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QUESTION: 1

Raju had to divide 1080 by N, a two-digit number. Instead, he performed the division using M which is obtained by reversing the digits of N and ended up with a quotient which was 25 less than what he should have obtained otherwise. If 1080 is exactly divisible both by N and M, find the sum of the digits of N.

Solution:

1080 = 2^{3} x 3^{2} x 5^{1}.

Now, N x Q = 1080 and M x (Q - 25) = 1080

Do a bit of hit and trial.

N = 27

QUESTION: 2

Ajay took a 4-digit number in base 5 notation. He subtracted the sum of the digits of the numbers from the number. From the result, he struck off one of the digits. The remaining 3 digits were 1, 0 and 2. Then the digit struck off by Ajay was:

Solution:

QUESTION: 3

A teacher wrote a number on the blackboard and the following observations were made by the students. The number is a four-digit number.The sum of the digits equals the product of the digits. The number is divisible by the sum of the digits.The sum of the digits of the number is

Solution:

Using options, the only possible value is 4112. The key here is: The sum of the digits equals the product of the digits.

QUESTION: 4

The sum of the factorials of the three-digits of a 3-digit number is equal to the three-digit number formed by these three digits, taken in the same order. Which of the following is true of the number of such three-digit numbers, if no digit occurs more than once?

Solution:

There is only one number, 145, which exhibits this property.

QUESTION: 5

Let S be a two-digit number such that both S and S^{2} end with the same digit and none of the digits in S equals zero. When the digits of S are written in the reverse order, the square of the new number so obtained has the last digit as 6 and is less than 3000. How many values of S are possible?

Solution:

**The correct option is Option A.**

**In this question, several restrictions are operating: If S and S2 are ending with the same unit digit, then it can be 0, 1,5,6, but it is given that none of the digits is equal to zero, so the unit digit can be only 1, 5, 6. Next, the unit digit of the square of the number written in reverse order is 6, so the tens place digit of the actual number should be either 4 or 6.**

**So, the actual numbers could be 41, 45, 46, 61, 65, 66.**

**Now, this square is less than 3000, so the only possibilities are 41, 45, 46.**

QUESTION: 6

Let N be a positive integer not equal to 1. Then none of the numbers 2, 3,...., N is a divisor of (N! - 1). Thus, we can conclude that

Solution:

Eliminate the options.

For example, option (1) can be eliminated by assuming N = 2

QUESTION: 7

16 students were writing a test in a class. Rahul made 14 mistakes in the paper, which was the highest number of mistakes made by any student. Which of the following statements is definitely true?

Solution:

The number of mistakes made by all the students will be between 0 and 14, i.e., students are having a total of 15 options to make mistakes. Since the number of students = 16, at least two students will have the same number of mistakes (that can be zero also, i.e., two students are making no mistakes). Hence, option 1 is the answer.

QUESTION: 8

What is the least number of soldiers that can be drawn up in troops of 12, 15, 18 and 20 soldiers and also in form of a solid square?

Solution:

In this type of question, We need to find out the LCM of the given numbers.

LCM of 12, 15, 18 and 20;

12 = 2 × 2 × 3;

15 = 3 × 5;

18 = 2 × 3 × 3;

20 = 2 × 2 × 5

Hence, LCM = 2 × 2 × 3 × 5 × 3 × 2 × 2 × 3 × 5 × 3

Since, the soldiers are in the form of a solid square.

Hence, LCM must be a perfect square. To make the LCM a perfect square, We have to multiply it by 5,

hence,

The required number of soldiers

= 2 × 2 × 3 × 3 × 5 × 5 × 2 × 2 × 3 × 3 × 5 × 5

= 900

QUESTION: 9

What is the remainder when (10^{3} + 9^{3})^{752} is divided by 123?

Solution:

A remainder can never be greater than the no. which is the dividing factor.

hence, the remainder < 123

which leaves us with only one option, i.e. (a)

QUESTION: 10

When 7179 and 9699 are divided by another natural number N , remainder obtained is same. How many values of N will be ending with one or more than one zeroes?

Solution:

9699 - 7129 = 2520.

Prime Factors of 2520: 1, 23, 32, 5, 7.

