A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?

We can check divisibility of 19⁵ + 21⁵ by 10 by adding the unit digit of 9⁵ + 1⁵ which is equal to 9 + 1 = 10.
So it must be divisible by 10.

Now, for divisibility by 20 we add 19 and 21 which is equal to 40. So, it is clear that it is also divisible by 20.

So 19⁵ + 21⁵ is divisible by both 10 and 20

Let the length of the train be x meter and its speed be y m/sec

Then, (x + 480 m) / 24 sec = y ....(i)
and , (x + 720 m) / 36 sec = y ....(ii)

From Eqs (i) and (ii)
(x + 480) / 24 = (x + 900) / 36
? 3x + 1440 = 2x + 1800
? Length of train, x= 360m

Let us assume the ratio term is r.
then the number will be 3r and 4r.
According to question,
(4r)² + 224 = 8 x (3r)²
16r² + 224 = 8 x 9r²
16r² + 224 = 72r²
72r² - 16r² = 224
56r² = 224
r² = 224/56
r² = 4
r = 2

First number = 3r = 3 x 2 = 6
Second number = 4r = 4 x 2 = 8

Explanation:

The digit in the unit's place should be greater than that in the tens' place.
Hence, if digit 5 occupies the unit place, then remaining four digits need not to follow any order,hence required number = 4!
However, if digit 4 occupies the unit place then 5 cannot occupy the ten;s position. Hence, digit at the ten's place and it will be filled by the digit 1, 2 or 3. This can happen in 3 ways. The remaining 3 digit can be filled in the remaining three place in 3! ways.
Hence, in all, we have (3 x 3!) numbers ending in 4. Similarly, if we have 3 in the unit's place and it will be either 1 or 2. this can happen in 2 ways. Hence, we will have (2 x 3! ) number ending in 3 . Similarly, we can find that there will be 3! numbers ending in 2 and no number with 1. Hence, total number of numbers
= 4! + (3) x 3! + (2 x 3!) + 3!
= 4! + 6 x 3! = 24 + (6 x 6) = 60

To construct 2 roads, three towns can be selected out of 4 in 4 x 3 x 2 = 24 ways. Now, if third road goes from the third town to the first town, a triangle is formed and if it goes to the fourth town, a triangle is not formed, So there are 24 ways to form a triangle and 24 ways of avoiding the formation of triangle.

The available digits are 0, 1, 2,...., 9. The first digit can be chosen in 9 ways (0 not acceptable), the second digit can be accepted in 9 ways (digit repetition not allowed). Thus, the code can be made in 9 x 9 = 81 ways.
Now, there are only 4 digits which can create confusion 1, 6, 8, 9. The same can be given in the following ways
Total number of ways confusion can arise
= 4 x 3 = 12
Thus, the ways in which no such confusion arise = 81-12 =69