Playing With Numbers

Option (a) ⇒ We express number as the sum of the product of its digits with their respective place values

1000 a + 100 b + 10 c + d is the generalised form of four digit number abcd but we want generalised form of four digit number abdc

Option (b) ⇒ 1000 a + 100 c + 10 b + d is the generalised form of acbd. Hence this option is incorrect.

Option (c) ⇒ 1000 a + 100 b + 10 d + c is the generalised form of the four digit number abdc. Therefore this option is correct.

Option (d) ⇒ a x b x c x d is simply the product of a,b,c,d .It is not a number it is the product of digits of the number. Therefore (d) option is incorrect.

Option (a) ⇒ The generalised form of the number is expressed as the sum of the product of its digits with their respective place values x + y is simply the sum of the digits of the number xy Therefore option (a) is incorrect.

Option (b) ⇒ 10x + y = 10 × X + 1 x Y clearly it is the sum of the product of digits with their respective places (Tens and ones). Hence 10x + y is the generalised form of xy.

Option (c) ⇒ 10x – y is the difference of product of digits with their respective places but in generalised form it is the sum. Therefore (c) is incorrect.

Option (d) ⇒ 10y + x = 10 x Y + 1 x X it is the general form of the number yx but question is asking about xy .Therefore (d) is incorrect.

Option (a) ⇒ usual form of abc is 100xa + 10xb + c.

Option (b) ⇒ The usual form of abco is 1000xa + 100xb + 10xc + 0x1 = 1000 a + 100 b + 10c. Therefore (b) is not correct option

Option (c) ⇒ The usual form of aobc is 1000xa + 100xo + 10xb + 1xc = 1000 a + 10 b + c. Therefore option (c) is correct.

Option (d) ⇒ The usual form of aboc is 1000xa + 100xb + 10xo + 1xc = 1000 a + 100 b + c. Therefore option (d) is incorrect.

General form of abc is 100a + 10b + c and general form of cba is 100c + 10b + a .

Now abc – cba = (100a + 10b + c) – (100c + 10b + a)

⇒ abc – cba = 100a – a + 10b – 10b + c – 100c.

⇒ abc – cba = 99a -99c = 99(a-c)

⇒ abc – cba is divisible by 99 because 99 is factor of abc – cba.

Hence all numbers which are factors of 99 will also be divisible by abc – cba.

Option (a) ⇒ 9 is the factor of 99 ( 11 x 9 = 99). Therefore abc-cba is divisible by 9.

Option (b) ⇒ 11 is the factor of 99 ( 11 x 9 = 99). Therefore abc-cba is divisible by 11

Option (c) ⇒ 18 is not the factor of 99 because does not give 0 remainder. Therefore abc - cba is not divisible by 18.

Option (d) ⇒ 33 is the factor of 99 (33 x 3 = 99). Therefore abc-cba is divisible by 33.

Three numbers can be formed by the digits x,y and z they are xyz, yzx and zxy.

⇒ xyz = 100x + 10y + z.

⇒ yzx = 100y + 10z + x.

⇒ zxy = 100z + 10x + y.

⇒ xyz + yzx + zxy = (100x + 10y + z) + (100y + 10z + x) + (100z + 10x + y).

⇒ xyz + yzx + zxy = 100x + 10x + x + 100y + 10y + y + 100z + 10z + z

⇒ xyz + yzx + zxy = 111x + 111y + 111z.

⇒ xyz + yzx + zxy = 111( x + y + z).

⇒ xyz + yzx + zxy is divisible by 111 because 111 is the common factor .

Also xyz + yzx + zxy will be divisible by factors of 111.

Option (a) ⇒ 11 is not the factor of 111 (becausedoes not give 0 as remainder).

Option (b) ⇒ 33 is not the factor of 111 (becausedoes not give 0 as remainder).

Option (c) ⇒ 37 is the factor of 111 gives 0 as remainder and ( 37 x 4 = 111). Therefore it is correct option.

Option (d) ⇒ 74 is not the factor of 111 (becausedoes not give 0 as remainder).

If a number is divisible by 55 than it must alsobe divisible by factors of 55 that is 5 and 11.

∵ aabb is divisible by 5

∴ By divisibility test of 5 b must be 0 or 5.

Therefore (c) option is correct.

abc = 100a + 10b + c.

bca = 100b + 10c + a.

cab = 100c + 10a + b.

⇒ abc + bca + cab = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b).

