Revision Notes
1. The factorised form
Factorisation is writing the number as a product of prime numbers. All like primes are collected together and written in the index form. Thus, the factorised form is:
n = p1a × p2b × p3c × ……
Where each of p1, p2, p3 , …. are distinct prime numbers and a, b, c, … are natural numbers.
Common Errors
i. 54 × 63 × 72 is not the factorised form because 6 is not a prime number. The factorised form will be 23 × 33 × 54 × 72 .
ii. 23 × 32 × 24 × 53 is not the factorised form since there are two terms with base being the prime 2. Both need to be collected together and the correct factorised form is 27 × 32 × 53
2. Convenient generalised Factorised form
A simpler working form of factorised form is 2a × 3b × 5c × 7d × 11e × ……, where instead of representing the base as p1, p2, p3 , …., we use all the prime numbers, 2, 3, 5, 7, 11, ….. In this representation, if a particular prime number is not a factor of the number, its power will simply be written as 0 i.e. a, b, c, d, e, …. are whole numbers.
E.g. 45 = 20 × 32 × 51 × 70 × 110 ×….
This simpler working form will help use write the conditions for the number being even/odd, or multiples/factors more easily, as given in the rest of the points below.
3. If the number is prime/composite …
2a × 3b × 5c × 7d × 11e × ……
… will be a prime number only when exactly one of a, b, c, d, e, … is 1 and all the others are 0
… will be a composite number when atleast two out of a, b, c, d, e, … are non-zero; OR any one of them is 2 or more while the others are 0.
4. If the number is even/odd
2a × 3b × 5c × 7d × 11e × ……
…… is even when a ≥ 1
…… is odd when a = 0
5. If the number is perfect square/cube/n-th power
2a × 3b × 5c × 7d × 11e × ……
…… is a perfect square, if each of a, b, c, d, e, …. is a multiple of 2 (in other words is even)
…… is a perfect cube, if each of a, b, c, d, e, …. is a multiple of 3
…… is a perfect n-th power, if each of a, b, c, d, e, …. is a multiple of n
5.1 When will a number be a perfect square as well as a perfect cube?
When each of a, b, c, d, e, …. is a multiple of 2 as well as a multiple of 3 i.e. a multiple of 6.
It would be useful to know the first few numbers which are both squares and cubes.
20 × 30 × 50 × 70 × 110 × …… = 1
26 × 30 × 50 × 70 × 110 × …… = 64 = 8² = 4³
20 × 36 × 50 × 70 × 110 × …… = 729 = 27² = 9³
212 × 30 × 50 × 70 × 110 × …… = 4096 = 64² = 16³
20 × 30 × 56 × 70 × 110 × …… = 15625 = 125² = 25³
Thus, the first few values which are both squares and cubes are: 1, 64, 729, 4096, 15625.
6. When will a number be a multiple of another number?
Factorise the other number as well, say 2p × 3q × 5r × ……
2a × 3b × 5c × 7d × 11e × …… will be a multiple of 2p × 3q × 5r × …… if and only if
a ≥ p, b ≥ q, c ≥ r, ……
7. When will a number be a factor of another number?
Factorise the other number as well, say 2p × 3q × 5r × ……
2a × 3b × 5c × 7d × 11e × …… will be a factor of 2p × 3q × 5r × …… if and only if
a ≤ p, b ≤ q, c ≤ r, ……
8. When will a number have atleast/exactly/atmost k trailing zeroes
Trailing zeroes here refer to the number of zeroes to the right of the first non-zero digit, from the right end.
2a × 3b × 5c × 7d × 11e × …… will have …
… atleast k trailing zeroes if both, a ≥ k and c ≥ k
… exactly k trailing zeroes if min(a, c) = k i.e. atleast one of a and c has to be exactly k and the other can be k or more than k.
… atmost k trailing zeroes if both, a ≤ k and c ≤ k
9. Finding the highest power of any number that divides a given number.
If we need to find the highest power of a prime number that divides a given number, then it is directly the index of that prime number in the factorised form of the given number.
E.g. the highest power of 2 that divides 211 × 35 × 56 × 72 × … is 11 itself.
However if we need to find the highest power of composite number, then some students find it difficult, so let’s take an intermediate step, finding the highest power of some power of a prime number …
E.g. the highest power of 4 = 22 that divides 211 × 35 × 56 × 72 × … is 11/2 = 5.5. And hence the highest power of 4 that divides the given number is 5.
The highest power of 8 = 23 that divides 211 × 35 × 56 × 72 × … is 11/3 = 3.66…. And hence the highest power of 8 that divides the given number is 3.
And now we can find the highest power of any composite number that divides a given number.
E.g. the highest power of 72 = 23 × 32 that divides 211 × 35 × 56 × 72 × … is
11/3 = 3.66
5/2 = 2.5
and the highest power of 72 that divides the given number is 2.