Number System- 2 - Free MCQ Practice Test with solutions, CAT Quant Aptitude

Calculation:

Let, 2nd number be m,

⇒ m × 77 = 11 × 616

⇒ m = 616/7

⇒ m = 88

∴ The 2nd number is 88.

It is given that a^m. bⁿ = 144¹⁴⁵, where a > 1 and b > 1.

144 can be written as 144 = 2⁴ x 3²

Hence, a^m . bⁿ = 144¹⁴⁵ can be written as a^m . bⁿ = (2⁴ x 3²) = 2⁵⁸⁰ x 3²⁹⁰ 

We know that 3²⁹⁰ is a natural number, which implies it can be written as a¹, where a > 1

Hence, the least possible value of m is 1. Similarly, the largest value of n is 580.

Hence, the largest value of (n-m) is (580-1) = 579

For 1421:

• 1421 ÷ 12 = 118 remainder 5
• The remainder when 1421 is divided by 12 is 5.

For 1423:

• 1423 ÷ 12 = 118 remainder 7
• The remainder when 1423 is divided by 12 is 7.

For 1425:

• 1425 ÷ 12 = 118 remainder 9
• The remainder when 1425 is divided by 12 is 9.

Step 2: Multiply the remainders

• Now, multiply the remainders from each of the divisions:
• 5 × 7 × 9
• First, multiply 5 and 7:
• 5 × 7 = 35
• Now, divide 35 by 12:
• 35 ÷ 12 = 2 remainder 11
• So, 5 × 7 gives a remainder of 11.
• Next, multiply 11 by 9:
• 11 × 9 = 99
• Now, divide 99 by 12:
• 99 ÷ 12 = 8 remainder 3
• Final Answer:
• The remainder when N is divided by 12 is 3.

Step by Step Solution:

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

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

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

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.

The number of choices for a (from 0 to 4) is 5 and for b (from 0 to 4) is 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.

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

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 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.

 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.

This is an alternating addition and subtraction series. In the first pattern, 10 is subtracted from each number to arrive at the next. In the second, 5 is added to each number to arrive at the next.

The series is a geometric progression where each term is multiplied by 3.

• 2 * 3 = 6
• 6 * 3 = 18
• 18 * 3 = 54
• 54 * 3 = 162

Thus, the next number in the series is 162.

This is a simple subtraction series.
Each number is 35 less than the previous number.
(544 - 35 = 509) (509 - 35 = 474) (474 - 35 = 439) (439 - 35 = 404)...

The series is 14, 28, 20, 40, 32, 64
Take three terms at a time. The notation be as follows:-
Let, First term = a = 14
It is evident that Second term= 2*a = 2 * 14 = 28 and,
The Third term = Second term - 8 OR ( 2*a - 8 ) = 28 - 8 = 20
Now consider the third, fourth and the fifth term.
Let, Third term = a = 20
It is evident that Fourth term= 2*a = 2 * 20 = 40 and,
The Fifth term = Second term - 8 OR ( 2*a - 8 ) = 40 - 8 = 32
Similarly consider fifth, sixth and the seventh(term to be found).
Let, Fifth term = a = 32
It is evident that Sixth term = 2*a = 2 * 32 = 64 and,
The Seventh term ( term to calculated) = Second term - 8 OR ( 2*a - 8 ) = 64 - 8 = 56
 

Let the digits be x and y respectively.
Therefore, x + y = 5 ----------- (1)
Original Number : 10x + y
Reversed Number : 10y + x
10x + y - (10y + x) = 27
10x + y - 10y - x = 27
9x - 9y = 27
9 (x - y) = 27
x - y = 3 ------------ (2)
Adding equation (1) and (2)
2x = 8
x = 4
Substituting value for x in equation (1), we get
4 + y = 5
y = 1
The numbers are 41 and 14.

• Define the Digits:

  • Let the digit in the ten’s place be x.
  • Then, the digit in the unit’s place is x + 5.

• Form the Two-Digit Number:

  • The two-digit number can be expressed as 10x + (x + 5) = 11x + 5.

• The sum of the Digits:

  • The sum of the digits is x + (x + 5) = 2x + 5.

• Set Up the Ratio:

  • According to the problem, the ratio of the number to the sum of its digits is 3:1.
  • Therefore, we have:
    11x + 5 / 2x + 5 = 3

• Solve for x:

  • Multiply both sides by 2x + 5:
    11x + 5 = 3(2x + 5)
  • Expand the right side:
    11x + 5 = 6x + 15
  • Subtract 6x from both sides:
    5x + 5 = 15
  • Subtract 5 from both sides:
    5x = 10
  • Divide by 5:
    x = 2

• Find the Unit’s Place Digit:

  • The unit’s place digit is x + 5 = 2 + 5 = 7.

• Form the Number:

  • The two-digit number is 10x + (x + 5) = 10(2) + 7 = 27.

LCM of 4,6 and 8 is 24
Divide 1000 by 24, we get quotient = 41 and 16 as remainder
so 41 numbers are there which are divisible by 4,6 and 8 together.

The number would be in the form of (7X+4) as when this number is divide by 7, will give remainder 4.
Now, we will try hit and trial method to obtained the number.
Put, X=17, then
7X+4=7×17+4=119+4=123
Now, when 123 divided by 7, gives quotient 17 , remainder =4
17 divided by 5, quotient =3, remainder =2
3 divide by 4 gives remainder 3.
So for first condition satisfied. 

Now, 123 divided by 8, quotient =15, remainder =3
15 divided by 5, quotient =3, remainder =0
3 divided by 6, remainder =3.

Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec 

Now we have to take LCM of time interval

⇒ LCM of (12, 15, 20, 30) = 60

No. of Times bell rings in 1 min (60 seconds) = 1 .

No. of times bell rings in 1 hr (60 minutes) = 1 x 60 = 60

No. of Times bell rings in 8 hr = 8 x 60 = 480 

No. of Times bell rings in 8 hr + the first time all bell rings also needs to be counted

⇒ 480 + 1 

∴ The bell ringing 481 times in 8 hours.

Here, the number which divides 14, 20, and 32 leaves the same remainder.
∴ We will be using HCF model 2
The required number will be the HCF of (20 - 14), (32 - 20), and (32 - 14).
i.e. HCF (6, 12, 18)
which will be 6.
Therefore, the required number is 6.