Numbers

1.

12112 को 11 से और 13223 को 13 से भाग देने पर प्राप्त शेषफलों का योग है -

1.
3
2.
4
3.
5
4.
2

Answer:Option 1

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2.Find the remainder when 73 × 75 × 78 × 57 × 197 × 37 is divided by 34.

1. 32
2. 30
3. 15
4. 28

Answer:Option 1

Explanation:

Solution:

Remainder,

\frac{73 \times 75 \times 78 \times 57 \times 197 \times 37}{34}

$\frac{{73 \times 75 \times 78 \times 57 \times 197 \times 37}}{{34}}$ ⇒

\frac{5 \times 7 \times 10 \times 23 \times 27 \times 3}{34}

$\frac{{5 \times 7 \times 10 \times 23 \times 27 \times 3}}{{34}}$
[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.]

\frac{5 \times 7 \times 10 \times 23 \times 27 \times 3}{34}

$\frac{{5 \times 7 \times 10 \times 23 \times 27 \times 3}}{{34}}$ ⇒

\frac{35 \times 30 \times 23 \times 27}{34}

$\frac{{35 \times 30 \times 23 \times 27}}{{34}}$ [Number Multiplied]

\frac{35 \times 30 \times 23 \times 27}{34}

$\frac{{35 \times 30 \times 23 \times 27}}{{34}}$ ⇒

\frac{1 \times - 4 \times - 11 \times - 7}{34}

$\frac{{1 \times - 4 \times - 11 \times - 7}}{{34}}$

[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.]

\frac{1 \times - 4 \times - 11 \times - 7}{34}

$\frac{{1 \times - 4 \times - 11 \times - 7}}{{34}}$ ⇒

\frac{28 \times - 11}{34}

$\frac{{28 \times - 11}}{{34}}$ ⇒

\frac{- 6 \times - 11}{34}

$\frac{{ - 6 \times - 11}}{{34}}$ ⇒

\frac{66}{34}

$\frac{{66}}{{34}}$ ⇒

R

${\text{R}}$ ⇒ 32
Required remainder = 32.

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3.Find the remainder when 67⁹⁹ is divided by 7.

1. 4
2. 6
3. 1
4. 2

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Remainder of \frac{67^{99}}{7} == R \Rightarrow \frac{{(63 + 4)}^{99}}{7} \\ [63 is divisible by 7 for any power, so required remainder will depend on the poer of 4] \\ Required remainder : \\ \frac{4^{99}}{7} == R == \frac{4^{(96 + 3)}}{7} \\ \frac{4^{3}}{7} \Rightarrow \frac{64}{7} \Rightarrow \frac{(63 + 1)}{7} == R \Rightarrow 1 \\ Note : \\ \frac{4}{7} remainder = 4 \\ \frac{(4 \times 4)}{7} = \frac{16}{7} remainder = 2 \\ \frac{(4 \times 4 \times 4)}{7} = \frac{64}{7} = 1 \\ \frac{(4 \times 4 \times 4 \times 4)}{7} = \frac{256}{7} remainder = 4 \\ \frac{(4 \times 4 \times 4 \times 4 \times 4)}{7} = 2 \end{aligned}

$\eqalign{ & {\text{Remainder}}\,{\text{of}}\,\frac{{{{67}^{99}}}}{7} = = R \Rightarrow \frac{{{{\left( {63 + 4} \right)}^{99}}}}{7} \cr & [{\text{63}}\,{\text{is}}\,{\text{divisible}}\,{\text{by}}\,{\text{7}}\,{\text{for}}\,{\text{any}}\,{\text{power,}}\,{\text{so}}\,{\text{required}}\,{\text{remainder}}\,{\text{will}}\,{\text{depend}}\,{\text{on}}\,{\text{the}}\,{\text{poer}}\,{\text{of}}\,{\text{4}}] \cr & {\text{Required}}\,{\text{remainder}}: \cr & \frac{{{4^{99}}}}{7} = = R = = \frac{{{4^{\left( {96 + 3} \right)}}}}{7} \cr & \frac{{{4^3}}}{7} \Rightarrow \frac{{64}}{7} \Rightarrow \frac{{\left( {63 + 1} \right)}}{7} = = R \Rightarrow 1 \cr & \cr & {\text{Note}}: \cr & \cr & \frac{4}{7}\,{\text{remainder}} = 4 \cr & \frac{{\left( {4 \times 4} \right)}}{7} = \frac{{16}}{7}\,{\text{remainder}} = 2 \cr & \frac{{\left( {4 \times 4 \times 4} \right)}}{7} = \frac{{64}}{7} = 1 \cr & \frac{{\left( {4 \times 4 \times 4 \times 4} \right)}}{7} = \frac{{256}}{7}\,{\text{remainder}} = 4 \cr & \frac{{\left( {4 \times 4 \times 4 \times 4 \times 4} \right)}}{7} = 2 \cr & \cr}$
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. A^x/B.

