Practice Test: Number System- 2 - Civil Engineering (CE) MCQ

Calculation:

Let the second number be m.

Using the relationship between HCF, LCM, and the numbers, we have:

• m × 77 = 11 × 616
• Solving for m: m = 616 / 7
• m = 88

Therefore, the second number is 88.

In each set of 100 numbers, there are 10 numbers whose tens digit and unit digit are the same. Again, in the same set, there are 10 numbers whose hundreds and tens digits are the same.

But one number in each set of 100 numbers whose Hundreds, Tens and Unit digit are the same as 111, 222, 333, 444, etc.
Hence, there are exactly (10 + 10 - 1) = 19 numbers in each set of 100 numbers. Further, there are 9 such sets of numbers.
Therefore such total numbers = 19 × 9 = 171

Alternatively,
9 × 10 × 10 - 9 × 9 × 9 = 900 - 729 = 171

Step 1. Count the Trailing Zeros

Trailing zeros come from factors of 10 in n!. Since 10 = 2 × 5, and there are always more factors of 2 than 5, the number of trailing zeros is determined by the number of factors of 5.

For 96!:

  ⌊96/5⌋ = 19
  ⌊96/25⌋ = 3
  ⌊96/125⌋ = 0

Thus, the total number of factors of 5 (and hence factors of 10) is 19 + 3 = 22. In other words, 96! ends in 22 zeros. To find the last nonzero digit, we “remove” these factors of 10 from the product.

Step 2. Use the Recursive Formula

A recursive formula for the last nonzero digit of n!, denoted d(n), is:

  d(n) = { (last nonzero digit of n! computed directly)     if n < 5,
       d(⌊n/5⌋) × f(n mod 5) × 2^(⌊n/5⌋) (mod 10)  if n ≥ 5 }

Here, the function f(r) gives the last nonzero digit of r! for 0 ≤ r < 5. Specifically, we have:

  f(0) = 1,  f(1) = 1,  f(2) = 2,  f(3) = 6,  f(4) = 4.

Step 3. Compute d(96)

Write 96 in the form 96 = 5q + r. Here:

  q = ⌊96/5⌋ = 19
  r = 96 mod 5 = 1

Thus, the formula gives:  d(96) = d(19) × f(1) × 2⁽¹⁹⁾ (mod 10).

Since f(1) = 1, this simplifies to:  d(96) = d(19) × 2⁽¹⁹⁾ (mod 10).

Step 4. Compute d(19)

Now, write 19 = 5q′ + r′. Here:

  q′ = ⌊19/5⌋ = 3
  r′ = 19 mod 5 = 4

Thus: d(19) = d(3) × f(4) × 2⁽³⁾ (mod 10).

For n = 3 (which is less than 5), we directly have:

  d(3) = f(3) = 6.

Also, f(4) = 4 and 2⁽³⁾ is represented as 2³ = 8. Therefore:

  d(19) = 6 × 4 × 8 = 192.

Taking this modulo 10:  192 (mod 10) = 2.

Step 5. Finish Calculating d(96)

as :  d(96) = d(19) × 2⁽¹⁹⁾ (mod 10).

We found d(19) = 2, so: d(96) = 2 × 2⁽¹⁹⁾ (mod 10), 

Now, compute 2¹⁹ (mod 10). Notice that powers of 2 modulo 10 cycle every 4:

  2¹ = 2,
  2² = 4,
  2³ = 8,
  2⁴ = 16 ≡ 6 (mod 10),
  the cycle repeats.

Since 19 mod 4 = 3, we have:

  2¹⁹ ≡ 2³ ≡ 8 (mod 10).

Thus: d(96) = 2 × 8 = 16 (mod 10).

Taking modulo 10 gives: 16 (mod 10) = 6.

The last nonzero digit of 96! is 6.

A and B are playing a game with integers from 1 to 100, placing signs between them. A starts by putting a plus sign. The result is determined by whether the final sum is even or odd. If even, A wins; if odd, B wins.

The analysis of the game is as follows:

• All numbers from 1 to 100 include 50 odd numbers.
• Adding any two odd numbers results in an even number.
• The sum of 50 odd numbers will also be even.

Therefore, no matter how the signs are placed, the result will always be even. Thus, A will always win the game.

Step by Step Solution:

Step 1

Identify the prime factorization of 10: 10 = 2¹ · 5¹.

Step 2

Determine the prime factorization of 10⁵:
10⁵ = (2¹ · 5¹)⁵ = 2⁵ · 5⁵.

Step 3

A factor of 10⁵ can be expressed as 2^a · 5^b where 0 ≤ a ≤ 5 and 0 ≤ b ≤ 5.

Step 4

For a factor to end with exactly one zero, it must be of the form 10¹ · k, where k is a factor of 10⁴. This means we need a ≥ 1 and b ≥ 1.

Step 5

The number of choices for a (from 1 to 5) is 5 and for b (from 1 to 5) is also 5. Therefore, the total number of factors is 5 × 5 = 25.

