Question Detail

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

Answer: Option C

Explanation:

3*6*9 = 162

1. 217 * 217 + 183 * 183

70578
82578
80578
80568

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}

2. \begin{aligned}
46^2 - (?)^2 = 4398 - 3066
\end{aligned}

Answer: Option D

3. 6897 is divisible by

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

Answer: Option C

4. There are four prime numbers written in ascending order. The product of first three is 385 and that of last three is 1001. The last number is:

Answer: Option B

Explanation:

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

5. 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.

Question Detail

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

1. 217 * 217 + 183 * 183

2. \begin{aligned} 46^2 - (?)^2 = 4398 - 3066 \end{aligned}

3. 6897 is divisible by

4. There are four prime numbers written in ascending order. The product of first three is 385 and that of last three is 1001. The last number is:

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

2. \begin{aligned}
46^2 - (?)^2 = 4398 - 3066
\end{aligned}