What least value should be replaced by * in 223*431 so the number become divisible by 9

Answer: Option C

Explanation:

\begin{aligned}
\frac {1}{2}\times 2(a^2 + b^2) = \frac {1}{2} \times [(a + b)^2 + (a - b)^2]
\end{aligned}
\begin{aligned}
= \frac {1}{2} \times [(217 + 183)^2 + (217 - 183)^2] = 80578
\end{aligned}

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 C

Answer: Option B

Explanation:

Trick: when multiplying with \begin{aligned} 5^n
\end{aligned} then put n zeros to the right of multiplicand and divide the number with \begin{aligned} 2^n
\end{aligned}
So using this we can solve this question in much less
time.
\begin{aligned} 1939392 \times 5^4 = \frac {19393920000}{16} = 1212120000
\end{aligned}

Answer: Option D