 4⁹⁶/6

We can write it in this form
(6 - 2)⁹⁶/6
Now, the Remainder will depend only on the powers of -2. So,
(-2)⁹⁶/6, It is same as
([-2]⁴)²⁴/6, it is the same as
(16)²⁴/6
Now,
(16 * 16 * 16 * 16..... 24 times)/6
On dividing 16 individually, we always get a remainder 4.
So,
(4 * 4 * 4 * 4............ 24 times)/6.
Hence, Required Remainder = 4
NOTE: When 4 has even number of powers, it will always give remainder 4 on dividing by 6.

1. Let:

   • E = number of problems solved by all 3 (easy problems),
   • M = number of problems solved by exactly 2 of them,
   • D = number of problems solved by exactly 1 of them (difficult problems).

2. Because there are 100 distinct problems total: E + M + D = 100.

3. Each person solved 60 problems, so the total number of “solves” is 3 × 60 = 180.

   • Each easy problem (solved by all 3) contributes 3 “solves.”
   • Each problem solved by exactly 2 people contributes 2 “solves.”
   • Each difficult problem (solved by exactly 1) contributes 1 “solve.”

   Hence: 3E + 2M + D = 180.

4. From E + M + D = 100, we can write M = 100 − E − D.
   Substitute into 3E + 2M + D = 180:

   3E + 2(100 − E − D) + D = 180
   3E + 200 − 2E − 2D + D = 180
   (3E − 2E) + (−2D + D) + 200 = 180
   E − D + 200 = 180
   E − D = −20
   D − E = 20.

So, the difference between the number of difficult problems and easy problems (D − E) is 20.

The smallest integer in the series is 1 and the greatest is 50. To ensure that every element of S1 is greater than or equal to each element of S2, we need to add a number x to each element of S1. The minimum value of x must be:

• Calculated as follows: the smallest element 1 plus the difference between the greatest and smallest integers, which is (G - L).
• This results in 49 since 50 - 1 = 49.

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

For example, if the integers from a1 to a50 are 1, 2, 3, ... , 50:

• S1 consists of 1, 2, 3, ..., 24,
• S2 consists of 25, 26, ..., 50.

His roll number must meet specific criteria: it starts with 267 and is divisible by both 11 and 13, as well as being a multiple of 7.

To find the unit digit of this number, we can follow these steps:

• The full roll number can be expressed as 267xxx, where xxx represents the last three digits.
• The number 267xxx must be divisible by 1001 (since 1001 = 7 × 11 × 13).
• Next, we check the multiples of 1001 that start with 267:

Calculating the first few multiples of 1001:

• 1001 × 267 = 267267
• 1001 × 268 = 268268

Only 267267 fits the criteria of starting with 267.

Finally, the unit digit of 267267 is 7.

We are given that when 7179 and 9699 are divided by a natural number N, the remainder obtained is the same. This means that 7179 - 9699 must be divisible by N.

First, calculate the difference between 9699 and 7179:

9699 - 7179 = 2520

Now, we need to find the divisors of 2520 that end with one or more zeros. ie multiples of 10.

Step 1: Find the divisors of 2520.

2520 = 2^3 × 3^2 × 5 × 7

l number of divisors is given by the product of one plus each of the exponents in the prime factorization:

(3 + 1) × (2 + 1) × (1 + 1) × (1 + 1) = 4 × 3 × 2 × 2 = 48

So, 2520 has 48 divisors in total.

Step 2: Find divisors of 2520 that are multiples of 10

.To be a multiple of 10, the divisor must include at least one factor of 2 and one factor of 5.

From the prime factorization of 2520, we know that there are 3 factors of 2 and 1 factor of 5, so the divisors of 2520 that are multiples of 10 must include at least one 2 and one 5.

the remaining factors after we take out one factor of 2 and one factor of 5 from the prime factorization of 2520:

2520 / 10 = 2^2 × 3^2 × 7

The number of divisors of 2^2 × 3^2 × 7 is:

(2 + 1) × (2 + 1) × (1 + 1) = 3 × 3 × 2 = 18

So, there are 18 divisors of 2520 that are multiples of 10.

Step 3: Conclusion The number of values of N that end with one or more zeros is 18.

Thus, the correct answer is:

c) 18.

Remainder,

To find the remainder when 73 * 75 * 78 * 57 * 197 * 37 is divided by 34, we can calculate the individual remainders:

• 73 mod 34 = 5
• 75 mod 34 = 7
• 78 mod 34 = 10
• 57 mod 34 = 23
• 197 mod 34 = 27
• 37 mod 34 = 3

Now, we multiply these remainders:

(5 * 7 * 10 * 23 * 27 * 3) mod 34

Calculating step-by-step:

• 5 * 7 = 35 mod 34 = 1
• 1 * 10 = 10
• 10 * 23 = 230 mod 34 = 28
• 28 * 27 = 756 mod 34 = 6
• 6 * 3 = 18

Finally, we find:

18 mod 34 = 18

However, we need to recalculate to ensure accuracy.

