## Number System Questions and Answers Part-10

1. Find the remainder when 73 × 75 × 78 × 57 × 197 × 37 is divided by 34.
a) 32
b) 30
c) 15
d) 28

Explanation: 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{{28 \times - 11}}{{34}} \cr & = \frac{{ - 6 \times - 11}}{{34}} \cr & = \frac{{66}}{{34}} \cr & {\text{R}} = 32 \cr}
Required remainder = 32

2. Find the remainder when 6799 is divided by 7.
a) 4
b) 6
c) 1
d) 2

Explanation:
\eqalign{ & {\text{Remainder of}}\frac{{{{67}^{99}}}}{7} \cr & 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 & {\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. $$\frac{{{{\text{A}}^{\text{x}}}}}{{\text{B}}}$$

3. Let N = 1421 × 1423 × 1425. What is the remainder when N is divided by 12?
a) 0
b) 9
c) 3
d) 6

Explanation: 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.]
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.

4. Three numbers are in ratio 1 : 2 : 3 and HCF is 12. The numbers are:
a) 12, 24, 36
b) 11, 22, 33
c) 12, 24, 32
d) 5, 10, 15

Explanation: 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.
Here HCF = 12,
The numbers are 12, 24 and 36.

5. What is the least number of soldiers that can be drawn up in troops of 12, 15, 18 and 20 soldiers and also in form of a solid square?
a) 900
b) 400
c) 1600
d) 2500

Explanation: 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}
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,
The required number of soldiers
= $$2 \times 2 \times 3 \times 3 \times 5 \times 5$$
= 900

6. Find the remainder when 65203 is divided by 7.
a) 4
b) 2
c) 1
d) 6

Explanation:
\eqalign{ & \frac{{{{65}^{203}}}}{7} \cr & \frac{{{{\left( {63 + 2} \right)}^{203}}}}{7} \cr & 63\,{\text{is}}\,{\text{divisible}}\,{\text{by}}\,7, \cr & {\text{so}}\,{\text{remainder}}\,{\text{will}}\,{\text{depend}}\,{\text{on}}\,{\text{the}}\,{\text{powers}}\,{\text{of}}\,2 \cr & \frac{{{2^{203}}}}{7} \cr & {\text{Its}}\,{\text{remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}} \cr & \frac{{{2^3}}}{7} \cr & {\text{Required}}\,{\text{Remainder}}\, \cr & \frac{8}{7} = 1\cr & {\text{Required}}\,{\text{remainder}} = 1 \cr}
Note: We have manipulated the powers in the form of (4x + n). It means 203 is taken as,
203 = 4x + n = 4 × 50 + 3.
We neglect power which is in the multiple of 4.

7. Find the remainder when 67107 is divided by 7.
a) 4
b) 2
c) 1
d) 6

Explanation:
\eqalign{ & \frac{{{{67}^{107}}}}{7} \cr & \frac{{{{\left( {7 \times 9 + 4} \right)}^{107}}}}{7} \cr & {\text{The}}\,{\text{remainder}}\,{\text{will}}\,{\text{be}}\,{\text{same}}\,{\text{as}} \cr & \frac{{{4^{107}}}}{7} \cr & \frac{{{4^3}}}{7} \cr & \frac{{64}}{7} \cr & {\text{Required}}\,{\text{Remainder}} = 1 \cr}

8. Find the remainder when 54124 is divided by 17.
a) 8
b) 13
c) 16
d) 9

Explanation:
\eqalign{ & \frac{{{{54}^{124}}}}{{17}} \cr & \frac{{{{\left( {17 \times 3 + 3} \right)}^{124}}}}{{17}} \cr & {\text{The}}\,{\text{remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}}, \cr & \frac{{{3^{124}}}}{{17}} \cr & \frac{{{3^{4 \times 31}}}}{{17}} \cr & {\text{Remainder}}\,{\text{will}}\,{\text{be}}\,{\text{same}}\,{\text{as}}, \cr & \frac{{{3^4}}}{{17}} \cr & \frac{{81}}{{17}} \cr & \text{Remainder} = 13 \cr}

9. Find unit digit of product (173)45 × (152)77 × (777)999.
a) 4
b) 2
c) 8
d) 6

\eqalign{ & \frac{{ {{{\left( {173} \right)}^{45}} \times {{\left( {152} \right)}^{77}} \times {{\left( {777} \right)}^{999}}} }}{{10}} \cr & {\text{Remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}}, \cr & \frac{{ {{3^{45}} \times {2^{77}} \times {7^{999}}} }}{{10}} \cr & \frac{{ {3 \times 2 \times {7^3}} }}{{10}} \cr & \frac{{ {6 \times 343} }}{{10}} \cr & {\text{Remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}}\,\frac{{ {6 \times 3} }}{{10}} \cr }