Number System - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option A

Remainder,
$$\frac{{73 \times 75 \times 78 \times 57 \times 197 \times 37}}{{34}}$$
$$ = \frac{{5 \times 7 \times 10 \times 23 \times 27 \times 3}}{{34}}$$
[We have taken individual remainder, which means if 73 is divided by 34 individually, it will give remainder 5, 75 divided 34 gives remainder 7 and so on.]

$$\eqalign{ & \frac{{5 \times 7 \times 10 \times 23 \times 27 \times 3}}{{34}} \cr & = \frac{{35 \times 30 \times 23 \times 27}}{{34}} \cr & = \frac{{1 \times - 4 \times - 11 \times - 7}}{{34}} \cr} $$

[We have taken here negative as well as positive remainder at the same time. When 30 divided by 34 it will give either positive remainder 30 or negative remainder -4. We can use any one of negative or positive remainder at any time.]

$$\eqalign{ & \frac{{1 \times - 4 \times - 11 \times - 7}}{{34}} \cr & = \frac{{28 \times - 11}}{{34}} \cr & = \frac{{ - 6 \times - 11}}{{34}} \cr & = \frac{{66}}{{34}} \cr & {\text{R}} = 32 \cr} $$
Required remainder = 32

Answer: Option C

$$\eqalign{ & {\text{Remainder of}}\frac{{{{67}^{99}}}}{7} \cr & {\text{or, }}R = \frac{{{{\left( {63 + 4} \right)}^{99}}}}{7} \cr} $$
63 is divisible by 7 for any power, so required remainder will depend on the power of 4
Required remainder
$$\eqalign{ & \frac{{{4^{99}}}}{7} = = R = = \frac{{{4^{\left( {96 + 3} \right)}}}}{7} \cr & \frac{{{4^3}}}{7} \Rightarrow \frac{{64}}{7} \Rightarrow \frac{{\left( {63 + 1} \right)}}{7} = = R \Rightarrow 1 \cr & \cr & {\text{Note}}: \cr & \frac{4}{7}{\text{remainder}} = 4 \cr & \frac{{\left( {4 \times 4} \right)}}{7} = \frac{{16}}{7}{\text{remainder}} = 2 \cr & \frac{{\left( {4 \times 4 \times 4} \right)}}{7} = \frac{{64}}{7} = 1 \cr & \frac{{\left( {4 \times 4 \times 4 \times 4} \right)}}{7} = \frac{{256}}{7}{\text{remainder}} = 4 \cr & \frac{{\left( {4 \times 4 \times 4 \times 4 \times 4} \right)}}{7} = 2 \cr} $$
If we check for more power we will find that the remainder start repeating themselves as 4, 2, 1, 4, 2, 1 and so on. So when we get A number having greater power and to be divided by the other number B, we will break power in (4n + x) and the final remainder will depend on x i.e. $$\frac{{{{\text{A}}^{\text{x}}}}}{{\text{B}}}$$

Answer: Option C

Remainder,
$$\eqalign{ & \frac{{1421 \times 1423 \times 1425}}{{12}} = R \cr & R \Rightarrow \frac{{5 \times 7 \times 9}}{{12}} \cr} $$

[Here, we have taken individual remainder such as 1421 divided by 12 gives remainder 5, 1423 and 1425 gives the remainder as 7 and 9 on dividing by 12.]

Now, the sum is reduced to,
$$\frac{{5 \times 7 \times 9}}{{12}} = \frac{{35 \times 9}}{{12}}$$
$$\frac{{35 \times 9}}{{12}}$$  = Remainder ⇒ -1 × -3 = 3 [Here, we have taken negative remainder.] So, required remainder will be 3.

Note: When, $$\frac{9}{{12}}$$ it gives positive remainder as 9 and it also give a negative remainder -3. As per our convenience,we can take any time positive or negative remainder.

Answer: Option A

Since, the numbers are given in the form of ratio that means their common factors have been cancelled.
Each one's common factor is HCF.
And here HCF = 12,
hence, the numbers are 12, 24 and 36.

Alternatively, Let the numbers be x, 2x and 3x.
The HCF in x, 2x and 3x is x because 1, 2, 3 are prime.
Hence,
x = 12; then the other numbers are 24 and 36.