To get a zero at end: (5*2), (5*22), (5*23)-> 3 ways

& in the remaining numbers 1, 32, 7 -> possible combinations are 1,3,9,7, 21, 63 -> 6 ways.

possible ways of N => 6*3 ways= 18 ways

QUESTION: 11

Every element of S1 is made greater than or equal to every element of S2 by adding to each element of SI an integer x. Then, x cannot be less than

Solution:

x must be equal to the greatest difference in the value of number of S1 and S2

QUESTION: 12

The History teacher was referring to a year in the 19th century. Rohan found an easy way to remember the year. He found that the number, when viewed in a mirror, increased 4.5 times. Which year was the teacher referring to?

Solution:

Going through the options, 8181/1818 = 4.5

QUESTION: 13

N is a number which when divided by 10 gives 9 as the remainder, when divided by 9 gives 8 as the remainder, when divided by 8 gives 7 as the remainder, when divided by 7 gives 6 as the remainder, when divided by 6 gives 5 as the remainder, when divided by 5 gives 4 as the remainder, when divided by 4 gives 3 as the remainder, when divided by 3 gives 2 as the remainder, when divided by 2 gives 1 as the remainder.What is N?

Solution:

Check it through the options.

Alternatively, answer will be LCM (2,3,4,..., 9) - 1 = 2520 - 1 = 2519

QUESTION: 14

How many different four digit numbers are there in the octal (Base 8) system, expressed in that system?

Solution:

The total number of numbers of four digits in octal system = 7 x 8 x 8 x 8 = 3584 When we convert this number into octal system, this is equal to 7000.

QUESTION: 15

Solution:

Using options, the only possible value is 4112. The key here is: The sum of the digits equals the product of the digits.

QUESTION: 16

Find the unit digit:

346 ^{765} * 768 ^{983} * 987 ^{599}

Solution:

In this type of problem

__Step 1:__ we find the unit digit of each term

__Step 2__: we find the product of the unit digits of each term

__Step 3: __ The unit digit of the product will be the product of whole number

The unit digit of 346 ^{765} = 6

The unit digit of 768 ^{983 }= 2

The unit digit of 987^{599 }= 3

6 * 2 * 3 = 36

Hence, the unit digit is 6.

QUESTION: 17

If a number when successively divided by 9, 7, and 5 leaves the remainders 8, 5, and 1 respectively. Find the smallest such 4-digit number.

Solution:

This is a problem of successive division.

Divisors 9 7 5

Remainders 8 5 1

The smallest such numbers will be {(1 * 7) + 5} * 9 + 8} Which will be equal to 116.

Now, the general form of the numbers will be:

(9 * 7 * 5) k + 116

= 315k + 116

When we put k = 3

Then the number we obtain is 315 * 3 + 116

The largest such 3-digit number is 1061.

QUESTION: 18

A number when divided by 841 gives a remainder of 87. What will be the remainder when we divide the same number by 29?

Solution:

Let the number be N and its quotient be k.

Then the number N can be written in the form of:

N = 841k + 87

Now, we have to find out the what will be the remainder when it is divided by 29.

The number is (841k + 87)

Let’s divide it by 29

(841k + 87)/ 29

841 and 87 both are completely divisible by 29.

Therefore, the remainder when the number N is divided by 29 is 0.

QUESTION: 19

A number when divided by 703 gives a remainder of 75. What will be the remainder when we divide the same number by 37?

Solution:

Let the number be N and its quotient be k.

Then the number N can be written in the form of:

N = 703k + 75

Now, we have to find out the what will be the remainder when it is divided by 37.

The number is (703k + 75)

Let’s divide it by 37

(703k + 75)/ 37

703 is divisible by 37 hence, remainder will be 0 whereas, 75 when divided by 37 leaves remainder 1.

Therefore, the remainder when the number N is divided by 37 will be (0 + 1) i.e. 1.

QUESTION: 20

Find the remainder when a number is divided by 49. The same number when divided by 147 leaves 53 as remainder.

Solution:

Let the number be N and its quotient be k.

Then the number N can be written in the form of:

N = 147k + 53

Now, we have to find out the what will be the remainder when it is divided by 49.

The number is (147k + 53)

Let’s divide it by 49

(147k + 53)/ 49

147 is divisible by 49 hence, remainder will be 0 whereas, 53 when divided by 49 leaves 1 as remainder.

Therefore, the remainder when the number N is divided by 49 will be (0 + 4) i.e. 4.

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