A number when divided by 44, gives 432 as quotient and 0 as remainder. What will be the remainder when dividing the same number by 31

Answer: Option C

Explanation:

Required digit is = \begin{aligned}
(4)^{102} + (4)^{103} \end{aligned}
as \begin{aligned} (4)^2 \end{aligned} gives unit digit 6 so \begin{aligned} (4)^{102} \end{aligned} unit digit is 6 and \begin{aligned} (4)^{103} \end{aligned} unit digit is, unit digit of \begin{aligned} 6 \times 4 \end{aligned}= 4, so answer will be unit digit of 6 + 4 = 0

Answer: Option B

Answer: Option D

Explanation:

\begin{aligned}
(a + b)^{2} = (a^2 + b^2 + 2ab)
\end{aligned}
\begin{aligned}
1307^{2} = (1300 + 7)^2 = (1690000 + 49 + 18200)
\end{aligned}

Answer: Option D

Explanation:

It is a GP. with r = 2, ie \begin{aligned} 2^1, 2^2, 2^3... \end{aligned}
If number of terms is n. Then,
\begin{aligned} 2 \times 2^{n-1} = 1024 \end{aligned} =>
\begin{aligned} 2^{n-1} = 512 \end{aligned}
or \begin{aligned} 2 \times 2^{n-1} = 2^9 \end{aligned}
so n-1 = 9 => n = 10

Answer: Option A

Explanation:

Trick: Number is divisible by 9, if sum of all digits is divisible by 9, so (2+2+3+*+4+3+1) = 15+* should be divisible by 9,
15+3 will be divisible by 9,
so that least number is 3.