Answer: Option A

In this type of question, We need to find out the LCM of the given numbers.
LCM of 12, 15, 18 and 20;
$$\eqalign{ & 12 = 2 \times 2 \times 3; \cr & 15 = 3 \times 5; \cr & 18 = 2 \times 3 \times 3; \cr & 20 = 2 \times 2 \times 5; \cr} $$
Hence, LCM = $$2 \times 2 \times 3 \times 5 \times 3$$
Since, the soldiers are in the form of a solid square.
Hence, LCM must be a perfect square. To make the LCM a perfect square, We have to multiply it by 5,
hence,
The required number of soldiers
= $$2 \times 2 \times 3 \times 3 \times 5 \times 5$$
= 900

Answer: Option B

We have to find the Least number, therefore we find out the LCM of 8, 12, 16 and 20.
8 = 2 × 2 × 2;
12 = 2 × 2 × 3;
16 = 2 × 2 × 2 × 2;
20 = 2 × 2 × 5;
LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240;
This is the least number which is exactly divisible by 8, 12, 16 and 20
Thus,
Required number which leaves remainder 5 is,
240 + 5 = 245

Answer: Option D

$$\eqalign{ & {7^{6n}} - {6^{6n}} \cr & = {7^6} - {6^6} \cr & = {\left( {{7^3}} \right)^2} - {\left( {{6^3}} \right)^2} \cr & = \left( {{7^3} - {6^3}} \right)\left( {{7^3} + {6^3}} \right) \cr & = \left( {343 - 216} \right) \times \left( {343 + 216} \right) \cr & = 127 \times 559 \cr & = 127 \times 13 \times 43 \cr} $$
Clearly, it is divisible by 127, 13 as well as 559

Answer: Option D

As the Number gives a remainder of 4 when it is divided by 7, then the number must be in form of (7x + 4)

The same gives remainder 1 when it is divided 4, so the number must be in the form of {4 × (7x + 4) + 1}

Also, the number when divided by 3 gives remainder 2, thus number must be in form of [3 × {4 × (7x + 4) + 1} + 2]

Now, On simplifying,
[3 × {4 × (7x + 4) + 1} + 2]
= 84x + 53
We get the final number 53 more than a multiple of 84 Hence, if the number is divided by 84,
The remainder will be 53

Answer: Option A

$$\eqalign{ & {\text{Given}}, \cr & \frac{{{2^{256}}}}{{17}} \cr & {\text{We}}\,{\text{can}}\,{\text{write}}\,{\text{it}}\,{\text{as}}:\,{\left( {{2^4}} \right)^{64}} \cr & \frac{{{{16}^{64}}}}{{17}} \cr & {\text{Individually, when 16 is divided by 17,}} \cr & {\text{gives a negative reminder of - 1}}{\text{.}} \cr & {\text{Required Remainder}}, \cr & {\left( { - 1} \right)^{64}} = 1 \cr & {\text{Alternatively}}, \cr & \frac{{{{16}^{64}}}}{{17}},\,{\text{can}}\,{\text{be}}\,{\text{written}}\,{\text{as}} \cr & \frac{{\left( {16 \times 16 \times 16 \times 16 \times 16\,......\,64\,{\text{times}}} \right)}}{{17}} \cr & {\text{Now,}}\,{\text{we}}\,{\text{take}}\,{\text{the}}\,{\text{negative}}\,{\text{remainder}}\,{\text{of}}\,{\text{each,}} \cr & {\text{16}}\,{\text{divided}}\,{\text{by}}\,{\text{17}}\,{\text{gives}}\,{\text{negative}}\,{\text{remainder}}\,{\text{ - 1}}{\text{.}} \cr & {\text{So,}}\,{\text{the}}\,{\text{Remainder}}\,{\text{will}}\,{\text{be}} \cr & \left( { - 1 \times - 1 \times - 1 \times - 1 \times - 1\,.......\,64\,{\text{times}}} \right) \cr & = 1 \cr} $$

Answer: Option D

$$\frac{{{4^{96}}}}{6},$$ we can write it in this form
$$\frac{{{{\left( {6 - 2} \right)}^{96}}}}{6}$$
Now, Remainder will depend only the powers of -2. So,
$$\frac{{{{\left( { - 2} \right)}^{96}}}}{6},$$   it is same as
$$\frac{{{{\left( {{{\left[ { - 2} \right]}^4}} \right)}^{24}}}}{6},$$   it is same as
$$\frac{{{{\left( {16} \right)}^{24}}}}{6}$$
Now,
$$\frac{{\left( {16 \times 16 \times 16 \times 16{\kern 1pt} ......{\kern 1pt} 24{\kern 1pt} {\text{times}}} \right)}}{6}$$
On dividing individually 16 we always get a remainder 4.
$$\frac{{\left( {4 \times 4 \times 4 \times 4{\kern 1pt} ......{\kern 1pt} 24{\kern 1pt} {\text{times}}} \right)}}{6}$$
Hence, Required Remainder = 4
NOTE: When 4 has even number of powers, it will always give remainder 4 on dividing by 6.