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Test: Number System- 3 - UPSC MCQ


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20 Questions MCQ Test CSAT Preparation - Test: Number System- 3

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Test: Number System- 3 - 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.

Detailed Solution for Test: Number System- 3 - Question 1

According to the question, 

Given information is,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 that 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.

1080 = 23 x 32 x 51.
Now, N x Q = 1080 and M x (Q - 25) = 1080
By Doing hit and trial method, 
N = 27
Hence the answer is 9.

Test: Number System- 3 - Question 2

A rectangular floor is fully covered with square tiles of identical size. The tiles on the edges are white and the tiles in the interior are red. The number of white tiles is the same as the number of red tiles. A possible value of the number of tiles along one edge of the floor is :

Detailed Solution for Test: Number System- 3 - Question 2

Let the rectangle has x and y tiles along its length and breadth respectively. 
The number of white tiles 
W = 2x + 2(y – 2) = 2 (x + y – 2) 
And the number of red tiles = R = xy – 2 (x + y – 2) 
Given that the number of white tiles is the same as the number of red tiles 
⇒ 2 (x + y – 2) = xy – 2 (x + y – 2) 
⇒ 4 (x + y – 2) = xy 
⇒ xy – 4x – 4y = –8 
⇒ (x – 4) (y – 4) = 8 = 8 ×1 or 4 × 2 
⇒ m – 4 = 8 or 4 
⇒m = 12 or 8 
Therefore, the number of tiles along one edge of the floor can be 12

Test: Number System- 3 - 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

Detailed Solution for Test: Number System- 3 - Question 3

Let the four-digit number be abcd, where a, b, c, and d are the digits of the number from the thousands place to the ones place.

Since the sum of the digits equals the product of the digits, we have:
a + b + c + d = abcd

And since the number is divisible by the sum of the digits, we have:
abcd / (a + b + c + d) = integer

Combining the two equations, we get:
abcd / (abcd) = integer
1 = integer

This means that the sum of the digits of the number must be equal to 1, which is only possible if the digits are 1, 1, and two 0s. However, this can't be a four-digit number since the thousands place can't be 0.

We also know that the sum of the digits is a factor of the four-digit number. Let's try combinations of numbers that multiply to a four-digit number.

Since the number is divisible by the sum of the digits, we can try to find a sum of digits that is a factor of a four-digit number. For example, if we try a sum of 8, we can find a four-digit number that is divisible by 8.

Let's try a sum of 8:
a + b + c + d = 8
abcd / 8 = integer

We need the product of the digits to also be 8, so the possible sets of digits are:
1, 1, 2, 4 (product = 8)
1, 1, 1, 8 (product = 8)

We can eliminate the second option (1, 1, 1, 8) since the maximum product we can get from three 1s and an 8 is 8, which is not a four-digit number.

Now let's try the first option (1, 1, 2, 4). We can arrange these digits to form different four-digit numbers and check if any of them are divisible by 8:

1124: not divisible by 8
1142: not divisible by 8
1214: not divisible by 8
1241: not divisible by 8
1412: not divisible by 8
1421: not divisible by 8
2114: not divisible by 8
2141: not divisible by 8
2411: not divisible by 8
4112: divisible by 8
4121: not divisible by 8

The number 4112 is divisible by 8 and has a sum of digits equal to 8 (1 + 1 + 2 + 4 = 8). Therefore, the sum of the digits of the number is 8.

Test: Number System- 3 - 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?

Detailed Solution for Test: Number System- 3 - Question 4

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

Test: Number System- 3 - Question 5

Let S be a two-digit number such that both S and S2 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?

Detailed Solution for Test: Number System- 3 - Question 5

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.

Test: Number System- 3 - 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

Detailed Solution for Test: Number System- 3 - Question 6

Eliminate the options.
For example, option (1) can be eliminated by assuming N = 2

Test: Number System- 3 - 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?

Detailed Solution for Test: Number System- 3 - Question 7

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.

Test: Number System- 3 - 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?

Detailed Solution for Test: Number System- 3 - Question 8

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

Test: Number System- 3 - Question 9

What is the remainder when (103 + 93)752 is divided by 123?

Detailed Solution for Test: Number System- 3 - Question 9

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)

Test: Number System- 3 - 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?

Detailed Solution for Test: Number System- 3 - Question 10

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

Test: Number System- 3 - Question 11

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

Detailed Solution for Test: Number System- 3 - Question 11

Test: Number System- 3 - 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?

Detailed Solution for Test: Number System- 3 - Question 12

Going through the options, 8181/1818 = 4.5

Test: Number System- 3 - 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?

Detailed Solution for Test: Number System- 3 - Question 13

Check it through the options.
Alternatively, answer will be LCM (2,3,4,..., 9) - 1 = 2520 - 1 = 2519

Test: Number System- 3 - Question 14

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

Detailed Solution for Test: Number System- 3 - Question 14

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.

Test: Number System- 3 - Question 15

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

Detailed Solution for Test: Number System- 3 - Question 15

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

Test: Number System- 3 - Question 16

Find the unit digit:
346 765 * 768 983 * 987 599

Detailed Solution for Test: Number System- 3 - Question 16

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   (for unit digit  remainder of (power)/4 is checked and periodicity is checked as per base no ) like r(remainder)=983/4 is 3 so 83 unit digit is 2
The unit digit of 987599 = 3
6 * 2 * 3 = 36
Hence, the unit digit is 6.

Test: Number System- 3 - Question 17

A certain number when successively divided by 4, 5 and 7 leaves remainders 2, 3 and 4 respectively. Find such a least number.

Detailed Solution for Test: Number System- 3 - Question 17

Here,

d1 = 4, d2 = 5 and d3 = 7

 r1 = 2, r2 = 3 and r3 = 4

Start from last digits

The number gives a remainder of 4 when it is divided by 7

⇒ (7x + 4)
The number gives remainder 3 when it is divided 5

⇒ [5 × (7x + 4) + 3]
Also, the number when divided by 4 gives remainder 2

⇒ [4 × {5 × (7x + 4) + 3} + 2]

⇒ [4 × (35x + 20 + 3) + 2]

⇒ 140x + 94  [∵ Least quotient is 1]

⇒ 140 + 94 = 234

∴ Such least number is 234.

Test: Number System- 3 - 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?

Detailed Solution for Test: Number System- 3 - Question 18

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.

Test: Number System- 3 - Question 19

A number when divided by 48 leaves a remainder of 31. Find the remainder if the same number is divided by 24

Detailed Solution for Test: Number System- 3 - Question 19

Correct option is B

Let the number be N
Since the number when divided by 48 leaves a remainder of 31,  we have N=48(n)+31 where 'n' is the quotient got by dividing number by 48

We can write N=24×2×n+31 = 24×(2n)+24+7 = 24(2n+1)+7

So, when N is divided by 24, remainder left is 7

Test: Number System- 3 - Question 20

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

Detailed Solution for Test: Number System- 3 - Question 20

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.

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