Putting n = 1, 2, 3, 4, 5 . . . .
n = 1 in the given equation
(-1)1 + (-1)4 × 1 = 0
Or, -1 + 1 = 0

n = 2
(-1)2 + (-1)4 × 2 = 0
Or, 1 + 1 ≠ 0

n = 3,
-1 + 1 = 0
We will see that the given condition will be satisfy only for ANY ODD NATURAL NUMBER.

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

Let the tower height will be x ft.
∴ Black part will be = $$\frac{1}{3}x$$
Remaining part = $$x - \frac{1}{3}x = \frac{2}{3}x$$
∴ Red part will be = $$\frac{2}{3}x \times \frac{5}{{11}} = \frac{{10}}{{33}}x$$
White part will be = $$\frac{2}{3}x - \frac{{10}}{{33}}x = \frac{{12x}}{{33}}$$
i.e. $$\frac{{12x}}{{33}}$$ = 60
x = 165 ft.

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

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

LCM = 45 HCF LCM + HCF = 1150 45 × HCF + HCF = 1150 46 × HCF = 1150 HCF = 25 Then, LCM = 45 × 25 = 1125 Now, use the formula, 1st number × second Number = LCM × HCF 125 × second number = 1125 × 25 Second Number = 225

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

First of all,we find the LCM of 4, 6, 10 and 15 LCM of 4, 6, 10, 10 = 60Now, The smallest four digit no. is 1000. We divide it by 60 $$\frac{{1000}}{{60}}$$ It leaves remainder 40 Now, the smallest four digit no which is divisible by 4, 6, 10, 15 is, 1000 + (60 - 40) = 1020 So, required number (as it gives remainder 2 always) would be = 1020 + 2 = 1022

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{two}}\,{\text{numbers}}\,{\text{be}}\,X\,{\text{and}}\,Y \cr & {\text{According}}\,{\text{to}}\,{\text{question}}, \cr & X \times {\frac{3}{5}} = 70\% \,of\,Y \cr & \frac{{3X}}{5} = \frac{{70Y}}{{100}} \cr & \frac{X}{Y} = \frac{{ {70 \times 5} }}{{ {100 \times 3} }} = \frac{7}{6} \cr & X:Y = 7:6 \cr} $$

1st no. = 2 2nd no. = x (let) 3rd no. = x + 2 4th no. = x + x + 2 = 2x + 2 5th no. = x + 2 + 2x + 2 = 3x + 4 6th no., 3x + 4 + 2x + 2 = 26 5x = 20 x = 4 7th no. = 26 + 3x + 4 = 26 + 3 × 4 + 4 = 42

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.

Let number be X According to question, 7X - 3X = 36 4X = 36 X = 9 Required Number is 9

The difference between just getting an A grade and just getting a B grade would be equal to (number of quizzes × 3).
The difference between the required score to manage an A grade and the score achieved to just manage a B grade as given in the problem’s information is equal to 97 – 70 = 27
Thus, number of quizzes × 3 = 27
Hence, number of quizzes is 9

Assume x is the required number. Step 1 Amitabh 2x Sashi 2x + 50 Step 2 Amitabh 4x + 100. Sashi 4x + 150 Step 3 Amitabh 8x + 300. Sashi 8x + 350 Step 4 Amitabh 16x + 700 Sashi 16x + 750 Now, 16x + 750 > 1000 And, x can be 16 So 7 is the sum

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

In every 30 minutes, Virus doubles it self. It means in 49.5 hours, it would be half empty and it will be full in 50 hours.

Required Number can be given by: HCF of (3026 - 11) and (5053 - 13) HCF of 3015 and 5040 = 15 To find HCF, we break the given numbers in their prime Factors 3015 = 3 × 3 × 5 × 67 5040 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7 We take common multiples in these two given numbers to get required HCF And Common multiples are: 3 × 5 So, Required HCF = 15

Let he has solved correctly X no. of sums. Therefore incorrect no. of sums = 2X Now, X + 2X = 48 3X = 48 X = 16 sums he has done correctly.

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

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