Number system digit problems are very important type questions. A sample problem goes like this:

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Take a number 256. In this number 6 is in units place, 5 is in tenth's place, 2 is in 100th place. In decimal system, the place value increases by 10 times while going to the left.

So 256 can be written as 100 × 2 + 10 × 5 + 1 × 6

Similarly a four digit number abcd can be written as 1000 × a + 100 × b + 10 × c + 1 × d = 1000a + 100b + 10c + d

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1. A two digit number when 18 added becomes another two digit number with reversed digits. How many such two digit numbers are possible?

a. 2

b. 3

c. 7

d. 8

Correct Option: C

Explanation:

Let the two digit number be 'ab' and the number formed after adding 18 to it is 'ba'

So ab + 18 = ba

\( \Rightarrow \) (10a + b) + 18 = 10b + a

\( \Rightarrow \) 18 = 9b – 9a

\( \Rightarrow \) 2 = b – a

So we know that ab, ba both are two digit numbers so a, b \( \ne \) 0.

Also when b = 9, a = 7; b = 8, a = 6; b = 7, a = 5; b = 6, a = 4; b = 5, a = 3; b = 4, a = 2; b = 3, a = 1

Total 7 combinations are possible. So correct option is c.

2. For a positive integer n, let \({P_n}\) denote the product of the digits of $n$ and \({S_n}\) denote the sum of the digits of $n$. The number of integers between 10 and 100 for which \({P_n} + {S_n} = n\)

a. 2

b. 3

c. 9

d. 12

Correct option: c

Explanation:

Let the two digit number be 'ab'

Given, a × b + a + b = 10a + b

a × b = 9a

b = 9

Therefore, for b = 9, the above condition satisfies. So all the two digit numbers for which units digit is 9 are our solutions. They range from 19 to 99. Total 9 numbers.

3. Ray writes a two digit number. He sees that the number exceeds 4 times the sum of its digits by 3. If the number is increased by 18, the result is the same as the number formed by reversing the digits. Find the sum of the digits of the number.

a. 8

b. 9

c. 10

d. 12

Correct option: a

Explanation:

Let the two digit number be xy.

4(x + y) +3 = 10x + y - - - (1)

10x + y + 18 = 10 y + x - - - (2)

Solving 1st equation,

4x + 4y + 3 = 10x + y

3y + 3 = 6x

6x – 3y = 3

Therefore, 2x – y = 1 - - - (3)

Solving 2nd equation,

xy + 18 = yx

\( \Rightarrow \) (10x + b) + 18 = 10y + x

\( \Rightarrow \) 18 = 9y – 9x

\( \Rightarrow \) 2 = y – x

we get y - x = 2 - - - (4)

adding 3 and 4, we get x = 3

By substituting in (4), we get y = 5

So the given number is 35. Sum of the digits = 8.

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4. The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B>A and B-A is perfectly divisible by 7, then which of the following is necessarily true?

a. 100<A<299

b. 106<A<305

c. 112<A<311

d. 118<A<317

Correct option: B

Explanation:

Let A = abc the B = cba

Given, B – A is exactly divisible by 7.

Therefore, 100c + 10b + a – (100a + 10b + c) = 99c – 99a = 99(c – a)

So 99(c – a) should be divisible by 7.

We know that 99 is not divisible by 7. So (c – a) should be divisible by 7 which implies c – a is 0, 7, 14...

c – a = 0 is not possible as c = a implies A = B but given that B > A

c – a = 14 is not possible as the maximum difference between c and a = 9 – 1 = 8 only.

So c – a = 7.

If c = 9, a = 2

c = 8, a = 1

b can take any value from 0 to 9

Therefore, minimum value of abc = 109, maximum value = 299

From the given options, option B satisfies this.

4. M = abc is a three digit number and N = cba, if M > N and M - N + 396c = 990. Then how many values of M are more than 300.

a. 20

b. 30

c. 40

d. 200

Correct option: a

Explanation:

From the given data,

abc – cba + 396c = 990

100a + 10b + c – (100c + 10b + a) + 396c = 990

99a – 99c + 396c = 990

Observe that each term is divisible by 99. So on dividing the above expression by 99, we get

a – c + 4c = 10

a + 3c = 10

For c = 1, a = 7

c = 2, a = 4

c = 3, a = 1

'b' can take any value from 0 to 9

We have to find the value of M more than 300. So minimum value of 'a' should be 4.

So total possibilities are 402, 412, ...., 492 = 10 values

701, 711, ....., 791 = 10 values

So total values = 20.

5. Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect square?

a. 2

b. 4

c. 0

d. 1

Correct option: d

Explanation:

Let the given number by xxyy

So 1000x + 100x + 10y + y = 1100x + 11y = 11(100x + y)

So 11(100x + y) has to be perfect square.

But the given expression is a multiple of 11 so 100x + y should also be a multiple of 11 and a multiple of perfect square.

Therefore, 11(100x + y) = 11 × (11 × k

Now give values for k and check the first two digits and last two digits are equal or not.

For k = 1, 11(100x + y) = 121 × 1

For k = 2, 121 × 2

. . . .

. . . .

For k = 8, 121 × 8

For k = 9, you get 9801. Ruled out.

If you further increase value of k, the given product becomes 5 digit number. So only one value is possible. That is 7744.

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*A two digit number when 18 added becomes another two digit number with reversed digits. How many such two digit numbers are possible?*"Take a number 256. In this number 6 is in units place, 5 is in tenth's place, 2 is in 100th place. In decimal system, the place value increases by 10 times while going to the left.

