Number System - 2 - Free MCQ Practice Test with solutions, SSC CGL Quant

Factors of composite number = (P₁ +1)(P₂+1)(P₃+1)......(P_n+1).
Where P₁,P₂,P₃... P_n are the powers of respective prime factors.
So,
120= 2³ x 3¹ x 5¹
Factors=(3+1)(1+1)(1+1)=16.

For n =1
7⁶ⁿ - 6⁶ⁿ = 7⁶ - 6⁶
⇒ ( 7³ )² - ( 6³ )²
⇒ ( 7³ - 6³ )( 7³ + 6³ )
= ( 343 - 216 )( 343 + 216 )
= 127 x 559
=127×13×43
∴ it is clearly divisible by 127.

Sum of the square of first n natural numbers = n(n+1)(2n+1) / 6
Given n = 35
∴ Required sum = ( 35 x 36 x 71 ) / 6
= 14910

Sum of first n even numbers = n( n +1 )
Given n = 84
∴ Required sum = 84 ( 84 + 1 )
= 84 x 85
= 7140

First 15 multiple of 8 are 8, 16, 24, ......, 120
Sum = 8 (1 + 2 + 3 + 4 ......+ 15)
= 8 [ n(n + 1) / 2]
= 8 x 15 x 8
960

Unit digit in 3⁴ = 1
∴ Unit digit in ( 3⁴ )¹⁶ = 1
∴ Unit digit in 3⁶⁵ = 3
Unit digit in 6⁵⁹ = 6
Unit digit in 7⁷¹ = 3
∴ Required unit digit = Unit digit in ( 3 x 6 x 3 )
= Unit digit in 54
= 4

1. Let the number be N. We know that when N is divided by 357, the remainder is 39, so
   N = 357·k + 39 for some integer k.

2. We want the remainder when N is divided by 17. Since 357 = 17·21, we have
   357·k is always a multiple of 17, so 357·k ≡ 0 (mod 17).

3. Therefore
   N mod 17 ≡ (357·k + 39) mod 17
   ≡ 0 + (39 mod 17)
   ≡ 39 mod 17.

4. Now compute 39 ÷ 17:
   17·2 = 34, and 39 – 34 = 5.

5. So the remainder when N is divided by 17 is 5.

Answer: 5 (option a).

Let the divisor be x and quotient be y.
Then, the number = xy + 24
Twice the number = 2xy + 48
Now 2xy is completely divisible by x.
On dividing 48 by x remainder is x
∴ x = 48 - 11
= 37

Number between 0 and 11 which are multiple of 2 or 3
= 11/2 + 11/3 - 11/6 = 5 + 3 - 1
= 7
Number between 0 and -11 which are multiple of 2 or 3
= 11/2 + 11/3 - 11/6 = 5 + 3 - 1
= 7
∴ Number be 15, including 0

We know that, ( x^m - a^m ) is divisible by ( x + a ), for even value of m.
∴ 17²⁰⁰ - 1²⁰⁰ is divisible by 17 +1
⇒ 17²⁰⁰ - 1 is divisible by 18.
Hence when 17²⁰⁰ divided by 18 the remainder is 1.

We know that when m is a odd number ( x^m + a^m ) is divisible by ( x + a ).
∴ Each one is divisible by (41 + 43).
So Common factor = (41 + 43).

19¹⁰⁰ = ( 20 -1 )¹⁰⁰
⇒ ( -1 )¹⁰⁰ / 20
⇒ 1 / 20
∴ Required Remainder = 1

For integers x and y, and for any even number n, the expression xⁿ - yⁿ is divisible by both x - y and x + y. This is a standard result in algebra. In this case, 7¹² - 4¹² is divisible by 7 - 4 = 3 and 7 + 4 = 11. Because 3 and 11 are coprime, their product 33 will also exactly divide 7¹² - 4¹². Therefore, the correct answer is 33.

We can check divisibility of 19⁵ + 21⁵ by 10 by adding the unit digit of 9⁵ + 1⁵ which is equal to 9 + 1 = 10.
So it must be divisible by 10.
Now, for divisibility by 20 we add 19 and 21 which is equal to 40. So, it is clear that it is also divisible by 20.
So 19⁵ + 21⁵ is divisible by both 10 and 20

When N is a natural number, then there is only one possible case that N, N + 2, N + 4 are prime numbers.
When N = 3, then N, N + 2, N + 4 = 3, 5, 7 all are primes.
so for only 1 value of n this is posiible