Question Detail

6897 is divisible by

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

Answer: Option C

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

Answer: Option C

Explanation:

3*6*9 = 162

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. How many terms are there in 2,4,8,16,....,1024

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

4. What is the unit digit in
\begin{aligned}
(6324)^{1797} \times (615)^{316} \times (341)^{476}
\end{aligned}

Answer: Option D

5. What least number should be added to 1056, so that the sum is completely divisible by 23

Answer: Option C

Explanation:

(1056 / 23) gives remainder 21

21 + 2 = 23, So we need to add 2

Question Detail

6897 is divisible by

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

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

3. How many terms are there in 2,4,8,16,....,1024

4. What is the unit digit in \begin{aligned} (6324)^{1797} \times (615)^{316} \times (341)^{476} \end{aligned}

5. What least number should be added to 1056, so that the sum is completely divisible by 23

4. What is the unit digit in
\begin{aligned}
(6324)^{1797} \times (615)^{316} \times (341)^{476}
\end{aligned}