⇒ abc + bca + cab = 100a + 10a + a + 100b + 10b + b + 100c + 10c + c

⇒ abc + bca + cab = 111a + 111b + 111c.

⇒ abc + bca + cab = 111( a + b + c).

⇒ abc + bca + cab is divisible by 111 and (a + b + c) .

Also abc + bca + cab will be divisible by factors of 111

Option (a) ⇒ abc + bca + cab is divisible by (a + b + c) because abc + bca + cab = 111(a + b + c).

Option (b) ⇒ 3 is divisible by 111 because sum of digits of 111 is 3 and 3 is divisible by 3( Divisibility test of 3).

Option (c) ⇒ 37 is the factor of 111 gives 0 as remainder and ( 37 x 4 = 111) therefore 111 is divisible by 37

Option (d) ⇒ 9 is not factor of 111 because not gives 0 as remainder. Therefore 111 is not divisible by 9 .

If a number is divisible by 55 than it must alsobe divisible by factors of 55 that is 5 and 11.

∵ 4ab5 is divisible by 11

∴ By divisibility test of 11 difference of the sum of the alternate digits must be a multiple of 11.

⇒ (4 + b) – (a + 5) = 0,11,22…

⇒ 4 + b – a – 5 = 0,11,22…

⇒ b – a -1 = 0,11,22…

⇒ b – a – 1 = 0 ( least value of b – a).

⇒ b – a = 1

Therefore option (b) is correct.

Since abc is a three digit number

∴ General form of abc is abc = 100a + 10b + c

Now; abc – a – b – c = 100a + 10b + c – a – b – c.

⇒ abc – a – b – c = 100a – a + 10b – b + c – c.

⇒ abc – a – b – c = 99a + 9b.

⇒ abc – a – b – c = 9( 11a + b).

∴ abc – a – b – c is divisible by 9 because 9 is one of its factor.

Hence the correct option is (a).

According to the question the number can be of form abcabc.

The general form of number abcabc is

Abcabc = 100000a + 10000b + 1000c + 100a + 10b + c.

Abcabc = 100000a + 100a + 10000b + 10b + 1000c + c.

Abcabc = 100100a + 10010b + 1001c.

Abcabc = 1001(100a + 10b + c).

Clearly we can see that abcabc is divisible by 1001 because 1001 is one of the factor of it.

∴ Option (d) is correct.

Option (a) ⇒ If the number is even meaning that if last digit is 0,2,4,6,8 than it is divisible by three.

Option (b) ⇒ The divisibility test of 3 says that if the sum of all the digits is divisible by three than the number is divisible by three. Therefore this option is correct.

Option (c) ⇒ If the number is divisible by both 3 and 2 than it is divisible by 6.

Option (d) ⇒ if the sum of all digits is divisible by 9 than the number is divisible by 9.

Since the sum of the digits of xyz i.e x + y + z = 6

∴ By divisibility test of three the number xyz is divisible by 3.

Also the last digit (z) is odd digit.

∴ xyz is the odd multiple of three.

Hence the option (a) is correct.

According to the question;

In one’s column A + 3 = 5

⇒ A is the number whose addition with 3 gives 5 in one’s place

∵ A is a digit hence its value is between 0 and 9

∴ A = 2 (2 + 3 = 5).

In ten’s column 5 + B + 0 = 6 (there is no carry from A + 3)

⇒ 5 + B = 6

⇒ B = 6 – 5 = 1;

Hence A = 2 and B = 1

∴ Option (c) is correct.

According to the question;

In one’s place B + 3 = 0

⇒ B is the number whose addition with 3 gives 0 in a one’s place.

⇒ B = 7 (7 + 3 = 10 and one’s place of 10 is 0 )

In ten’s place A + 8 + 1 = 15 ( 1 is the carry from B + 3)

⇒ A + 9 = 15

⇒ A = 15 – 9 = 6

∴ A + B = 6 + 7 = 13.

Hence the option (a) is correct.

According to the question

A x A = 9

⇒ The one’s place in the product of A and A is 9

∴ A is the number whose multiplication with itself gives a two digit number having 9 in it’s one’s place.

Hence A = 7 because A x A = 49 and one’s place of 49 is 9

∴ Option (c) is correct.

6 A × B = A 8 B

AB = B

⇒ A = 1

6B = A8

⇒ 6B = 18

⇒ B = 3

A – B = 1 – 3 = -2

For a number to be divisible by 99 it should be divisible by both 11 and 9.

