Number System Test - 3

n(A) = number of numbers divisible by 2 = 50
n(B) = number of numbers divisible by 3 = 33
n(C) = numbers of numbers divisible by 5 = 20

Using n(A ∪ B ∪ C) - (n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + a(A ∩ B ∩ C)), we get

100 - (50 + 33 + 20 - 16 - 10 - 6 + 3)
= 100 - 74
= 26

The smallest number is 10 which when divided by 7 gives a remainder 3.
And the largest such number is 94.
So, 94 = 10 + (n - 1)7
∴ n = 13
S₁₃ = 13/2 (10 + 94) = 676

Now, Q + R = 8 or Q = 8 – R
Q² + R² = 34
(8 – R)² + R² = 34

64 + R² – 16R + R² = 34
2R² – 16R + 30 = 0
Or R² – 8R + 15 = 0

Or R² – 5R – 3R + 15 = 0
Or (R – 5) (R – 3) = 0
∴ R = 5, 3

Case I
When R = 5, Q = 3
3 × 4 + 5 = 17
Number = 17

Case II
When R = 3
So, Q = 5 and 5 × 4 + 3 = 23
Number = 23

Considering the unit digit only in number like 509, 519, 529,...,589, 9 occurs 9 times .
Considering the ten digit from 590 - 598, (590,591,592,...,598) 9 occurs 9 times.

Hence, there are 9 + 9 = 18 numbers between 500 to 600 that contain the digit nine exactly.

All the number in this series except 2! are divisible by 3, therefore, reminder will be 2.

If n is even, then even + even + 1 = odd
If n is odd, then odd + odd + 1 = even

Therfore, n to the power 4 + n to the power 2 + 1 is always odd.

D = abcd / 9999, 49995 = 9999 * 5. Hence, once we multiply D with 9999 we will get a natural number.