So 256 can be written as 100 × 2 + 10 × 5 + 1 × 6

Similarly a four digit number abcd can be written as 1000 × a + 100 × b + 10 × c + 1 × d = 1000a + 100b + 10c + d

**Very important note:**When you represent a number in this format, Left most digit can take any value from 1 to 9 but not**zero**. Remaining digits can take any value from 0 to 9.##

Solved Examples

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__Level 1 Problems:__

__Level 1 Problems:__

1. A two digit number when 18 added becomes another two digit number with reversed digits. How many such two digit numbers are possible?

a. 2

b. 3

c. 7

d. 8

Correct Option: C

Explanation:

Let the two digit number be 'ab' and the number formed after adding 18 to it is 'ba'

So ab + 18 = ba

\( \Rightarrow \) (10a + b) + 18 = 10b + a

\( \Rightarrow \) 18 = 9b – 9a

\( \Rightarrow \) 2 = b – a

So we know that ab, ba both are two digit numbers so a, b \( \ne \) 0.

Also when b = 9, a = 7; b = 8, a = 6; b = 7, a = 5; b = 6, a = 4; b = 5, a = 3; b = 4, a = 2; b = 3, a = 1

Total 7 combinations are possible. So correct option is c.

2. For a positive integer n, let \({P_n}\) denote the product of the digits of $n$ and \({S_n}\) denote the sum of the digits of $n$. The number of integers between 10 and 100 for which \({P_n} + {S_n} = n\)

a. 2

b. 3

c. 9

d. 12

Correct option: c

Explanation:

Let the two digit number be 'ab'

Given, a × b + a + b = 10a + b

a × b = 9a

b = 9

Therefore, for b = 9, the above condition satisfies. So all the two digit numbers for which units digit is 9 are our solutions. They range from 19 to 99. Total 9 numbers.

3. Ray writes a two digit number. He sees that the number exceeds 4 times the sum of its digits by 3. If the number is increased by 18, the result is the same as the number formed by reversing the digits. Find the sum of the digits of the number.

a. 8

b. 9

c. 10

d. 12

Correct option: a

Explanation:

Let the two digit number be xy.

4(x + y) +3 = 10x + y - - - (1)

10x + y + 18 = 10 y + x - - - (2)

Solving 1st equation,

4x + 4y + 3 = 10x + y

3y + 3 = 6x

6x – 3y = 3

Therefore, 2x – y = 1 - - - (3)

Solving 2nd equation,

xy + 18 = yx

\( \Rightarrow \) (10x + b) + 18 = 10y + x

\( \Rightarrow \) 18 = 9y – 9x

\( \Rightarrow \) 2 = y – x

we get y - x = 2 - - - (4)

adding 3 and 4, we get x = 3

By substituting in (4), we get y = 5

So the given number is 35. Sum of the digits = 8.

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__Level 2 Problems:__

4. The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B>A and B-A is perfectly divisible by 7, then which of the following is necessarily true?

a. 100<A<299

b. 106<A<305

c. 112<A<311

d. 118<A<317

Correct option: B

Explanation:

Let A = abc the B = cba

Given, B – A is exactly divisible by 7.

Therefore, 100c + 10b + a – (100a + 10b + c) = 99c – 99a = 99(c – a)

So 99(c – a) should be divisible by 7.

We know that 99 is not divisible by 7. So (c – a) should be divisible by 7 which implies c – a is 0, 7, 14...

c – a = 0 is not possible as c = a implies A = B but given that B > A

c – a = 14 is not possible as the maximum difference between c and a = 9 – 1 = 8 only.

So c – a = 7.

If c = 9, a = 2

c = 8, a = 1

b can take any value from 0 to 9

Therefore, minimum value of abc = 109, maximum value = 299

From the given options, option B satisfies this.

4. M = abc is a three digit number and N = cba, if M > N and M - N + 396c = 990. Then how many values of M are more than 300.

a. 20

b. 30

c. 40

d. 200

Correct option: a

Explanation:

From the given data,

abc – cba + 396c = 990

100a + 10b + c – (100c + 10b + a) + 396c = 990

99a – 99c + 396c = 990

Observe that each term is divisible by 99. So on dividing the above expression by 99, we get

a – c + 4c = 10

a + 3c = 10

For c = 1, a = 7

c = 2, a = 4

c = 3, a = 1

'b' can take any value from 0 to 9

We have to find the value of M more than 300. So minimum value of 'a' should be 4.

So total possibilities are 402, 412, ...., 492 = 10 values

701, 711, ....., 791 = 10 values

So total values = 20.

5. Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect square?

a. 2

b. 4

c. 0

d. 1

Correct option: d

Explanation:

Let the given number by xxyy

So 1000x + 100x + 10y + y = 1100x + 11y = 11(100x + y)

So 11(100x + y) has to be perfect square.

But the given expression is a multiple of 11 so 100x + y should also be a multiple of 11 and a multiple of perfect square.

Therefore, 11(100x + y) = 11 × (11 × k

^{2}) = 121 × k^{2}Now give values for k and check the first two digits and last two digits are equal or not.

For k = 1, 11(100x + y) = 121 × 1

^{2 }= 121 which is a 3 digit number so ruled out.For k = 2, 121 × 2

^{2 }= 484 also ruled out.. . . .

. . . .

For k = 8, 121 × 8

^{2 }= 7744For k = 9, you get 9801. Ruled out.

If you further increase value of k, the given product becomes 5 digit number. So only one value is possible. That is 7744.