Number System Test - 2

Given that t_n+1 = 2t_n +1 and t₀ = 0

Put n = 0.

So, t_{0+1 =} 2t₀ + 1 or t_{1 =} 0+1

So, t_{1 =} 1

Now, put n = 1, so t₁₊₁₌ 2t₁ +1, t₂ = 2(1) + 1 = 2 + 1 = 3

Similarly, when n = 2, then t₂₊₁ = 2t₂ +1 = t₃ = 2(3) + 1 = 7

Continuing these steps, we get t₁₀ = 1023

Since n is an even number, so n² + n will also be even. But when 1 is added to n² + n, it becomes odd because even + odd = odd.

Let two digit number = 10x + y and number on reversing the digits = 10y + x
Sum of number and its reverse is 10x + y + 10y + x = 11 (x + y)
Now, the number will be perfect square when x + y = 11
The combinations with sum 11 are (2, 9) (3, 8) (4, 7) (5, 6) (6, 5) (7, 4) (8, 3) and (9, 2).

|a| + |b| > |a + b| is greater when a and b are of opposite signs. When a and b are of opposite signs, |a| and |b| both will become positive and add. But in |a + b|, they would subtract and we know sum of the given two numbers is always greater than their difference. 

Divide the power of 7, i.e. 84 in parts of 3 because 7³ when divided by 342 leaves 1 as a remainder.
Now, 84 is divisible by 3.

Therefore, 7⁸⁴, when divided by 342, leaves 1 as a remainder.

Although x is a negative integer, yet x² is a positive integer. Since z is greater than 1, so, z² is always positive. But if z² is greater than x², x² - z² will be negative. Therefore, it cannot be positive every time.

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