Option (a) ⇒ Divisibility by 9: sum of digits = 9 + 1 + 3 + 4 + 6 + 2 = 25 and we know that 25 is not divisible by 9. Therefore the number 913462 is not divisible by 9. Hence it is not divisible by 99.

Option (b) ⇒ Divisibility by 9: sum of digits = 1 + 1 + 4 + 3 + 4 + 5 = 18. And 18 is divisible by 9 .Hence the number 114345 is divisible by 9.

Divisibility by 11 : difference of the sum of alternate digits = (1 + 4 + 4) – (1 + 3 + 5) = 9 – 9 = 0 .And 0 is multiple of 11 (11 x 0 = 0) .Therefore the number 114345 is divisible by 11

Since the number is divisible by both 11 and 9

∴ The number is divisible by 99.

Option (c) ⇒ Divisibility by 9: sum of digits = 1 + 3 + 5 + 7 + 9 + 2 = 27. And 27 is divisible by 9 .Hence the number 114345 is divisible by 9.

Divisibility by 11 : difference of the sum of alternate digits = (1 + 5 + 9) – (3 + 7 + 2) = 15-12 = 3. And 3 is not a multiple of 11.Therefore the number 114345 is not divisible by 11.

∴ the number 135792 is not divisible by 99.

Option (d) ⇒ Divisibility by 9: sum of digits = 3 + 5 + 7 + 2 + 4 + 0 + 6 = 27. And 27 is divisible by 9 .Hence the number 114345 is divisible by 9.

Divisibility by 11 : difference of the sum of alternate digits = (3 + 7 + 4 + 6) – (5 + 2 + 0) = 20-7 = 13. And 13 is not a multiple of 11.Therefore the number 114345 is not divisible by 11.

∴ the number 3572406is not divisible by 99.

We know that a number is a multiple of 3 and 9 if the sum of digits is divisible by 3 and 9 respectively.

Adding all the digits in the number we get 3 + 1 + 3 + 4 + 6 + 7 + 3 = 27 which is divisible by 9.

We know that a number is a multiple of 3 if the sum of all the digits in that number is a multiple of 3.

2 + 0 + x + 3 = x + 5

Since x>0, x + 5’s the closest multiples of 3 are 6,9 and 12.

⇒ x + 5 = 6 gives x = 1

⇒ x + 5 = 9 gives x = 4

⇒ x + 5 = 12 gives x = 7

⇒ Therefore, x is 1, 4 or 7.

We know that a number is a multiple of 9 if the sum of its digits is divisible by 9.

⇒ 3 + x + 5 = x + 8

⇒ x + 8 is divisible by 9 if x + 8 = 9, which gives x = 1.

Let x and y be the digits in a two-digit number A = xy.

Reversing the digits, we get B = yx.

⇒ Sum of the digits in A is (10x + y).

⇒ Sum of the digits in B is (10y + x).

⇒ Sum S = A + B = (10x + y) + (10y + x).

⇒ S = 10x + x + 10y + y = 11x + 11y = 11(x + y)

⇒ The number is divisible by 11.

Let x and y be the digits in a two-digit number A = xy.

Reversing the digits, we get B = yx.

Sum of the digits in A is (10x + y).

Sum of the digits in B is (10y + x).

Difference S = A + B = (10x + y) - (10y + x).

S = 10x-x-10y + y = 9x-9y = 9(x-y)

The number is divisible by 9.

Let x, y and z be the digits in a three-digit number A = xyz.

Reversing the digits, we get B = zyx.

Sum of the digits in A is (100x + 10y + z).

Sum of the digits in B is (100z + 10y + x).

Difference S = A + B = (100x + 10y + z) - (100z + 10y + x).

S = 100x-x + 10y-10y + z-100z = 99x-99z

99(x-z) = 9x11(x-z)

The number is divisible by 9 and 11.

Adding all the digits according to their places,

20 + B + 10A + B = 80 + A

9A + 2B = 60

9A = 60-2B

Since A and B both are whole numbers, 60-2B should be divisible by 9.

60-2B = 54 so B = 3.

Substituting B = 3 in 9A + 2B = 60,

we get A = 6.

Since the units place contains a 6, BxB should give a number with 6 in units place. That is possible when B = 4 or 6.

If B = 4, (10A + B)x(B) = 96 becomes (10A + 4)x4 = 96

40A + 16 = 96, gives A = 2.

If B = 6, (10A + B)x(B) = 96 becomes (10A + 6)x6 = 96

60A + 36 = 96, gives A = 1.

