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

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.

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 A

Explanation:

Although it is a simple question, but the trick is to save time in solving this.
Rather than multiplying it we can do as follows:
586645 * (10000 - 1) = 5866450000 - 586645 = 5865863355

Answer: Option D

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}