Question Detail

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.

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 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

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. 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

5. 587 * 999 = ?

586411
586413
486411
486413

Answer: Option B

Explanation:

Trick: 587 * (1000 - 1) = 587000 - 587 = 586413

Question Detail

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

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

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

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

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

5. 587 * 999 = ?

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