# Number System

#### Unit’s and Ten’s Digit: 1

Divisibility

Divisibility:

Source: eazynotes

This chapter on divisibility governs the basic rules to identify whether a given number is divisible by a particular divisor. It may also be intended to find the remainder obtained when a particular number is divided by a particular divisor. It is the most fundamental lesson that will govern the other aspects of Number System that we will study in the later lessons.

Divisiblity test means finding out whether a particular whole number is divisible by another without actually performing the division. The number to be divided is known as Dividend while the number by which dividend gets divided is known as Divisor. One by one we’ll look at the divisibility rules for some of the most important divisors.

Divisibility by 2,4,8,16,32,…

a. Divisiblity by 2: To check whether a number is divisible by 2, we only need to check the last digit of the number. If the last digit is divisible by 2, then the number is divisible by 2. So, if the unit digit is either 0,2,4,6 or 8, then the number is completely divisible by 2.

e.g. 344445256 , 68837392, 99999999730 etc are all divisible by 2 as the last digit of all these numbers is even.

b. Divisiblity by 4: If the number formed by taking the last two digits of a number is a multiple of 4, then the entire number is divisible by 4.

e.g. Consider the number 44563373672, the last two digits are 72 and as 72 is divisible by 4, so the entire number 44563373672 is divisible by 4.

Logic: Consider any arbitrary six digit number : abcdef. This number can be written as:  abcd00 + ef = 100(abcd) + ef

Now 100(abcd) is completely divisible by 4, so for the entire number to be divisible by 4, the part ef must be divisible by 4. This means the number formed by the last two digits of abcdef must be divisible by 4.

c. Divisiblity by 8: If the number formed by taking the last three digits of a number is a multiple of 8, then the entire number is divisible by 8.

e.g. Consider the number 63458880, the last three digits are 880 which is a multiple of 8, so the above number 63458880 is divisible by 8.

Consider another number 712988, the last three digits are  988 which is not the multiple of 8, so the number 712988 is not divisible by 8.

Logic : Almost same as the one discussed for 4.

abcdef = abc000 + def  = 1000(abc) + def

Here the first part is completely divisible by 8, so if the second part is divisible by 8, then the complete number abcdef is divisible by 8.

d. Divisiblity by 16: If the number formed by taking the last four digits of a number is a multiple of 16, then the entire number is divisible by 16.

And so on goes the divisibility rule for 32,64,…….

Divisibility by 5,25,125,……..

1. Divisiblity by 5: A number is divisible by 5, if the last digit is divisible by 5. i.e. if the unit digit is either ‘5’ or ‘0’, then the number is divisible by 5.

e.g. 766667785, 23444490 etc are divisible by 5.

b. Divisiblity by 25: A number is divisible by 25, if the number formed by the last two digits is a multiple of 25 i.e. if the last two digits are 00,25,50 and 75, then the entire number will be divisible by 25.

Logic: consider a six digit number abcdef.

abcdef = abcd00 + ef = 100(abcd) + ef

Now the first part is divisible by 25. If the second part is divisible by 25, then the entire number is divisible by 25.

E.g. consider the number 8346575, the last two digits are 75 which is a multiple of 25. So, the number 8346575 is divisible by 25.

c. Divisiblity by 125: A number is divisible by 125, if the number formed by the last two digits is a multiple of 125.

Logic same as that applied for the rule of 25.

Divisiblity by 3:

A number is said to be divisible by 3, if the sum of the digits of that number is a multiple of 3.

e.g. 1246572 is a number whose sum of the digits is 1+2+4+6+5+7+2 = 27 which is a multiple of 3, hence the number is divisible by 3.

Logic: Let us consider a four digit number abcd. The given number can be written as: abcd = 1000a + 100b+10c + d = 999a +99b +9c + a +b+c+d .

Now 999a +99b+9c is a part divisible by 3, so if the part a+b+c+d is divisible by 3, then the complete number abcd is divisible by 3, i.e. if the sum of the digits is a multiple of 3, then the number is divisible by 3.

Divisiblity by 9:

A number is said to be divisible by 9, if the sum of the digits of that number is a multiple of 9.

e.g. 124657236 is a number whose sum of the digits is 1+2+4+6+5+7+2 +3+9= 36 which is a multiple of 9, hence the number is divisible by 9.

Logic: Let us consider a four digit number abcd. The given number can be written as: abcd = 1000a + 100b+10c + d = 999a +99b +9c + a +b+c+d .

Now 999a +99b+9c is a part divisible by 9, so if the part a+b+c+d is divisible by 9, then the complete number abcd is divisible by 9, i.e. if the sum of the digits is a multiple of 9, then the number is divisible by 9.

Note: If a number is divisible by 9, then it necessarily will be divisible by 3 but the vice versa is not true i.e. if a number is divisible by 3, then it might not be necessarily

Divisibility by a Composite Number:

To check the divisibility by a composite number (A number that has more than two factors), break the divisor into product of two coprime or relatively prime (numbers that have the H.C.F. as 1) numbers and check the divisibility individually by these two numbers.