Therefore, A = 2 B = 4 or A = 1 B = 6 (Both are valid)

The answer contains 3 digits, 4 9 and B.

From the problem, it is clear that the value of B should be greater than 4 and less than 9.

If the value of B is less than 4, the answer will contain only two digits instead of 3.

If the value of B is greater than 9, the solution will contain more than 3 digits.

Since the units place contains 1, and B ranges anywhere between 4 and 9, The tenth place when multiplied by B gives 49.

⇒ BxB = 49.

⇒ B = 7.

We know that a number is a multiple 9 if the sum of digits is divisible by 9.

Adding all the digits in the number we get 1 + x + 3 + 5 = 9 which gives x + 9 = 9 so x = 0.

From the divisibility by 11 rule, if abcd is divisible by 11,

Then a-b + c-d = 0,

gives a + c = b + d, or each should be 0.

It also follows that a + c or b + d should be divisible by 11.

Let abcd be a four-digit number.

The digits at odd places are a and c, even places are b and d. (a + c)-(b + d) = 0.

It also follows that (a + c) – (b + d) should be divisible by 11 if (a + c)-(b + d)≠0.

This is a direct consequence from the divisibility by 11 rule.

Adding the corresponding places to the number we have 3A = 10 + A

⇒ 2A = 10

⇒ A = 5.

B can take values from 4 to 9. (Squares with more than 2 digits)

Say B = 4, 4x4 = 16 so A = 1, B = 6.

The number initially was tu or 10t + u, but after adding 1 in its units place, tu gets shifted one unit higher.

This implies, t shifts to 100^th place, u shifts to 10^th place, 1 is allocated in units place.

The new number is tu1 or 100t + 10u + 1.

True(T)

If we have a two digit number and at the units place we have an even number ,then that number becomes an even number and every even number is divisible by 2(test of divisibility by 2)

For example: 26 and 34 are the two digit even numbers .

False(F)

If we have a three digit number abc and if c is even then it becomes an even number. Even numbers are those numbers which end with a digit like 0,2,4,6,8.

as c is an even number, it can be any number among 0,2,4,6,8.When c is 0 only in that case abc will be divisible by 5. Only those numbers are divisible by 5 whose units digit is either 0 or 5(test of divisibility by 5).

For example:240,360

False(F)

If there is a four digit number abcd and if ab is divisible by 4 then its not necessary that abcd will also be divisible by 4.If a number is divisible by 4,then the number formed by it’s digits in unit’s place and ten’s place is divisible by 4.

For example: Let us suppose the number abcd is 2463 , here 24 is divisible by 4 but 2463 is not divisible by 4.

True(T)

If a number is divisible by 6,then it is divisible by both 2 and 3.

Since abc is divisible by 6,it is also divisible by 2 and 3.Therefore c is an even number and sum of it’s digits is divisible by 3.

For example: The three digit numbers are 234,138

Hence the three –digit number is divisible by 6

True(T)

When the number is in the form 3N + 2 ,

which can also be written as ,

3×N + 2

⇒ (multiple of 3) + 2

Therefore , when we divide 3N + 2 by 3,then we will get a remainder 2.

For example:11, 17

11 can be written as ,

3×3 + 2

and

17 can be written as ,

3×5 + 2

Thus,11 and 17 will leave the remainder 2 when divided by 3.

True(T)

When the number is in the form 7N + 1 ,

which can also be written as ,

7×N + 1

⇒ (multiple of 7) + 1

Therefore , when we divide 7N + 1 by 7,then we will get a remainder 1.

For example:22, 29

22 can be written as,

7×3 + 1

and

29 can be written as ,

7×4 + 1

Thus,22 and 29 will leave the remainder 1 when divided by 7.

True(T)

Here it’s given a is divisible by b,

Let b = x1.x2 ,where x1 and x2 are prime numbers.

Since a is divisible by b,a is a multiple of b.

i.e. a = p.b

⇒ a = p.x1.x2

Or a = cx2 = dx1 where c = px1 and d = px2

⇒ a is a multiple of p1 as well as p2.

Hence a is divisible by each factor of b.

False(F)

Here we have,

So,B×4 is a two digit number whose units digit is 2.

Therefore, the value of B is either 3 or 8.

But if B = 3,B×4 = 12 and A×4 + 1 ≠19,for any value of A between 0 to 9.

