Question Detail

6897 is divisible by

11 only
19 only
both 11 and 19
neither 11 nor 19

Answer: Option C

1. Find the unit digit in \begin{aligned}
(544)^{102} + (544)^{103}
\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

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

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.

3. What least number must be subtracted from 427398 so that remaining number is divisible by 15

Answer: Option A

Explanation:

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

4. Simplify 586645 * 9999

5865863355
5665863355
4865863355
4665863355

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