Number System - Aptitude MCQ Questions and Solutions with Explanations

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

$$\eqalign{ & {\text{Remainder of}}\frac{{{{67}^{99}}}}{7} \cr & {\text{or, }}R = \frac{{{{\left( {63 + 4} \right)}^{99}}}}{7} \cr} $$
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. $$\frac{{{{\text{A}}^{\text{x}}}}}{{\text{B}}}$$

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.

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.

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

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

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

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

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

$$\frac{{{4^{96}}}}{6},$$ 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.