Introduction
- A number 'x' is said to be completely divisible by another number ‘y’ if it leaves no remainder.
- In general, it can be said that any integer I, when divided by a natural number N, there exist a unique pair of numbers Q and R which are called the quotient and Remainder respectively.
Thus, I = QN + R, where Q is an integer and N is a natural number or zero and 0 < R < N (i.e. remainder has to be a whole number less than N).
- If the remainder is zero, we say that the number I is divisible by N.
- When R > 0, we say that the number I is divisible by N with a remainder.
Example: 25/8 can be written as: 25 = 3 * 8 + 1, where 3 is the quotient and 1 is the remainder. - Suppose, we divide – 32 by 7. Contrary to what you might expect, the remainder, in this case, is + 3 (and not – 4). This is because the remainder is always non-negative. Thus, –32/7 gives quotient as – 5 and remainder as + 3.
Example: –25/7 can be written as –25 = 7 * (-4) + 3, where -4 is the quotient and 3 is the remainder.
Theorems of Divisibility
- If a is divisible by b, then ac is also divisible by b.
- If a is divisible by b and b is divisible by c, then a is divisible by c.
- If a and b are natural numbers such that a is divisible by b and b is divisible by a, then a = b.
- If n is divisible by d and m is divisible by d then (m + n) and (m – n) are both divisible by d. This has an important implication.
Suppose 28 and 742 are both divisible by 7. Then (742 + 28) as well as (742 – 28) are divisible by 7. (and in fact so is + 28 – 742). - If a is divisible by b and c is divisible by d then ac is divisible by bd.
- The highest power of a prime number p, which divides n! exactly is given by [n/p] + [n/p^{2}] + [n/p^{3}] +... where [x] denotes the greatest integer less than or equal to x.
Factors
- The number of factors based questions has regularly appeared in various competitive exams, including CAT. Primarily, these questions are based on the prime factorization of a number.
- Generally, factors based questions are of the following types:
(i) The number of factors.
(ii) The sum of factors.
(iii) Product of factors.
(iv) Odd and Even Factors.
(v) Number of Factors which are perfect squares for certain kinds of questions. - We already have the direct formula and shortcut techniques which we apply to get the answer. However, if an aspirant wants to master this topic, must also understand the conceptual approach to solve these types of questions. This article will focus more on the concepts and approach to tackle question based on factors.
Prime Factorisation
We all know that every composite number can be written as a product of some prime numbers.
Example: We can write 90 as 2 × 3^{2} × 5. This process is called prime factorisation, and it is the first step to solve any questions related to factors.
Number of Factors of a Number
- Factors are those numbers that divide the given number completely.
Example: Factors of 72 - 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 48 and 72. - Observe that the number 1 is always a factor of every number, and the number itself will be the factor of the number.
- Now let us understand the formula for finding the number of factors of any composite number. We will take the above example of 72.
72 = 2^{3} × 3^{2} - Clearly, the number 72 is divisible by each of 2^{0},2^{1}, 2^{2}, 2^{3}...^{ }but not by any higher power of 2 like 2^{4}, 2^{5}, …
- Similarly, the number 72 is divisible by each of 3^{0}, 3^{1}, 3^{2},..... but not by 3^{3},3^{4},….
- We should also observe that the number can be divisible by any ‘combination’ of one of 2^{0},2^{1}, 2^{2}, 2^{3} and one of 3^{0}, 3^{1}, 3^{2}.....by numbers of type 2^{1} × 3^{2} or 2^{2 }× 3^{1 }and so on.
How to Find the Sum of Factors of a Number?
- You might have encountered with such questions in earlier CAT papers or any Test series. Well, questions of such types require prior knowledge of solving it. Otherwise, you will waste time trying to solve such questions and end up with no results.
- There is a complete algorithm to solve such questions.
- Example: Find the sum and number of Factors of 720.
Write the number in Standard Form i.e. 720 = 2^{4} x 3^{2} x 5^{1}
Create a number of brackets as the number of prime factors.
Fill each bracket with the sum of all the powers of the respective prime number starting from 0 to the highest power of that number.
Sum of the factors = (2^{0} + 2^{1} + 2^{2} + 2^{3} + 2^{4}) x (3^{0} + 3^{1} + 3^{2}) x (5^{0} + 5^{1})
= (1 + 2 + 4+ 8 +16) x (1 + 3 + 9) x (1+ 5)
= 31 x 13 x 6 = 2418
How to Find Number of Factors of a Number?
- If you will check the sum of factors in step 3, and expand the bracket, you will find that all are actually factors of the number.
- Lets take a small number for this say 80:
80 = 2^{4} x 5^{1}
Sum of the factors = (2^{0} + 2^{1} + 2^{2} + 2^{3} + 2^{4}) x (5^{0} + 5^{1}) = 31 x 6 = 186 - So to calculate the number of factors of 80:
Add 1 to the powers of prime factors and multiply it
No. of Factors of 80 = (4 + 1) x (1 + 1) = 10. - Example: Find the number of Factors of 720.
