Numbers
1.

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

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

Explanation:

Solution:
Remainder,
$\frac{73×75×78×57×197×37}{34}$
⇒
$\frac{5×7×10×23×27×3}{34}$

[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.]

$\frac{5×7×10×23×27×3}{34}$
⇒
$\frac{35×30×23×27}{34}$
[Number Multiplied]
$\frac{35×30×23×27}{34}$
⇒
$\frac{1×-4×-11×-7}{34}$

[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.]

$\frac{1×-4×-11×-7}{34}$
⇒
$\frac{28×-11}{34}$
⇒
$\frac{-6×-11}{34}$
⇒
$\frac{66}{34}$
⇒
$\text{R}$
⇒ 32
Required remainder = 32.

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

Explanation:

Solution:
$\begin{array}{rl}& \text{Remainder}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\frac{{67}^{99}}{7}==R⇒\frac{{\left(63+4\right)}^{99}}{7}\\ & \left[\text{63}\phantom{\rule{thinmathspace}{0ex}}\text{is}\phantom{\rule{thinmathspace}{0ex}}\text{divisible}\phantom{\rule{thinmathspace}{0ex}}\text{by}\phantom{\rule{thinmathspace}{0ex}}\text{7}\phantom{\rule{thinmathspace}{0ex}}\text{for}\phantom{\rule{thinmathspace}{0ex}}\text{any}\phantom{\rule{thinmathspace}{0ex}}\text{power,}\phantom{\rule{thinmathspace}{0ex}}\text{so}\phantom{\rule{thinmathspace}{0ex}}\text{required}\phantom{\rule{thinmathspace}{0ex}}\text{remainder}\phantom{\rule{thinmathspace}{0ex}}\text{will}\phantom{\rule{thinmathspace}{0ex}}\text{depend}\phantom{\rule{thinmathspace}{0ex}}\text{on}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{poer}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{4}\right]\\ & \text{Required}\phantom{\rule{thinmathspace}{0ex}}\text{remainder}:\\ & \frac{{4}^{99}}{7}==R==\frac{{4}^{\left(96+3\right)}}{7}\\ & \frac{{4}^{3}}{7}⇒\frac{64}{7}⇒\frac{\left(63+1\right)}{7}==R⇒1\\ & \\ & \text{Note}:\\ & \\ & \frac{4}{7}\phantom{\rule{thinmathspace}{0ex}}\text{remainder}=4\\ & \frac{\left(4×4\right)}{7}=\frac{16}{7}\phantom{\rule{thinmathspace}{0ex}}\text{remainder}=2\\ & \frac{\left(4×4×4\right)}{7}=\frac{64}{7}=1\\ & \frac{\left(4×4×4×4\right)}{7}=\frac{256}{7}\phantom{\rule{thinmathspace}{0ex}}\text{remainder}=4\\ & \frac{\left(4×4×4×4×4\right)}{7}=2\\ & \end{array}$

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. Ax/B.

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

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.

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?

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;
$\begin{array}{rl}& 12=2×2×3;\\ & 15=3×5;\\ & 18=2×3×3;\\ & 20=2×2×5;\end{array}$

Hence, LCM =
$2×2×3×5×3$

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 =
$2×2×3×3×5×5$
= 900.

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

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.

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

Explanation:

Solution:
$\begin{array}{rl}& {7}^{6n}-{6}^{6n}={7}^{6}-{6}^{6}\\ & ={\left({7}^{3}\right)}^{2}-{\left({6}^{3}\right)}^{2}\\ & =\left({7}^{3}-{6}^{3}\right)\left({7}^{3}+{6}^{3}\right)\\ & =\left(343-216\right)×\left(343+216\right)\\ & =127×559\\ & =127×13×43\end{array}$

Clearly, it is divisible by 127, 13 as well as 559.

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?

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.

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

Explanation:

Solution:

NOTE: When 4 has even number of powers, it will always give remainder 4 on dividing by 6.

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

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,
$\begin{array}{rl}& =\left\{\text{n}×\frac{\text{a}+\text{L}}{2}\right\}\\ & =13×52\\ & =676\end{array}$

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