Divisibility by 6- Divide 6 into two parts i.e. 6 = 2 * 3

Check by 2 and by 3

Divisibility by 12- Divide 12 into product of two co-prime numbers, 12 = 4*3

Now, check individually by 4 and 3.

Divisibility by 18: 18 = 9*2, Now, check individually by 9 and 2.

Divisibility by 24: 24 = 8*3, Now, check individually by 8 and 3.

Divisibility by 72: 72= 8*9, Now, check individually by 8 and 9.

Divisibility by 48: 48 = 16*3, Now, check individually by 16 and 3.

Example 1: If a number 774958A96B is to be divisible by 8 and 9, the respective values of A and B will be

1. 7 and 8       2. 8 and 0          3.5 and 8           4. 0 and 6

Solution:  For a number to be divisible by 8, the last three digits of the number must be a multiple of 8.

774958A96B has the last two digits as 96B which must be a multiple of 8.

So, B is either 0 or 8.

774958A96B is divisible by 9, so sum of the digits must be a multiple of 9.

7+7+4+9+5+8+A+9+6+B = 55 + A + B

55+A+B must be a multiple of 9. Also, B =0 or 8.

If B =0, A =8

If B = 8, A = 0 .

Thus, option (2) is correct.

Example 2 : If the number 3462AB is divisible by 36 and A and B represent the digits of the number, then how many pairs of (A,B) are possible?

Solution: The number 3462AB is divisible by 36

36 = 9*4, so for the number to be divisible by 36, it must be divisible by both 9 and 4.

Sum of the digits of 3462AB = 15 + A + B

For it to be divisible by 9, the sum of the digits must be a multiple of 9

So, either  A + B = 3            or A +B = 12

Case1: A+B = 3

Possible pairs (3,0),(2,1),(1,2),(0,3). But as3462AB is divisible by 4 as well, so AB must be a multiple of 4. Out of the given pairs, only (1,2) satisfy the condition.

Case2: A+B = 12

Possible pairs (9,3),(8,4),(7,5),(6,6),(5,7),(4,8) and (3,9) . But as3462AB is divisible by 4 as well, so AB must be a multiple of 4. Out of the given pairs, only (8,4) , (4,8) satisfy the condition.

So, a total of 3 pairs of (A,B) are possible i.e. (1,2), (4,8) and (8,4).

Example 3: Given that  a15 * 4b6 is perfectly divisible by 36. How many pairs of (a,b) are possible where a ≠ 0?

Solution:  The product of a15 and 4b6 must be perfectively divisible by 36.

Clearly a15 has 5 as its unit digit, so the number cannot be divisible by 4.

This implies that 4b6 must necessarily be a multiple of 4. The possible values of b become 1,3,5,7 and 9.

Now there are three cases:

1. a15 is a multiple of 9 while 4b6 is a multiple of 4

2. a15 is a multiple of 3 while 4b6 is a multiple of both 4 and 3, i.e. a multiple of 12

3. a15 can be anything but 4b6 is a multiple of 36.

Case 1: a15 is a multiple of 9, so a+1+5 i.e. a+6 must be a multiple of 9. So a = 3

Possible pairs = (3,1)(3,3)(3,5)(3,7) and (3,9)

Case 2: a15 is a multiple of 3, a+6 is a multiple of 3. Thus a = 3,6,9

Also 4b6 must be divisible by both 4 and 3, out of the values that b can take only 5.

So, possible pairs are (3,5), (6,5) and (9,5). As we have already counted the pair (3,5) once in case 1, so only unique pairs = (6,5) and (9,5)

Case3: There is no such pair possible for this case.

So, the total number of pairs = 7 i.e. (3,1)(3,3)(3,5)(3,7),(3,9),(6,5) and (9,5).

Divisibility by 11:

For checking whether a number is divisible by 11, add the digits at the odd positions and add the digits at the even positions. If the difference between the two sum obtained is a multiple of 11, then the number is divisible by 11.

e.g.  let’s consider a number: 24376,

The sum of the digits at the odd places is 2+3+6 = 11.

The sum of the digits at the even places is 4 +7 = 11

The difference = 0, so the number is divisible by 11.

Divisibility Rule for 7,11 & 13:

To understand the common rule for 7,11 and 13, consider a six digit number abcdef.

abcdef = abc000 + def

= 1000abc + def

Adding and subtracting abc we get

abcdef = 1000abc + abc + def –abc

= abc(1001) + def –abc

As 1001 = 7*11*13, so abcdef = (7*11*13)abc + def-abc

Now, the first part is divisible by 7, 11 and 13. So, if def-abc is also divisible by 7 or 11 or13, then we can say that the entire number abcdef is divisible by 7, 11 and 13.

Now def-abc is the difference of the numbers formed by the last three digits and first three digits of the number abcdef. So, if this difference is a multiple of 7 or 11 or 13, then we can say that the number is divisible by these divisors.

But how do we apply this rule to a large number?

Consider the number 7856698710923367.

