The greatest number by which the product of three consecutive multiples of 3 is always

Answer: Option A

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 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 C

Explanation:

3*6*9 = 162

Answer: Option C

Explanation:

469157 * (10000 - 1)
= 4691570000 - 469157
= 4691100843