Therefore,B should be equal to 8,then B×4 = 32,

A × 4 + 3 = 19(when we multiply AB by 4, refer the above multiplication)

Hence,A×4 = 19-3 = 16

⇒ A = 4

Therefore,A = 4 ,B = 8,and A + B = 4 + 8 = 12

True(T)

Here, we have

This implies B + C is either 2 or two digit number whose one’s

digit is 2.

Its given that B ≠ 0, C ≠ 0,so B + C is a two digit number whose one’s digit is 2.

When B = C = 1, then A should be 3 so that we have

But in this case A + B + C ≠ 14,so B = C = 1 cannot happen.

(I)Here either B or C should be 5 or 7 and A should be 3 so that

A + B + C = 14.

(II)If B = C = 6 and A = 2, then

and A + B + C = 14

False(F)

Here it’s given 213x27 is divisible by 9,where x is a digit. As this number 213x27 is divisible by 9, then when we sum up the digits of this number it should be a multiple of 9.

So, when we add all the digits, of the number 213x27,we get

2 + 1 + 3 + x + 2 + 7 = 15 + x.

Then 15x should be equal to any multiple of 9.i.e.9,18,27,etc

When a number is divisible by 9 then when we sum up the digits of that number it results in a multiple of 9.

Here, lets assume 15 + x = 18

⇒ x = 3

False(F)

Here it’s given N ÷ 5 leaves remainder 3

⇒ N = 5n + 3,where n = 0,1,2,3,..

and N ÷ 2 leaves remainder 0

⇒ N is an even number

But N = 5n + 3,it’s sum of two terms whose second term is odd.

Therefore,5n should be an odd number.

5n can be odd when n = 1,3,5,…

So, in this case when N = 5n + 3

⇒ N = 5(1) + 3 = 8 when(n = 1)

Hence when we substitute n = 1,3,5,.. in N = 5n + 3,we get 8,18,28 etc

Now, when we divide N by 10 ,N can be written as

N = 10×n + 8,when n = 0,1,2,3,..

Therefore, when N ÷ 10 ,always leaves remainder 8.

If a number is divisible by 8,then the number obtained by the last three digits of the number should be divisible by 8.

So, 6a2 should be divisible by 8.

Here,a can be any value from 0,1,…9.

When a = 0,6a2 = 602,which is not divisible by 8.

When a = 1,6a2 = 612,which is not divisible by 8.

When a = 2,6a2 = 622, which is not divisible by 8.

When a = 3,6a2 = 632,which is divisible by 8.

Hence, the least value of a is 3,which makes the number 91876a2 divisible by 8.

Here we have

where Q - P = 3

⇒ P is 4 or 6 .

[ P×P has unit digit 6]

If P = 4 then we have,(1 × P) + 1 = Q(we are multiplying 1 P and P, refer the above multiplication(4×4 = 16))

⇒ (1 × 4) + 1 = Q

⇒ 5 = Q or Q = 5

If Q = 5,then Q - P = 5 - 4 = 1 ≠ 3

Therefore, P = 6.

When P = 6, then we have , (1×P) + 3 = Q(we are multiplying 1 P and P, refer the above multiplication,(6×6 = 36))

⇒ (1×6) + 3 = 6 + 3 = 9

When Q = 9,then Q – P = 9-6 = 3

Hence P and Q are 6 and 9 respectively.

Here we have,

As it’s mentioned that there is no carry-over addition, then

1 + C = 6

⇒ C = 5

A + C = 9

⇒ A + 5 = 9

⇒ A = 4

B + A = 7

⇒ B = 7-A = 7-4 = 3

Therefore, the value of A + B + C = 4 + 3 + 5 = 12

Given: A five-digit number AABAA is divisible by 33.

When the number AABAA is divisible by 33 then its divisible by 3 and 11 because when a number say x is divisible by y, then x is divisible by the factors of y as well.

As AABAA is divisible by 3,we can say that sum of all the digits is a multiple of 3.

Therefore A + A + B + A + A = 0,3,6,9,12…

⇒ 4A + B = 0,3,6,9,12…

Now as AABAA is divisible by 11,then according to the divisibility rule of 11,the difference of sum of digits of odd places and even places should be divisible by 11.

Therefore, for the number AABAA, we can write the difference of sum of digits of odd places and even places as

(A + B + A)-(A + A) = 0,11,22,…

⇒ (2A + B)-(2A) = 0,11,22,..

⇒ B = 0(As B is a digit in the number AABAA,it can be only 0)

So, A can be any digit like 3,6,9,…and B = 0.

Hence the numbers in the form of AABAA divisible by 33 are

33033,66066 and 99099.