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

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

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

Explanation:

3*6*9 = 162

Answer: Option A

Explanation:

On dividing 427398 by 15 we get the remainder 3, so 3 should be subtracted

Answer: Option B

Explanation:

$$\begin{aligned} \frac{abc}{bcd} = \frac{385}{1001} => \frac{a}{d} = \frac{5}{13} \end{aligned}$$ So d = 13