Practice Test: Number System- 3 - Interview Preparation MCQ

According to the question, Raju had to divide 1080 by N, a two-digit number. Instead, he used M, which is the result of reversing the digits of N. This led to a quotient that was 25 less than expected. Both N and M must divide 1080 exactly.

The prime factorisation of 1080 is:

• 1080 = 2³ × 3² × 5¹

From the conditions:

• N × Q = 1080
• M × (Q - 25) = 1080

Using a trial-and-error method:

• If N = 27, then M = 72 (the reverse of 27).
• Both 27 and 72 divide 1080 evenly.

The sum of the digits of N (27) is:

• 2 + 7 = 9

Therefore, the answer is 9.

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.

Let the rectangle have x and y tiles along its length and breadth, respectively.

• The number of white tiles is: W = 2x + 2(y - 2) = 2(x + y - 2)
• The number of red tiles is: R = xy - W = xy - 2(x + y - 2)

Given that the number of white tiles equals 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

The pairs of factors of 8 are:

• 8 and 1
• 4 and 2

From these pairs:

• If x - 4 = 8, then x = 12
• If x - 4 = 4, then x = 8

Thus, a possible value for the number of tiles along one edge of the floor is 12.

There is only one number, 145, that meets the condition where the sum of the factorials of its digits equals the number itself. The factorials of the digits are calculated as follows:

• 1! = 1
• 4! = 24
• 5! = 120

The sum of these factorials is:

1 + 24 + 120 = 145

Since no other three-digit number satisfies this condition, we conclude that:

• Exactly one such number exists: 145.

The correct option is Option A.

In this question, several restrictions apply:

• If S and S² end with the same digit, possible digits are 1, 5, or 6 (as 0 is not allowed).
• The unit digit of the square of the reversed number must be 6. Thus, the tens digit of the original number must be either 4 or 6.

The potential numbers are:

• 41
• 45
• 46
• 61
• 65
• 66

However, to meet the condition that the square is less than 3000, the valid options are:

• 41
• 45
• 46

Thus, the total number of possible values for S is three.

d) None of the foregoing statements is necessarily correct.

To understand why this is the case, consider the following points:

• When N is a positive integer greater than 1, the expression N! (the factorial of N) is the product of all positive integers up to N.
• The expression N! - 1 will not be divisible by any integer from 2 to N, as each of these integers divides N! and leaves a remainder of 1.
• This property does not guarantee that any of the four statements hold true for all values of N.
• Therefore, we can conclude that none of the proposed statements about divisors of N! minus 1 are universally valid.

Hence, the correct answer is D.

The number of mistakes made by the students ranges from 0 to 14, providing a total of 15 options for mistakes. Since there are 16 students, it is inevitable that at least two students must have made the same number of mistakes.

This could include scenarios where two students made no mistakes at all. Therefore, the first statement is definitely true:

• At least two students made the same number of mistakes.

A remainder can never be greater than the number that is the dividing factor.

Therefore, the remainder must be less than 123. This limits our options significantly.

• Option A: 10
• Option B: 729
• Option C: 752
• Option D: 1000

Among these options, only 10 is less than 123, making it the correct choice.

Thus, the answer is A.

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:

To determine the minimum value of x, consider the following:

• S1 represents a set of numbers.
• S2 is another set of numbers.
• For x to make every element in S1 greater than or equal to every element in S2, we need to ensure:
• Each element in S1 + x ≥ each element in S2.

This means that the smallest element in S1 plus x must at least equal the largest element in S2. Therefore:

• Let G be the largest value of S2.
• Let L be the smallest value of S1.
• The minimum value of x can be calculated as:
• x ≥ G - L.

Thus, the correct answer is (G - L).

The solution to the problem involves finding a year in the 19th century that, when viewed in a mirror, appears to be 4.5 times larger.

To solve this:

• Consider the year 1818.
• When viewed in a mirror, it appears as 8181.
• Calculating the ratio: 8181 divided by 1818 equals 4.5.

Thus, the year the teacher referred to is 1818.

To find the value of N, we can follow these steps:

• Start by determining the least common multiple (LCM) of the numbers from 2 to 10.
• The LCM of 2, 3, 4, 5, 6, 7, 8, 9, and 10 is 2520.
• Since N must give a remainder of 1 when divided by each of these numbers, we calculate:
• N = LCM - 1 = 2520 - 1 = 2519.

Thus, the value of N is 2519.

The total number of four-digit numbers in the octal system can be calculated as follows:

• The first digit can be any number from 1 to 7 (7 options).
• The remaining three digits can each be any number from 0 to 7 (8 options for each).

Therefore, the calculation is:

• 7 (first digit) × 8 × 8 × 8 = 3584.

In octal representation, this total is equal to 7000.

