Number System Digit problems

Number System Digit problems

Number system digit problems are very important type questions.  A sample problem goes like this:
"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 345.  In this number 5 is in units place, 4 is in tenth's place, 3 is in 100th place.  In decimal system, the place value increases by 10 times while going to the left.
So 345 can be written as 100 × 3  +  10 × 4 + 1 × 5
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

Level 1 Problems:


1.  A two digit number when 27 added becomes another two digit number with reversed digits.  How many such two digit numbers are possible?
a. 2
b. 3
c. 7
d. 6
Correct Option: D
Explanation:
Let the two digit number be 'ab' and the number formed after adding 18 to it is 'ba'
So ab + 27 = ba
(10a + b) + 27 = 10b + a
27 = 9b – 9a
3 = b – a
So we know that ab, ba both are two digit numbers so a, b ≠\ne 0.
Also when b = 9, a = 6; b = 8, a = 5; b = 7, a = 4; b = 6, a = 3; b = 5, a = 2; b = 4, a = 1;
Total 6 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.  Raj 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
(10x + y) + 18 = 10y + x
18 = 9y – 9x
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.

6. The ten’s digit in a two-digit number is 3 less than the one’s digit. If we interchange the digits in the number,
we obtain a new number that, when added to the original number, results in the sum 143. Find this number.

 

1. The digits in a two-digit number add up to 9. If we interchange the digits in the number, we obtain a new
number that is 63 greater than the original number. Find the original number.
2. The digits in a two-digit number add up to 17. If we interchange the digits in the number, we obtain a new
number that is 9 less than the original number. Find the original number.
3. The digits in a two-digit number add up to 7. If we interchange the digits in the number, we obtain a new
number that is 45 less than the original number. Find the original number.
4. The one’s digit in a two-digit number is 7 greater than the ten’s digit. If we interchange the digits in the
number, we obtain a new number that, when added to the original number, results in the sum 99. Find this
number.
5. The ten’s digit in a two-digit number is 2 greater than the one’s digit. If we interchange the digits in the
number, we obtain a new number that, when added to the original number, results in the sum 176. Find this
number.

 

Problem 1.

Thrice the square of a natural number decreased by four times the number is equal to 50 more than the number. The number is,

  1. 4
  2. 5
  3. 6
  4. 10

Problem 2.

The difference between two positive numbers is 3. If sum of their squares is 369, then the sum of the numbers is,

  1. 25
  2. 81
  3. 27
  4. 33

Problem 3.

A number consists of two digits such that the digit in the ten's place is less than the digit in the unit's place by 2. Three times the number added to 6/7 th of the reverse of the number is 108. The sum of the digits of the number is,

  1. 6
  2. 8
  3. 9
  4. 7

Problem 4.

Of three numbers, the second is twice the first and thrice the third. If the average of the numbers is 44, the difference between the first number and the third number is,

  1. 18
  2. 12
  3. 6
  4. 24

Problem 5.

A two digit number is five times the sum of its digits. If 9 is added to the number, the digits interchange their positions. The sum of the digits of the number is,

  1. 11
  2. 9
  3. 6
  4. 7

Problem 6.

In a three digit number, the digit at the hundred's place is two times the digit at the unit's place and sum of the three digits is 18. If the digits are reversed the number is reduced by 396. The difference of hundred's and ten's digits of the number is,

  1. 5
  2. 2
  3. 1
  4. 3

Problem 7.

If the sum of two numbers are multiplied by  each number separately, the products so obtained are 247 and 114. The sum of the numbers is,

  1. 20
  2. 19
  3. 21
  4. 23

Problem 8.

A two digit number is three times the sum of its digits. If 45 is added to the number, its digits are interchanged. Sum of the digits of the number is,

  1. 11
  2. 7
  3. 5
  4. 9

Problem 9.

The numbers 2272 and 875 are divided by a three digit number giving same remainders. The sum of the digits of this three digit number is,

  1. 12
  2. 13
  3. 10
  4. 11

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 × k2) = 121 × k2
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 × 12  = 121 which is a 3 digit number so ruled out.
For k = 2, 121 × 22  = 484 also ruled out.
. . . .
. . . .
For k = 8, 121 × 82  = 7744
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.