Test: Number System- 3 - UPSC

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.

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

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.

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

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.

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

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.

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

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)

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

For the least element L in S1 to be greater than the greatest element or equal to G in S2, the number that is added to L cannot be less than G-L.

Going through the options, 8181/1818 = 4.5

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

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.

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

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 987⁵⁹⁹ = 3
6 * 2 * 3 = 36
Hence, the unit digit is 6.

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.

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.

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

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.