Number System MCQ Quiz - Objective Question with Answer for Number System - Download Free PDF

Given:

Number = 6745K2

We need to find the value of K such that 6745K2 is divisible by 9

Formula used:

A number is divisible by 9 if the sum of its digits is divisible by 9

Calculation:

Sum of digits = 6 + 7 + 4 + 5 + K + 2 = 24 + K

We need 24 + K divisible by 9

Try K = 3

⇒ 24 + 3 = 27 (divisible by 9)

∴ The correct answer is option (3): 3

Given:

Original quotient (Q) = 24 Remainder = 0 Dividend was taken 10% less Divisor was taken 20% less

Formulas used:

dividend = Q × divisor

Calculation:

Let the original dividend (D) be 100 units. 

The correct dividend (D) and divisor (d) result in a quotient of 24 and remainder 0 (i.e., D = 24d),

Suhas erroneously took 90% of the dividend (0.9D) and 80% of the divisor (0.8d).

The quotient (Q) of these two erroneous values is:

Q = 0.9D / 0.8d.

Substitute D = 24d

Q = 0.9 × 24d / 0.8d Q = 21.6 / 0.8 Q = 27.

So, Suhas would have obtained a quotient of 27.

7¹⁰³ = 7 (7¹⁰²⁾ = 7 (343)³⁴ = 7 (345 – 2)³⁴

7¹⁰³ = 23K₁ + 7.2³⁴

Now 7.2³⁴ = 7 . 2² . 2³²

= 28 . (256)⁴

= 28 (253 + 3)⁴

∴ 28 × 81 ⇒ (23 + 5) (69 + 12)

23K₂ + 60

∴ Remainder = 14

Given:

A 6-digit number is formed using consecutive natural numbers.

Formula used:

Any number with digits as consecutive natural numbers (e.g., 123456, 234567, etc.) follows a pattern and must be tested for divisibility.

Calculation:

Take example: 123456

Check divisibility:

123456 ÷ 3 = 41152 → divisible

123456 ÷ 6 = 20576 → divisible

123456 ÷ 9 = 13717.33 → not divisible

Try another: 234567

234567 ÷ 3 = 78189 → divisible

234567 ÷ 6 = 39094.5 → not divisible

So only consistent divisibility for all such numbers is by 3

∴ The correct answer is option (2) 3.

Calculation

No. of 3 digits = 999 – 99 = 900

No. of 3 digit numbers divisible by 2 & 3 i.e. by 6 

\(\frac{900}{6}=150\)

No. of 3 digit numbers divisible by 4 & 9 i.e. by 36 

\(\frac{900}{36}=25\)

∴ No of 3 digit numbers divisible by 2 & 3 but not by 4 & 9 

150 – 25 = 125 

Given:

\((49^{15} - 1) \)

Concept used:

aⁿ​​​​​​ - bⁿ is divisible by (a + b) when n is an even positive integer.

Here, a & b should be prime number.

Calculation:

\((49^{15} - 1) \)

⇒ \(({(7^2)}^{15} - 1) \)

⇒ \((7^{30} - 1) \)

Here, 30 is a positive integer.

​According to the concept,

\((7^{30} - 1) \) is divisible by (7 + 1) i.e., 8.

∴ 8 is a divisor of \((49^{15} - 1) \).

Concept used:

If the last 2 digits of any number divisible by 4, then the number is divisible by 4

Calculation:

According to the question, the numbers are

2332, 2552, 4664, 2772, 6776, 4884, 2992, and 6996

So, there are 8 such numbers in the form abba, divisible by 4

∴ The correct answer is 8

Mistake Points

If you are considering an example ending with 20,

then, 'abba' will be '0220', and 0220 is not a four-digit number. 

Similarly in the case of the example ending with 40,60,80.

Given:

Five-digit number 750PQ is divisible by 3, 7 and 11

Concept used:

Concept of LCM

Calculation:

The LCM of 3, 7, and 11 is 231.

By taking the largest 5-digit number 75099 and dividing it by 231.

If we divide 75099 by 231 we get 325 as the quotient and 24 as the remainder.

Then, the five-digit number is 75099 - 24 = 75075.

The number = 75075 and P = 7, Q = 5

now,

P + 2Q = 7 + 10 = 17

∴ The value of P + 2Q is 17.

Calculation:

(265)4081 + 9 is divisible by 266

⇒ (266 - 1)⁴⁰⁸¹+ 9

Now when divided by 266, 

⇒ \( (266 - 1)^{4081}\over 266\) + \(9 \over 266\)

Remainder from first fraction will be (- 1)⁴⁰⁸¹and + 9 from second fraction

Remainder as whole = - 1 + 9 = 8

∴ Remainder when (265)4081+ 9 is divided by 266 is 8.

Calculation:

625 + 626 + 627 + 628

Taking 6²⁵ commons from the expression:

⇒ 625(6⁰ + 6¹ + 6² + 6³)

⇒ 6²⁵(1 + 6 + 36 + 216)

⇒ 625 × 259

After simplifying we get that the given expression is the multiple of 259.

Thus, the given expression is divisible by 259.

∴ The correct answer is option (4).

Given:

2727 + 27

Concept used:

Aⁿ + Bⁿ is divisible by (A + B) when n is odd.

Calculation:

Now, (2727 + 27)

⇒ (2727 + 1²⁷ + 27 - 1)

⇒ (2727 + 127) + 26

Here, according to the concept, (2727 + 127) is divisible by (27 + 1) i.e. 28.

Hence, the remainder = 26

∴ The remainder when 2727 + 27 is divided by 28 is 26.

Given:

New dividend = 110% × original dividend

New divisor = 125% × original divisor

Formula used:

Dividend = divisor × quotient + remainder 

Calculation:

Let the original dividend = 100x

Mistakenly increased dividend = 100 × 110% = 110x

Let original divisor = 100y

Mistankenly increased divisor = 100y × 125% = 125y

Dividend = divisor × quotient + remainder 

100x = 100y × 25 + 0

x/y = 25/1

Increased dividend = 110x = 110 × 25 = 2750

Increased divisor = 125y = 125 × 1 = 125

Quotient = 2750/125 = 22

∴ The correct answer is 22. 

Given:

A six-digit number Is divisible by 33

Formula used:

Dividend = divisor × quotient + remainder

Calculation:

Dividend = divisor × quotient + remainder

⇒ 33 × q + 0 = 33q

If 54 is added to the dividend then,

New number = 33q + 54

⇒ 3 × (11q + 18)

So, we can clearly say that the new number is divisible by 3.

∴ The correct option is 1.
Mistake Points 

Please note that this is the official paper of SSC and SSC has given the 3 as the correct answer, but 111111 is also the 6 digit number and if we add 54 it will be divisible by both 3 and 5.

Given:

The 8-digit number 123456xy is divisible by 8

Concept used:

If the last three digits of a number are divisible by 8, then the number is completely divisible by 8.

Calculation:

So, 6xy should be divisible by 8

Now,

Possible numbers are 600, 608, 616, 624, 632, 640, 648, 656, 664, 672, 680, 688, 696

So, total of 13 possible pairs can be made

∴ The required answer is 13.

Concept used:

If (n + 1)a is divided by n

Then,

Remainder is 1

Calculation:

742 = (7²)21

⇒ (49)21

⇒ (48 + 1)21

So, the remainder is 1

∴ The required answer is 1.