option c

The first number in this AP can be found as given above

The first term = ((((5*6) +4)*5) + 2

Common difference = 5*6*7=210

General form of the AP  is therefore 172+210k,

So the numbers will be 172, 382,592 and 802

option (c)

Let the prime factors be abc

Ways of writing

$a^{31}$ least value $=2^{31}$

$a{b}^{15}$ least value $=2^{15}3$

$a^3{b}^7$ least value $=2^7{3}^3$

$a{b}^3{c}^3$ least value $=2^3{3}^{3}5=1080$

$a{b}^{7}c$ least value $=2^7{3}^1{5}^1=1920$

option (d)

Using the last two digits technique

$(42{)}^{442}=(21\times 2{)}^{442}$

$=(21{)}^{442}\times (2{)}^{442}=(21{)}^{442}\times (2^{44}{)}^{10}\times 2^2=\_\_41\times \_\_76\times \_\_04=\_\_41\times \_\_04=\_\_64.$ Option (d)

option d

Any number of the form aaa (3 digits) or aaaaaa (multiple of 3 digits, i.e. 6, 9 .....) is divisible by 111

Therefore group the above numbers into groups of 3 or multiples of 3.

A 32 digit number can be broken up as (3*10 digits) + (2 digits). The first part is divisible by 111. the last two digits will thus be the remainder = 99

Answer=b

Approach 1

Go by frequency method

$\frac{17}{36}$ gives a remainder $=17$

$\frac{{17}^2}{36}$ gives a remainder $=1$

Frequency $=2$

Odd powers remainder $=17$, even powers $=1$

Approach 2

Euler’s number of $36$ is $36\times \frac{1}{2}\times \frac{2}{3}=12$

$36=12k.$ hence, from Fermat theorem $\frac{{17}^{36}}{36}{|}_R=1$

Note :
Euler's number is the number of co_primes of number which is less than that number.
A number N can be written as $a^m{b}^n$ (where a and b are the prime factors of N)
eg) $20=2^2\times 5^1$
Here $a=2$ and $b=5,m=2$ and $n=1$
Euler's number $=N[1-\left(\frac{1}{a}\right)][1-\left(\frac{1}{b}\right)]$
From fermat Theorem
$\frac{N^{\left(Eule{r}^{\prime }snumberof{y}^{+}k\right)}}{y}{\mid }_R=1$ (i.e When a number N is raised to the Euler's number of a number)
$"y"$ is divided by $"y"$, the remainder will be 1

option d

Use Divisibility Rules

34 = 17 *2

So we should use the divisibility test of 17 ( Compartmentalization method - taking 8 digits at a time (Sum of digits at odd places taken 8 at a time- Sum of digits at even places taken 8 at a time)

Combine 8 digits together, 9876......80 digits*100 + 98 :

There will be equal number of groups (of 8 digits taken at a time) at odd places and even

Places in 9876.......80 digits*100.

So Sum of groups at odd places - Sum of groups at even places = 0

Therefore first part is divisible by 17. It is also divisible by 2, as 100 is divisible by 2.Hence, we only need to find the remainder when 98 is divided by 34

Remainder= 30

option (e)

$7^1={077}^5=--07$

$7^2={497}^6=--49$

$7^3={437}^7=--43$

$7^4={017}^8=--01$

There is a cyclicity of 4 as 07, 49 , 43 ,01 is the last 2 digits for the powers of 7

Here $342=4k+2\Rightarrow$Required answer is $07+49=56$

$1+\frac{1}{3}=\frac{4}{3}=\mathrm{1.333.}$ the value of x will be slightly lesser than $1.333$ as the denominator keeps increasing

Look at the options

Option $a=\frac{5}{1.7}>2$ .this can never be the answer

Option $b=\frac{1}{1.7}<1.$ this also can never be the answer

Option $c=\frac{3}{1.4}>2.$ this cannot be the answer

Option $d=\frac{2.5}{1.7}$ which is the correct answer

Option $e=2,$ which can never be the answer

option (c )

A number ending in 5, with the tens digit an odd number and raised to an odd power, will

always end with 75. Answer is option (c)