Using the simplified method:

• Remainders: 5, 7, 10, 23, 27, 3
• Product: (1 * -4 * -11 * -7) mod 34
• Calculating gives us: 66 mod 34 = 32

Thus, the required remainder is 32.

Prime numbers less than 10 are 2, 3, 5, and 7.

If the difference between the largest and smallest number ends in 5, the prime numbers must be 7 and 2 in the last positions.

The smallest and largest numbers can be arranged as:

• Smallest: 257
• Largest: 752

Since the sum of the digits must be greater than 13, x must be 5.

Verifying:

• 752 - 257 = 495

Thus, the answer is option (B).

The product of the numbers is:

• 7 × 5 × 2 = 70

This number 1212121212... repeated 300 times is divisible by 9. We can express it as:

• 1212121212... 300 times = 9N, where N is the quotient obtained when divided by 9.

Now, we need to find the remainder when this number is divided by 99:

• 1212121212... 300 times / 99 = Remainder of 9 divided by 9 times 134680... written 50 times divided by 11.

[Note: 121212 = 13468 x 9]

Next, we find the remainder of 134680134680... (50 times) when divided by 11 using the divisibility rule:

• Applying the rule from the right-hand side:
• The remainder when 134680... (50 times) is divided by 11 is 2.

Thus, the total remainder is:

• 18.

Alternatively, we can use the divisibility rule of 10 to find the remainder in this case.

Number: ababab

This can be represented as:

• abab = ab × 10000 + ab × 100 + ab
• Which simplifies to: ab (10000 + 100 + 1)
• Thus, it becomes: ab × 10101

Since ab is a common factor, every six-digit number of the form ababab is divisible by 10,101.

Correct Answer: A

Explanation: The unit digit of a number is determined by its last digit.

For the number 76476756749, the last digit is 9. We can observe the pattern of the unit digits of powers of 9:

• 9¹ = 9
• 9² = 81 (unit digit is 1)
• 9³ = 729 (unit digit is 9)
• 9⁴ = 6561 (unit digit is 1)

The unit digit alternates between 9 and 1. Since 8754874878 is an even number, the unit digit will be 1.

Thus, the correct option is A.

Find the unit digit:

To determine the last digit of 17 raised to the power of 19 and 19 raised to the power of 13, follow these steps:

• The base is 17, but we only need the unit digit, which is 7.
• The unit digit of powers of 7 follows a cycle of 4:
• 7¹ = 7
• 7² = 49 (last digit 9)
• 7³ = 343 (last digit 3)
• 7⁴ = 2401 (last digit 1)
• Next, find 19¹³ modulo 4 to determine the position in the cycle:
• Calculating 19 modulo 4 gives a remainder of 3.
• Thus, we need to calculate 7³.

The unit digit of 7³ is 3.

The number can be expressed in the form of (7X + 4), since dividing by 7 yields a remainder of 4.

Using a trial and error method, we can determine the number:

• Let X = 17.
• Then, 7X + 4 = 7 × 17 + 4 = 119 + 4 = 123.

Now, let's check the conditions:

• Dividing 123 by 7: Quotient = 17, Remainder = 4.
• Dividing 17 by 5: Quotient = 3, Remainder = 2.
• Dividing 3 by 4: Remainder = 3.

All conditions are satisfied.

Now, let's find the remainders when 123 is divided by the new numbers:

• Dividing 123 by 8: Quotient = 15, Remainder = 3.
• Dividing 15 by 5: Quotient = 3, Remainder = 0.
• Dividing 3 by 6: Remainder = 3.

Four bells ringing together and ring at an interval of 12 sec, 15 sec, 20 sec, and 30 sec respectively. How many times will they ring together in 8 hours?

The bells ring at the following intervals:

• 12 seconds
• 15 seconds
• 20 seconds
• 30 seconds

To find out how many times they ring together in 8 hours, we first calculate the least common multiple (LCM) of their intervals:

LCM of (12, 15, 20, 30) = 60 seconds

This means the bells ring together once every 60 seconds, or once per minute.

Now, we can calculate the number of times the bells ring together:

• In 1 minute: 1 time
• In 1 hour (60 minutes): 1 x 60 = 60 times
• In 8 hours: 8 x 60 = 480 times

We must also include the initial instance when all the bells first ring together:

So, the total number of times they ring together is:

• 480 + 1 = 481

Therefore, the bells will ring together 481 times in 8 hours.

The greatest number that divides 14, 20, and 32 leaving the same remainder is determined as follows:

To find this number, we use the concept of the highest common factor (HCF).

The required number is the HCF of the differences between the numbers:

• 20 - 14 = 6
• 32 - 20 = 12
• 32 - 14 = 18

Now, we calculate the HCF of these differences:

• HCF(6, 12, 18) = 6

Thus, the greatest number that divides 14, 20, and 32, leaving the same remainder, is 6.