Question Detail

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

1. 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

2. 1939392 * 625

1212120010
1212120000
1212120011
1212121010

Answer: Option B

Explanation:

Trick: when multiplying with \begin{aligned} 5^n
\end{aligned} then put n zeros to the right of multiplicand and divide the number with \begin{aligned} 2^n
\end{aligned}
So using this we can solve this question in much less
time.
\begin{aligned} 1939392 \times 5^4 = \frac {19393920000}{16} = 1212120000
\end{aligned}

3. 469157 * 9999

1691100843
4591100843
4691100843
3691100843

Answer: Option C

Explanation:

469157 * (10000 - 1)
= 4691570000 - 469157
= 4691100843

4. 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.