Number System Solved Problems | Hamaraguru

Number System Solved Problems | Hamaraguru

1. Find the number of factors that exactly divide the number 32.
32 can be divided by 2, 4, 8, 16, and 32.
Therefore the number of factors that exactly divide 32 is 5.

2. Which of the following numbers are divisible by 198?
792

594

992 

1385 

2178 

Since 198 = 2 × 9 × 11
The numbers are divisible by 198 if they are divisible by 2, 9, and 11.
Now 792 is divisible by 2, 9, and 11
594 is divisible by 2, 9, and 11
992 is divisible by 2, but not by 9 and 11
1385 is odd. So it is not divisible by 2
1980 is divisible by 2, 9, and 11
Therefore, among the given numbers, 792, 594, and 1980 are divisible by 198.

3. Find the largest 5 digit number that is divisible by 99.
Largest 5 digit number = 99999
When we divide 99999 by 99, the remainder is 9.
Therefore, the largest 5 digit number exactly divisible by 99 is 99999 − 9 = 99990.

4. Find the smallest 4 digit number divisible by 56.
The smallest 4 digit number is 1000.
When 1000 is divided by 56, the remainder is 48.
Since 56 − 48 = 8.
Therefore, the smallest 4 digit number divisible by 56 = 1000 + 8 = 1008

5. Find the sum of the first 7 prime numbers.
Sum of first 7 prime numbers = 2 + 3 + 5 + 7 + 11 + 13 + 17 = 58

6. The number 34567*5432 is exactly divisible by 6. Find the smallest natural number that can replace the ‘*’ in the given number.
A number is perfectly divisible by 6 if it is divisible by 2 and 3.
Since the number is even, it is divisible by 2.
The number is perfectly divisible by 3 if the sum of the digits in the number is divisible by 3.
Let the required number be x.
Now, 3 + 4 + 5 + 6 + 7 + 5 + 4 + 3 + 2 = 39. 
So the smallest number that can replace the ‘*’ could be 0, but 0 is not a natural number.
The next possible number is 3, which is a natural number.
Therefore if 3 replaces the ‘*,’ it will be divisible by 3.

7. A certain number, when divided by 56, leaves 29 as a remainder. What will be the remainder when the same number is divided by 8?
Let the number be x
Let q be the quotient when x is divided by 56
Since dividend = quotient × divisor + remainder,
x = 56q + 29


Therefore, the remainder of the required division will be the remainder when 29 is divided by 8.
When 29 is divided by 8 the remainder is 5.
Therefore, the remainder of the required division is 5.

8. When a number is divided by 3 leaves remainder 2. What is the remainder when the square of this number is divided by 3?
Let the number be x
Let q be the quotient for the division x ÷ 3
Since dividend = quotient × divisor + remainder,
x = 3q + 2
Therefore, square of the number


We see that x^2 = (a multiple of 3) + 1
Therefore, the remainder of division x^2 ÷ 3 is 1.

9. When a three-digit number 5p2 is added to another three-digit number 873, the result is 13q5. Find the value of q – p.


From this addition, we can see that p + 7 = q, Or, q − p = 7

10. Find the smallest possible value of y such that the product of 3865 and 8y3 is divisible by 15.
A number is perfectly divisible by 15 if it is divisible by both 3 and 5.
Since 3865 is divisible by 5, the given product will be divisible by 15 if 8y3 is divisible by 3.
8 + 3 = 11.
12 is the closest number to 11 that is divisible by 3.
Therefore, the smallest possible value of y is 1.

11. Find the smallest possible number that must be subtracted from 20645 so that it is divisible by 43.
When 20645 is divided by 43, the remainder is 5. 
Therefore, if 5 is subtracted from 20645, the resulting number is divisible by 43.