Begin from the right and partition this number into groups of three as shown

Now, add I,III and V together i.e. 367 + 710 + 856 = 1933 = N1

Add II,IV and VI together i.e. 923 + 698+ 7 = 1628 = N2

Find N1 – N2, i.e. 1933 -1628 = 305

If N1 – N2 is divisible by 7, then the entire number is divisible by 7

If N1 – N2 is divisible by 11, then the entire number is divisible by 11

If N1 – N2 is divisible by 13, then the entire number is divisible by 13

Here 305 is neither divisible by 7 or 11 or 13.

Note: This method is also applicable to find the remainders by 7,11 and 13

Here 305/7 gives a remainder of 4, so the entire number would give a remainder of 4 by 7.

305/11 gives a remainder of 8, so the entire number would give a remainder 8 by 11.

305/13 gives a remainder of 6, so the entire number would give a remainder 6 by 13.

Property: A single digit when repeated 6k times will always be divisible by 7,11 and 13 .

For example : aaaaaa would be divisible by 7,11 and 13 as aaa-aaa = 0

aaaaaaaaaaaa   : aaa aaa aaa aaa   2(aaa) -2(aaa) = 0, hence divisible by 7,11 and 13.

This is true for a being written a multiple of 6 times because each  triplet gets cancelled out with the following triplet.

Example 4 : (2222…….)100 times + (333333) 96timesA is divisible by 13, what is the value of A where A is a single digit number.

Solution : Let’s check out the remainder when (2222…….)100 times  is divided by 13.

For divisibility rule of 13, we make pairs of three from the right, add the odd positioned pairs, similarly add the even positioned pairs and find the difference.

Now, (222…..)96 times will get cancelled out and the final difference is 222-2 = 220

Remainder (220/13) = 12

Similarly (333333) 96timesA  = 10(333…..)96times + A

The first part is divisible by 13, as any digit written 6n times is always divisible by 7,11 and 13.

The remainder when (333333) 96timesA  is divided by 13, is A

Final remainder is 12+A, and this should be further divisible by 13.

Value of A = 1

Divisibility Rule for 27 & 37:

Again let’s consider a six digit number abcdef

abcdef = abc000 + def

= 1000abc + def

Subtracting and Adding abc we get

abcdef = 1000abc - abc + def +abc

= abc(999) + def +abc

As 999 = 37*27 (breaking into product of co-primes)

, so abcdef = (37*27)abc + def+abc

The first part is divisible by both 37 and 27. If the part def+abc is also divisible by 37 or 27, we can say the entire number is divisible by 37 or 27.

If we have a large number, then divide the entire number into group of three starting from the right and adding these groups together. For e.g.

Consider the number 7856698710923367.

Adding together I,II,III,IV,V and VI we get

367+923+710+698+856+7 = 3561, if this number is divisible by 27 or 37, then the entire number is divisible by 37 or 27. In this case, 3561 is not divisible by 27 or 37. So, the given number is either divisible by 27 or 37.

Similarly if one wants to find the divisibility rule for 101 ,

abcd = ab00 + cd

= 100ab +cd

Abcd = 100ab + ab + cd-ab

= 101ab + cd -ab

Thus, first part is divisible by 101, if the second part is also divisible by 101, then the entire number becomes divisible by 101.

For a large number, make groups of two starting from the right. Add all the odd positioned groups and add together all the even positioned groups. Find the difference of these sums, if it comes out to be a multiple of 101, then the entire number is divisible by 101.

Similarly one can find the divisibility rule for 99.

These rules are enough to be able to solve divisibility problems appearing in CAT and other B-school entrance examinations.

Example 5: What is the remainder when (243642)100 times is divided by 999?

Solution: The rule for 37 and 27 will hold for 999 as well because 999=37*27

For the divisibility rule of 999, we partition the number into groups of three each starting from the right and add all such pairs.

(243642)100 times  when partitioned into groups of three and added, we’ll get

642*100 + 243*100 = 100(885) = 88500 (This is because each of the triplet 642 and 243 will appear 100 times)

Now we need to divide 88500/999 to get the final remainder as 588.

Example 6: abcde is the smallest number such that all the digits are different and their values are chosen from 2,3,4,5 and 6. Given cde is a multiple of 5, bcd is a multiple of 4 and abc is a multiple of 3. What is the value of abcde?

Solution: As cde is a multiple of 5, so e must definitely be 5.

Also, bcd is a multiple of 4. Thus, cd must be a multiple of 4. The following pairs of cd are possible- 24,32, 36 and 64.

Thus, cde could be 245 or 325 or 365 or 645

Accordingly ab could be 36245  or 63245 or 46325 or 64325 or 42365 or 24365 or 23645 or 32645.

Also, abc must be a multiple of 3, i.e. a+b+c must be a multiple of 3.

Only 42365 and 24365 satisfies all the conditions.

Smallest number would be 24365 satisfying all the conditions.

Example7 : 555…532timesP888832times  is a number perfectly divisible by 7. What is the value of P, if it is a single digit number?

Solution:  For divisibility of 7, partition into groups of three starting from the right. Remember an even pair gets subtracted from an odd number positioned pair.

Here A88 and 55 would be left and the other pairs get cancelled out. Now A88 is the 11th such pair while 55 is the 22nd such pair.

So, A88 -55 = A33 must be a multiple of 7.

A =1 makes A33 a multiple of 7.

Thus A = 1.