720 = 2^{4} x 3^{2} x 5^{1}
Number of Factors of 720 = (4 + 1) x (2 + 1) x (1 + 1) = 5 x 3 x 2 = 30
Question for Divisibility Rules
Try yourself:Find the sum the sums of divisors of 144 and 160.
Explanation
144 = 2^{4} X 3^{2}
⇒ Sum of divisors of 144 = (2^{0} +2^{1} +2^{2} +2^{3} +2^{4} )(3^{0} +3^{1} +3^{2}) = 31 X 13 = 403
160 = 2^{5} X 5^{1}
⇒ Sum of divisors of 160 = (2^{0} +2^{1} +2^{2} +2^{3} +2^{4} +2^{5})(5^{0} +5^{1}) = 63 X 6 =3 78
Sum of two = 403 + 378 = 781
Divisibility Rules
- As the name suggests, divisibility rules or tests are the procedures used to check whether a number is divisible by another number without necessarily performing the actual division.
- Since not all number is completely divisible by other numbers, therefore, the divisibility rules are actually the shortcuts of determining the actual divisor of a number just by examining the digits that make the number.
➤ Divisibility Test for 1
- The divisibility test for 1 does not have any condition for numbers. All numbers are divisible by 1 irrespective of how large the numbers are.
- When any number is divided by 1, the result is the number itself.
Example: 5/1= 5 and 100000/1 = 100000.
➤ Divisibility Test for 2
- A number is divisible by 2 if the last digit of the number is 2, 4, 6, 8 or 0.
Example: 102/2 = 51, 54/2 = 27, 66/2 = 33, 28/2 = 14 and 20/2 = 10.
Q. Check if 288 is divisible by 2.
Solution:
Given, 288 is a number.
If the last digit of 288 is divisible by 2, then 288 is also divisible by 2.
The last digit of 288 is 8, which is divisible by 2, such that;
8/2 = 4
Hence, 288 satisfy the divisibility rule for 2.
➤ Divisibility Test for 3
- The divisibility test for 3 states that a number is completely divisible by 3 if the digits of the number are divisible by 3 or is a multiple of 3.
Example: Consider two numbers 308 and 207.
(i) Check 308 by summing its digits: 3 + 0 + 8 = 11, since 11 is not divisible by 3, then 308 is also not divisible by 3.
(ii) Check 207 by summing its digits: 2 + 0 + 7 = 9, since 9 is a multiple of 3, then 207 is also divisible by 3.
➤ Divisibility Test for 4
- The divisibility test for 4 states that a number is divisible by 4 if the last two digits of the number are divisible by 4.
Example: Consider two numbers, 2508 and 2506.
(i) 2508 is divisible by 4 because the last two digits 08 are divisible by 4.
(ii) 2506 is not divisible by 4 because the last two digits 06 are not divisible by 4.
Q. Check if 195 is divisible by 4 or not.
Solution: As we can see, the last digit of 195 is 5, which is not divisible by 4.
Hence, 195 is not divisible by 4.
➤ Divisibility Test for 5
- All numbers with the last digit as 0 or 5 are divisible by 5.
Example: 100/5 = 20, 205/5 = 41.
➤ Divisibility Test for 6
- A number is divisible by 6 if its last digit is an even number or zero and the sum of the digits is a multiple of 3.
Example: 270 is divisible by 2 because the last digit is 0 and the sum of the digits is: 2 + 7 + 0 = 9 which is also divisible by 3.
Therefore, 270 is divisible by 6.
➤ Divisibility Rules for 7
- The divisibility test of 7 is explained in the following algorithm Consider a number 1073.
- To check whether the number is divisible by 7 or not? Eliminate number 3 and multiply it by 2, which becomes 6.
- Subtract 6 from the remaining number 107, therefore 107 – 6 = 101. Repeat again the process we have 1 x 2 = 2, and the remaining number is 10 – 2 = 8.
Since 8 is not divisible by 7, therefore the number 1073 is also not divisible by 7.
➤ Divisibility Test for 8
- The divisibility test for 8 states that a number is divisible by 8 if its last three digits are divisible by 8.
➤ Divisibility Test for 9
- The divisibility test for 9 is the same as the divisibility test for 3. If the sum of the digits of a number is divisible by 9, then the number is also divisible by 9.
Example: In a number like 78532, the sum of its digits is: 7+8+5+3+2 = 25. - Since 25 is not divisible by 9, the 78532 is also not divisible by 9. Considering another case of number: 686997, the sum of digits is: 6 + 8 + 6 + 9 + 9 + 7 = 45. Since the sum is divisible by 9, then the number 686997 is divisible by 9.