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

• The digits of 4112 are 4, 1, 1, and 2.
• Calculating the sum: 4 + 1 + 1 + 2 = 8.
• Calculating the product: 4 × 1 × 1 × 2 = 8.
• Both the sum and product equal 8.
• Additionally, 4112 is divisible by 8 (4112 ÷ 8 = 514).

Thus, the sum of the digits of the number is 8.

Solution:

To find the unit digit of the expression, follow these steps:

• Step 1: Determine the unit digit of each term.
• Step 2: Calculate the product of these unit digits.
• Step 3: The unit digit of the overall product will be the unit digit of the computed product.

The unit digits are as follows:

• The unit digit of 346⁷⁶⁵ is 6.
• The unit digit of 768⁹⁸³ is 2. (This is found by checking the remainder of the exponent divided by 4: 983 ÷ 4 gives a remainder of 3, and the unit digit of 8³ is 2.)
• The unit digit of 987⁵⁹⁹ is 3. (The exponent's remainder when divided by 4 is also 3, making the unit digit of 7³ equal to 3.)

Now, multiply the unit digits:

6 * 2 * 3 = 36.

Thus, the unit digit of the product is 6.

To find the least number that leaves specific remainders when divided by 4, 5, and 7, follow these steps:

Step 1: Division by 4

• Let the number be x.
• When x is divided by 4, the remainder is 2. This can be written as:
• x = 4 * q1 + 2, where q1 is the quotient.

Step 2: Division by 5

• The quotient q1 is divided by 5, leaving a remainder of 3:
• q1 = 5 * q2 + 3.
• Substituting back into the equation for x gives:
• x = 4 * (5 * q2 + 3) + 2
• x = 20 * q2 + 14.

Step 3: Division by 7

• The quotient q2 is divided by 7, leaving a remainder of 5:
• q2 = 7 * q3 + 5.
• Substituting into the equation for x gives:
• x = 20 * (7 * q3 + 5) + 14
• x = 140 * q3 + 114.

Finding the Least Number

• To find the smallest positive integer x, set q3 to 0:
• x = 140 * 0 + 114, thus x = 114.

Verification

1. Divide 114 by 4:
   • 114 divided by 4 gives a quotient of 28 with a remainder of 2.
2. Divide the quotient (28) by 5:
   • 28 divided by 5 gives a quotient of 5 with a remainder of 3.
3. Divide the next quotient (5) by 7:
   • 5 divided by 7 gives a quotient of 0 with a remainder of 5.

All conditions are satisfied with the number 114.

Answer: The least such number is 114.

Let the number be N and its quotient be k.

The number N can be expressed as:

• N = 841k + 87

We need to determine the remainder when N is divided by 29. This can be done as follows:

• Calculate N = (841k + 87)
• Now, divide by 29:
• Both 841 and 87 are completely divisible by 29.

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

Let the number be N.

Since the number divided by 48 leaves a remainder of 31, we can express this as:

• N = 48n + 31, where n is the quotient.

To find the remainder when N is divided by 24, we can rewrite N:

• N = 24(2n) + 31

Now, we need to calculate the remainder of 31 when divided by 24:

• 31 = 24 × 1 + 7

Thus, the remainder when N is divided by 24 is 7.

Solution:

To solve for n in the equation n² = 123456787654321, we can observe a pattern in the squares of numbers comprised solely of the digit 1:

• 11² = 121
• 111² = 12321
• 1111² = 1234321
• 11111² = 123454321
• 111111² = 12345654321
• 11111111² = 123456787654321

From this pattern, we conclude that n = 11111111.

Let the number be N and its quotient be k.

The number N can be expressed as:

• N = 703k + 75

We need to find the remainder when N is divided by 37. We can express this as:

• (703k + 75) / 37

Since 703 is divisible by 37, it contributes a remainder of 0. Now, we need to find the remainder of 75 when divided by 37:

• 75 ÷ 37 leaves a remainder of 1.

Therefore, the remainder when N is divided by 37 is:

• 0 + 1 = 1.

Assume the numbers are a and b, ab = 616

Let A = 100x + 10y + z and B = 100z + 10y + x. Given the condition that B - A is divisible by 7, we can express this as:

• B - A = 99(z - x)

This means that (z - x) must be divisible by 7. The possible pairs for z and x that satisfy this condition are:

• 8 and 1 (since 8 - 1 = 7)
• 9 and 2 (since 9 - 2 = 7)

The digit y can take any value from 0 to 9. Therefore:

• The lowest value of A occurs when x is 1, y is 0, and z is 8, resulting in A = 108.
• The highest value of A occurs when x is 2, y is 9, and z is 9, resulting in A = 299.

Thus, the values of A are restricted to the range:

• 108 ≤ A ≤ 299

Therefore, the range 106 < a=""><> is necessarily true.