Only, one value P can be assumed, which is at 2 at P = 2
P^{2} + 3 = 7 which is also prime
Again at P = 3, 5, 7, 11 .......
P^{2} + 3 = an even number which can not be prime
The required number = Number of numbers, which are(divisible by 3 + divisible by 7 - divisible by 21)Numberofnumberdivisibleby3, = [(198−3)3]+1=66Numberofnumberdivisibleby7,=[(196−7)7]+1=28Numberofnumberdivisibleby21,=[(189−21)21]+1=9Thus,thedivisiblevalue=(66+28−9)=85Thus, number of numbers which are not divisible by 3 or 7=200−85=115$$\begin{array}{rl}& \text{The required number}\\ & \text{= Number of numbers, which are}\\ & \text{(divisible by 3 + divisible by 7 - divisible by 21)}\\ & \text{Number}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{number}\phantom{\rule{thinmathspace}{0ex}}\text{divisible}\phantom{\rule{thinmathspace}{0ex}}\text{by}\phantom{\rule{thinmathspace}{0ex}}\text{3,}\\ & \text{=}\left[\frac{\left(198-3\right)}{3}\right]+1\\ & =66\\ & \text{Number}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{number}\phantom{\rule{thinmathspace}{0ex}}\text{divisible}\phantom{\rule{thinmathspace}{0ex}}\text{by}\phantom{\rule{thinmathspace}{0ex}}7,\\ & =\left[\frac{\left(196-7\right)}{7}\right]+1\\ & =28\\ & \text{Number}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{number}\phantom{\rule{thinmathspace}{0ex}}\text{divisible}\phantom{\rule{thinmathspace}{0ex}}\text{by}\phantom{\rule{thinmathspace}{0ex}}21,\\ & =\left[\frac{\left(189-21\right)}{21}\right]+1\\ & =9\\ & \text{Thus,}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{divisible}\phantom{\rule{thinmathspace}{0ex}}\text{value}\\ & =\left(66+28-9\right)\\ & =85\\ & \text{Thus, number of numbers which are not divisible by}3\text{or}7\\ & =\text{}200\text{}-\text{}85\\ & =\text{}115\end{array}$$
Since, (A^{n} - B^{n}) is divisible by (A - B), So, (27^{67} - 1^{3}) is divisible by (27 -1) = 26
Hence, Expression is also divisible by 13 as it is divisible by 26
Thus given expression is divisible by 13 so the remainder will be 0
Let S be the sum of the expression, Then
S = 4^{0} + 4^{1} + 4^{2} + 4^{3} + ........ + 4^{40}
S = (1 + 4 + 16 + 64) + 4^{4} (1 = 4 + 16 + 64) + ...... + 4^{36} + 4^{40}
Since, (1 + 4 + 16 + 64) = 85, is divisible by 17. Hence, except 4^{40} remaining expression is divisible by 17.
Thus,
For a no. to be divisible by 11,
Sum(odd digit nos) - Sum(even digit nos) = 0 or divisible by 11
If we look at 9876543210, the difference we get is 5
i.e. [(9 + 7 + 5 + 3 + 1) - (8 + 6 + 4 + 2 + 0) = 5]
The series is up to 1000 digit,
That means, we get
100010$$\frac{1000}{10}$$
= 100 time 5,
then the difference will be 5 × 100 = 500
In order for the difference to be divisible by 11, we need to add 5 and the no will become 505
505 is divisible by 11
From 1- 99 digit 1 is used 20 times. And From 100 - 199, 1 is used 120 times
So, from 1 to 199, 1 is used,
20 + 120 = 140 times
We need 136.So leave 199, 198, 197 and 196
Required pages = 195
Let the digits of x be
x = abcde
According to the question,
x = 1bcde [Given ten thousands place is 1.]
Now we can check the options as given that the sum of the all digit is divisible by 3.
10080 is the only number given in the option which satisfies all the given conditions.