➤ Divisibility Test for 10
- The divisibility rule for 10 states that any number whose last digit is zero, then the number I divisible by 10.
Example: The numbers: 30, 50, 8000, 20 33000 are divisible by 10.
➤ Divisibility Rules for 11
- This rule states that a number is divisible by 11 if the difference of the sum of alternative digits is divisible by 11.
Example: To check whether number 2143 is divisible by 11 or not.
The sum of alternative digits of each group: 2 + 4 = 6 and 1+ 3 = 4
Since 6 - 4 = 2, Therefore 2143 is not divisible by 11.
➤ Divisibility Rules for 13
- To check whether a number is divisible by 13, repeated addition of the last digit is done 4 times to the remaining number until a two-digit number is arrived at.
- If the two-digit number is divisible by 13, then the whole number is also divisible by 13.
Example: 2795 → 279 + (5 x 4) → 279 + (20) → 299 → 29 + (9 x 4) → 29 + 36 →65.
Since the two-digit number is found to be 65, which is divisible by 13.
Therefore, the number 2795 is also divisible by 13.
Solved Questions
Q.1. How many three-digit numbers are divisible by 5 or 9?
a) 260
b) 280
c) 200
d) 180 Ans: a) 260
Solution: Three digit numbers divisible by 5 or 9 = three digit numbers divisible by 5 + three digit numbers divisible by 9 – three digit numbers divisible by 5 and 9.
The three digit numbers divisible by 5 = 100, 105, 110….995
The sequence given is in A.P with common difference 5. Let 995 be the nth term of the A.P, then, 995 = 100 + (n – 1)5 = 100 + 5n – 5
Thus, n = 180 – (1)
The three digit numbers divisible by 9 = 108, 118, … 999
The sequence given is in A.P with common difference 9. Let 999 be the pth term of the A.P, then, 999 = 108 + (p – 1)9 = 108 + 9p – 9
Thus, p = 100 – (2)
The three digit numbers divisible by 45 = 135, 180, …990
The sequence given is in A.P with common difference 45. Let 990 be the qth term of the A.P, then
990 = 135 + (q – 1)45 = 135 + 45q – 45
Thus, q = 20 – (3)
Thus, from (1), (2) and (3) the three digit numbers divisible by 5 or 9 = 180 + 100 – 20 = 260 so, 260 three- digit numbers are divisible by 5 or 9.
Q.2. If 8A5146B is divisible by 88, then what is the value of AxB?
a) 4
b) 16
c) 8
d) 12
e)18
Ans: d) 12
Solution: Since the given number is divisible by 8, the last three digits should also be divisible by 8. Only when B = 4, 46B is a multiple of 8. Thus, B = 4.
As the given number is divisible by 11, the difference between the sum of its odd digits and even digits must be a multiple of 11.
Thus, (8 + 5 + 4 + 4) – (A + 1 + 6) = 14 – A should be divisible by 11. Only when A = 3, 14-A is divisible by 11.
Thus, the value of AxB = 4×3 = 12.
Q.3. What is the number of even factors of 36000 which are divisible by 9 but not by 36?
a) 20
b) 4
c) 10
d) 12
Ans: b) 4
Solution: 36000 = 2^{5} * 3^{2} * 5^{3}
Since we are talking of even factors, there must be at least one 2 in the required factors.
Since the number is divisible by 9, we must have both the threes.
We cannot have more than 1 two as it will make the number divisible by 36.
So we have 1 way of choosing 2, 1 way of choosing 3, 4 ways of choosing 5.
Thus the required number of factors are
1*1*4 = 4
Q.4. The number A39K2 is completely divisible by both 8 and 11. Here both A and K are single-digit natural numbers. Which is a possible value of A+K?
a) 8
b) 10
c) 12
d) 14
Ans: b) 10
Solution:
The number is divisible by 11, so the difference between the sum of the digits at the odd places and the digits at the even places is either 0 or a multiple of 11.
Let the difference be a 0, so
11 + A = 3 + K
=> K – A = 8, the only possible value is 9,1
Now we have to check if it satisfies the divisibility by 8 test.
K= 9 makes the last 3 digits 992. This is divisible by 8.
Let’s check for other cases when the difference is 11
11 + A – 3 – K = 11 => A – K = 3
The possible values in this case are (9,6), ( 8,5), (7,4), (6,3), (5,2), (4,1)
Among these cases, only (8,5) and (4,1) will be divisible by 8. So the possible values of the sum are 13, 5, and 10.
Now, the difference between the sum of odd and even places cannot be 22
11 + A – 3 – K = 22 => A – K = 14
Since, both A and K are single-digit natural numbers, this is not possible.
Thus the only possible values of sum are 5, 10, and 13.
In the given options only 10 is there. So it is the correct answer.