Practice Test: Number System- 2 - CAT MCQ

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:

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 2ª · 5ᵇ 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 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.

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 of the series is 1 and greatest integer is 50. If each element of S1 is made greater of equal to every element of S2, then the smallest element 1 should be added to (50 - 1) = 49. Hence option (G-L) is the correct answer.

As this is a variable based question: the word "ANY" can be used.

Let the series of integers a1,a2,.......,a50 be 1,2,3,4,5,.......,50.

S1=1,2,3,4,.........24, S2=25,26,27,..........50.

His roll number is divisible by 1001 and the first three digits are 267. Hence the last three digits will also be 267.

When 7179 and 9699 are divided by another natural number N, remainder obtained is same.
 Let remainder is R, then (7179 — R) and (9699 — R) are multiples of N and {(9699 — R) — (7179 — R)} is multiple of N. Then 2520 is multiple of N or the largest value of N is 2520. Total factors of N which are multiples of 10 is 18. 

In this problem, the lights are flashed at the intervals 2.5, 4.25 and 5.125 seconds and put off after one second
each.

The total duration of intervals of these lights are (2.5+1) = 3.5 s, (4.25+1) = 5.25 s and (5.125+1) = 6.125 s.

We have to find the minimum duration. It would be the LCM of thes three numbers.

Since each word is put after a second. So LCM [5/2 + 1) (17/4 + 1) (41/8 + 1)] = LCM of numerator / HCF of denominator = 49*3/2 = 73.5.

Hence they will glow for full one second after 73.5-1 =72.5 sec.

Alright, let's tackle this problem step by step. We need to find the remainder when the product of the numbers 73, 75, 78, 57, 197, and 37 is divided by 34. 

>> Understanding the Problem

Calculating the product of all these numbers directly would give us a very large number, which isn't practical. Instead, we can use modular arithmetic to simplify the problem. The idea is to find each number modulo 34 first, then multiply those results together, and finally take modulo 34 again to find the remainder.

Step 1: Find Each Number Modulo 34

Let's compute each number modulo 34:

1.73 mod 34:
   - 34 multiplied by 2 is 68.
   - 73 minus 68 is 5.
   - So, 73 mod 34 is 5.

2. 75 mod 34:
   - 34 multiplied by 2 is 68.
   - 75 minus 68 is 7.
   - So, 75 mod 34 is 7.

3.78 mod 34:
   - 34 multiplied by 2 is 68.
   - 78 minus 68 is 10.
   - So, 78 mod 34 is 10.

4.57 mod 34:
   - 34 multiplied by 1 is 34.
   - 57 minus 34 is 23.
   - So, 57 mod 34 is 23.

5.197 mod 34:
   - 34 multiplied by 5 is 170.
   - 197 minus 170 is 27.
   - So, 197 mod 34 is 27.

6.37 mod 34:
   - 34 multiplied by 1 is 34.
   - 37 minus 34 is 3.
   - So, 37 mod 34 is 3.

Now, the product modulo 34 is equivalent to multiplying these results together and then taking modulo 34 again:
(5 x7 x10 x23 x27 x3) mod 34.

>> Step 2: Multiply the Results Step by Step Modulo 34

Let's multiply these numbers step by step, taking modulo 34 at each step to keep the numbers manageable.

1. Multiply 5 and 7:
   5 x7 = 35.
   35 mod 34:
   - 35 - 34 = 1.
   - So, 35 mod 34 is 1.

2. Multiply by 10:
   1 x10 = 10.
   10 mod 34 is already 10.

3. Multiply by 23:
   10 x23 = 230.
   230 mod 34:
   - 34 x6 = 204.
   - 230 - 204 = 26.
   - So, 230 mod 34 is 26.

4. Multiply by 27:
   26 x27.
   First, calculate 26 x27:
   - 26 x20 = 520.
   - 26 x7 = 182.
   - 520 + 182 = 702.
   Now, 702 mod 34:
   - 34 x20 = 680.
   - 702 - 680 = 22.
   - So, 702 mod 34 is 22.

5. Multiply by 3:
   22 x3 = 66.
   66 mod 34:
   - 34 x1 = 34.
   - 66 - 34 = 32.
   - So, 66 mod 34 is 32.

>> Step 3: Final Remainder

After performing all these multiplications and reducing modulo 34 at each step, we find that the remainder is 32.

Prime numbers less than 10 = 2, 3, 5, 7.

If the difference between the largest and the smallest number is ending in 5, the prime numbers in the end position have to be 7 and 2.

The smallest and largest numbers are of form 2_7 and 7_2

Since it is given that the sum of the digits is >13, x will be 5.

Verifying, 752-257 = 495. Answer is option (b).

as 7*5*2 = 70

55³ + 17³ - 72³ = (55-72)k + 17³. This is divisible by 17

Remainder when 55³ is divided by 3 = 1

Remainder when 17³ is divided by 3 = -1

Remainder when 723 is divided by 3 = 0

So, 55³ + 17³ - 72³ is divisible by 3

So, the answer is d) 3 and 17

Number = ababab

=ab×10000+ab×100+ab

=ab(10000+100+1)

=ab(10101)

Correct Answer :- a

Explanation : The unit digit of the number will depend on the last digit.

As we know that 9¹ = 9

9²  = 81

9³ = 729

9⁴  = 6561

The unit digit of the number is 1 and 9, from the options we can pick the answer

Hence option a) is correct

17 is raised to the power of 19 and 19 is raised to the power of 13.
To find the last digit of the number of this kind we will start with the base, and the base here is 17.
To get the unit digit of a number our only concern is the digit at the unit place i.e.7.
The cyclicity of 7 is 4.
Dividing 19¹³  by 4.
Remainder will be 3.
7 raised to power 3 (7³), the unit digit of this number will be 3.

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.