Ques 1
When 3^333 is divided by 11, the remainder is
Solution
Finding the Remainder of \( 3^{333} \div 11 \) Using Cyclicity
Step 1: Identify the cycle of \( 3^n \mod 11 \)
\[ \begin{align*} 3^1 &\equiv 3 \mod 11 \\ 3^2 &\equiv 9 \mod 11 \\ 3^3 &\equiv 27 \equiv 5 \mod 11 \\ 3^4 &\equiv 15 \equiv 4 \mod 11 \\ 3^5 &\equiv 12 \equiv 1 \mod 11 \end{align*} \] The powers of 3 modulo 11 repeat every 5 terms.
Step 2: Reduce the exponent using the cycle
Since the cycle length is 5, we compute: \[ 333 \mod 5 = 3 \Rightarrow 3^{333} \equiv 3^3 \mod 11 \] From our earlier list: \[ 3^3 \equiv 5 \mod 11 \]
✅ Final Answer:
\[ \boxed{5} \]
Ques 2
A function f maps the set of natural numbers to whole numbers, such that f(xy) = f(x)f(y) + f(x) + f(y) for all x, y and f(p) = 1 for every prime number p. Then, the value of f(160000) is
1) 4095 2) 1023 3) 2047 4) 8191
Solution
f(xy) = f(x)f(y) + f(x) + f(y) = (f(x) + 1)(f(y) + 1) – 1
f(160000) = f(400 × 400) = (f(400) + 1)(f(400) + 1) – 1
Now, f(400) = f(20 × 20) = (f(20) + 1)(f(20) + 1) – 1
Now, f(20) = f(4 × 5) = (f(4) + 1)(f(5) + 1) – 1
Now, f(4) = f(2 × 2) = (f(2) + 1)(f(2) + 1) – 1
Since f(p) = 1 where p is a prime number, so
| Expression | Calculation | Value |
| f(4) | (1 + 1)(1 + 1) – 1 | 3 |
| f(20) | (3 + 1)(1 + 1) – 1 | 7 |
| f(400) | (7 + 1)(7 + 1) – 1 | 63 |
| f(160000) | (63 + 1)(63 + 1) – 1 | 4095 |
