of 3. Hence, 9528 is divisible by 3. But the sum of the digits of 3725 is 3 +
7 + 2 + 5 = 17, which is not a multiple of 3. Hence, 3725 is not divisible by 3.
 A number is divisible by 6, if it is divisible by both 2 and 3. For example, 39612 is
divisible by 2. Sum of the digits of 39612 is 3 + 9 + 6 + 1 + 2 = 21, which is a
multiple of 3. So, 39612 is divisible by 3. Now, 39612 is divisible by both 2 and 3.
So, it is divisible by 6.
3513 is not divisible by 2, but it is divisible by 3 (3 + 5 + 1+ 3 = 12, which is a
multiple of 3). So, 3513 is not divisible by 6.
Example
Write the smallest digit and the greatest digit in the blank space of the following
number, so that the number is divisible by 6.
931
Solution:
Since the number is divisible by 6, it has to be divisible by 2. So, the unit’s digit may
be 0, 2, 4, 6 or 8. Also, the number has to be divisible by 3. For this, the sum of the
digits should be a multiple of 3.
Now, 9 + 3 + 1 = 13
If we add 2, 5, or 8 to 13, then we get a number which is a multiple of 3. If the digit
in blank space is 2 or 8, then the obtained number will be divisible by 2 as well as 3.
So, the required smallest number is 2 and the largest number is 8.
 A number is divisible by 9, if the sum of the digits in the number is a multiple of 9.
For example, the sum of the digits in the number 9567 is 9 + 5 + 6 + 7 = 27,
which is a multiple of 9. So, 9567 is a multiple of 9.
 A number with 3 or more digits is divisible by 4, if the number formed by its last
two digits (one’s and ten’s) is divisible by 4. For example, the last two digits of 9584
is 84, which is divisible by 4. So, 9584 is divisible by 4.
 A number with 4 or more digits is divisible by 8, if the number formed by its last
three digits (one’s, ten’s and hundred’s) is divisible by 8. For example, the last three
digits of 9368 is 368, which is divisible by 8. So, 9368 is divisible by 8.
 A number is divisible by 11, if the difference between the sum of the digits at odd
places (from the right) and the sum of the digits at even places (from the right) is
either 0 or divisible by 11. For example, for the number 82918, the sum of the
 digits at odd places = 8 + 9 + 8 = 25 and the sum of the digits at even places = 1 +
2 = 3. Now, 25 – 3 = 22, which is a multiple of 11. Hence, 82918 is a multiple of
11.
 Two numbers are called co-prime, if they have 1 as the only common factor. For
example, the factors of 8 are 1, 2, 4 and 8. The factors of 15 are 1, 3, 5, and 15.
The common factor to both 8 and 15 is 1. So, they are co-prime numbers.
 If a number is divisible by another number, then it is divisible by each factor of that
number. For example, we know 40 is divisible by 20. So, 40 is divisible by each
factor of 20 (i.e., by 1, 2, 4, 5, 10 and 20).
 If a number is divisible by two co-prime numbers, then it is divisible by their
product as well. For example, 70 is divisible by two co-prime numbers 5 and 7.
Also, 70 is divisible by 5 × 7 = 35.
 If two numbers are divisible by a number, then their sum is also divisible by that
number. For example, 8 and 10 are divisible by 2. So, their sum, that is, 8 + 10 =
18 is also divisible by 2.
 If two numbers are divisible by a number, then their difference is also divisible by
the number. For example, 12 and 33 are divisible by 3. Now, their difference, that
is, 33 – 12 = 21 is also divisible by 3.
 We can express a number as the product of prime numbers. Expressing a number as
the product of prime numbers is known as prime factorisation. To find the prime
factorisation of a number, we need to divide it by prime numbers that are factors of
the given number, till we get 1.
Example: Find the prime factorisation of 840.
We can proceed as follows:

2 840
2 420
2 210
3 105
5 35
7 7
1

So, the prime factorisation of 840 is

840 = 2 × 2 × 2 × 3 × 5 × 7
 The highest common factor (HCF) of two or more given numbers is the highest of
their common factors. To find the HCF of given numbers, first of all we need to
prime factorise the given numbers.
For example, the HCF of 90, 120, 150 can be found as:
2 90
3 45
3 15
5 5
1
 2 120
2 60
2 30
3 15
5 5
1

2 150
3 75
5 25
5 5
1

Thus, the HCF of 90, 120 and 150 = 2 × 3 × 5 = 30

 The lowest common multiple (LCM) of two or more given numbers is the least of
their common multiples. To find the LCM of 90, 120 and 150, we may proceed as
follows.

2 90, 120, 150

2 45, 60, 75
2 45, 30, 75
3 45, 15, 75
3 15, 5, 25
5 5, 5, 25
5 1, 1,5
1, 1, 1

LCM of 60 and 120 = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

Contribute to this Revision Note:

If you find anything of importance missing from this note, email it to us at revision-notes@meritnation.com, and we’ll
add it to this note under your name!