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4.Let N = 1421 × 1423 × 1425. What is the remainder when N is divided by 12?

1. 0
2. 9
3. 3
4. 6

Answer:Option 1

Explanation:

Solution:

Remainder,
(1421*1423*1425)/12 ==R==>(5*7*9)/12

[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,
(5*7*9)/12 = (35*9)/12
(35*9)/12 = Remainder ==> -1 * -3 = 3 [Here, we have taken negative reminder.] So, required remainder will be 3.

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

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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?

1. 900
2. 400
3. 1600
4. 2500

Answer:Option 1

Explanation:

Solution:

In this type of question, We need to find out the LCM of the given numbers.
LCM of 12, 15, 18 and 20;

\begin{aligned} 12 = 2 \times 2 \times 3; \\ 15 = 3 \times 5; \\ 18 = 2 \times 3 \times 3; \\ 20 = 2 \times 2 \times 5; \end{aligned}

$\eqalign{ & 12 = 2 \times 2 \times 3; \cr & 15 = 3 \times 5; \cr & 18 = 2 \times 3 \times 3; \cr & 20 = 2 \times 2 \times 5; \cr}$
Hence, LCM =

2 \times 2 \times 3 \times 5 \times 3

$2 \times 2 \times 3 \times 5 \times 3$
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

$2 \times 2 \times 3 \times 3 \times 5 \times 5$ = 900.

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6.Find the least number which will leaves remainder 5 when divided by 8, 12, 16 and 20.

1. 240
2. 245
3. 265
4. 235

Answer:Option 1

Explanation:

Solution:

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.

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7.7⁶ⁿ- 6⁶ⁿ, where n is an integer >0, is divisible by

1. 13
2. 127
3. 559
4. All of these

Answer:Option 1

Explanation:

Solution:

\begin{aligned} 7^{6 n} - 6^{6 n} = 7^{6} - 6^{6} \\ = {(7^{3})}^{2} - {(6^{3})}^{2} \\ = (7^{3} - 6^{3}) (7^{3} + 6^{3}) \\ = (343 - 216) \times (343 + 216) \\ = 127 \times 559 \\ = 127 \times 13 \times 43 \end{aligned}

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

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8.After the division of a number successively by 3, 4 and 7, the remainder obtained is 2, 1 and 4 respectively. What will be remainder if 84 divide the same number?

1. 80
2. 75
3. 42
4. 53

Answer:Option 1

Explanation:

Solution:

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.

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9.Find the remainder when 4⁹⁶ is divided by 6.

1. 0
2. 2
3. 3
4. 4

Answer:Option 1

Explanation:

Solution:

\begin{aligned} \frac{4^{96}}{6}, we can write it in this form \\ \frac{{(6 - 2)}^{96}}{6} \\ Now, Remainder will depend only the powers of - 2 . So, \\ \frac{{(- 2)}^{96}}{6}, it is same as \\ \frac{{({[- 2]}^{4})}^{24}}{6}, it is same as \\ \frac{{(16)}^{24}}{6} \\ Now, \\ \frac{(16 \times 16 \times 16 \times 16 . . . . . . 24 times)}{6} \\ On dividing individually 16 we always get a remainder 4 . \\ So, \\ \frac{(4 \times 4 \times 4 \times 4 . . . . . . 24 times)}{6} \\ Hence, Required Remainder = 4 \end{aligned}

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

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10.What is the sum of all two digit numbers that gives a remainder of 3 when they are divided by 7?

1. 666
2. 676
3. 683
4. 777

Answer:Option 1

Explanation:

Solution:

The two digit number which gives a remainder of 3 when divided by 7 are:
10, 17, 24 ..... 94.
Now, these number are in AP series with
1st Term, a = 10;
Number of Terms, n = 13;
Last term, L = 94 and
Common Difference, d = 7.
Sum,