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

Answer: Option B

Explanation:

Trick: 587 * (1000 - 1) = 587000 - 587 = 586413

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 C

Answer: Option D

Explanation:

Largest 4 digit number is 9999
After doing 9999 ÷ 88 we get remainder 55
Hence largest 4 digit number exactly divisible by 88 = 9999 - 55 = 9944