Number System Quiz

The unit digit of 17 is 7. Cyclicity of 7 is 4 (7, 9, 3, 1). $222 \mod 4 = 2$. So, the unit digit is $7^2 = 49$, which ends in 9.

$3^{63}$: Cyclicity of 3 is 4. $63 \mod 4 = 3$. $3^3 = 27 \rightarrow 7$. $7^{52}$: $52 \mod 4 = 0$, so we take the 4th term in cycle: 1. $7 \times 1 = 7$.

$7 \equiv 1 \pmod 6$. Therefore, $7^{103} \equiv 1^{103} = 1 \pmod 6$.

$2^3 = 8 \equiv 1 \pmod 7$. $256 = 3 \times 85 + 1$. $2^{256} = (2^3)^{85} \times 2^1 \equiv 1^{85} \times 2 = 2 \pmod 7$.

$3^{100} \pmod 7$: $3^6 \equiv 1$, $100 = 6 \times 16 + 4$. $3^4 = 81 \equiv 4 \pmod 7$. $5^{100} \equiv (-2)^{100} = 2^{100} \pmod 7$. $2^3 \equiv 1$, $100 = 3 \times 33 + 1$. $2^1 = 2 \pmod 7$. $4 + 2 = 6$.

Alternating sum: $5 - 4 + 3 - 2 + 1 = 3$. Since 3 is positive and less than 11, the remainder is 3.

Sum of digits at odd positions: $5 + 7 = 12$. Sum of digits at even positions: $8 + 4 = 12$. Difference $12 - 12 = 0$. Since the difference is 0, the number is already divisible by 11.

Largest 4-digit number is 9999. $9999 \div 15$ gives remainder 9. $9999 - 9 = 9990$. 9990 is divisible by both 3 and 5, so it is divisible by 15.

The numbers are 3, 6, 9, ..., 99. This is an AP with $a=3, d=3, l=99$. $n = (99-3)/3 + 1 = 32 + 1 = 33$.

Prime factorization: $360 = 2^3 \times 3^2 \times 5^1$. Number of factors = $(3+1)(2+1)(1+1) = 4 \times 3 \times 2 = 24$.

Number of divisors = $(4+1)(3+1)(2+1) = 5 \times 4 \times 3 = 60$.

Number of trailing zeros = $\lfloor 100/5 \rfloor + \lfloor 100/25 \rfloor = 20 + 4 = 24$.

$\lfloor 100/3 \rfloor + \lfloor 100/9 \rfloor + \lfloor 100/27 \rfloor + \lfloor 100/81 \rfloor = 33 + 11 + 3 + 1 = 48$.

Number of digits = $\lfloor 100 \times \log 2 \rfloor + 1 = \lfloor 100 \times 0.3010 \rfloor + 1 = 30 + 1 = 31$.

$1 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 32 + 0 + 8 + 4 + 0 + 1 = 45$.

45 divided by 2 gives remainders: 1, 0, 1, 1, 0, 1 (from bottom up), so $45_{10} = 101101_2$.

$2 \times 16^1 + F \times 16^0 = 32 + 15 \times 1 = 47$.

$\text{LCM}(8, 12, 15) = 120$. The number is $\text{LCM} + \text{remainder} = 120 + 3 = 123$.

Total numbers = $^4P_3 = 4 \times 3 \times 2 = 24$.

$2^{15} = 32768$. Sum of digits = $3+2+7+6+8 = 26$.

Ending digit is 6. $80^2 = 6400$, $90^2 = 8100$. Try $84^2$: $84 \times 84 = 7056$. Yes.

$\text{Product of numbers} = \text{HCF} \times \text{LCM}$. $30 \times x = 6 \times 180 \rightarrow x = 1080/30 = 36$.

Numbers are $3 \times \text{HCF}$ and $5 \times \text{HCF}$, i.e., $3 \times 4 = 12$ and $5 \times 4 = 20$.

$999 \div 13$ leaves remainder 11 (or $-2$). $(-2)^{100} = 2^{100} \pmod{13}$. By Fermat's Little Theorem, $2^{12} \equiv 1 \pmod{13}$. $100 = 12 \times 8 + 4$. $2^{100} \equiv 2^4 = 16 \equiv 3 \pmod{13}$.