Number System - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \frac{{{{65}^{203}}}}{7} \cr & Or,\,\frac{{{{\left( {63 + 2} \right)}^{203}}}}{7} \cr & 63\,{\text{is}}\,{\text{divisible}}\,{\text{by}}\,7, \cr & {\text{so}}\,{\text{remainder}}\,{\text{will}}\,{\text{depend}}\,{\text{on}}\,{\text{the}}\,{\text{powers}}\,{\text{of}}\,2 \cr & Or,\,\frac{{{2^{203}}}}{7} \cr & {\text{Its}}\,{\text{remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}} \cr & \frac{{{2^3}}}{7} \cr & {\text{Now,}}\,{\text{Required}}\,{\text{Remainder}}\,{\text{wold}}\,{\text{be}}, \cr & \frac{8}{7} = 1\cr & {\text{Required}}\,{\text{remainder}} = 1 \cr} $$
Note: We have manipulated the powers in the form of (4x + n). It means 203 is taken as,
203 = 4x + n = 4 × 50 + 3.
We neglect power which is in the multiple of 4.

$$\eqalign{ & \frac{{{{67}^{107}}}}{7} \cr & Or,\,\frac{{{{\left( {7 \times 9 + 4} \right)}^{107}}}}{7} \cr & {\text{The}}\,{\text{remainder}}\,{\text{will}}\,{\text{be}}\,{\text{same}}\,{\text{as}} \cr & \frac{{{4^{107}}}}{7} \cr & Or,\,\frac{{{4^3}}}{7} \cr & Or,\,\frac{{64}}{7} \cr & {\text{Required}}\,{\text{Remainder}} = 1 \cr} $$

$$\eqalign{ & \frac{{{{54}^{124}}}}{{17}} \cr & Or,\,\frac{{{{\left( {17 \times 3 + 3} \right)}^{124}}}}{{17}} \cr & {\text{The}}\,{\text{remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}}, \cr & \frac{{{3^{124}}}}{{17}} \cr & Or,\,\frac{{{3^{4 \times 31}}}}{{17}} \cr & {\text{Remainder}}\,{\text{will}}\,{\text{be}}\,{\text{same}}\,{\text{as}}, \cr & \frac{{{3^4}}}{{17}} \cr & Or,\,\frac{{81}}{{17}} \cr & \text{Remainder} = 13 \cr} $$

To find unit digit of a number or an Expression, We have to divide the number or expression by 10 and the remainder obtained by this operation would be the required unit digit.
$$\eqalign{ & \frac{{ {{{\left( {173} \right)}^{45}} \times {{\left( {152} \right)}^{77}} \times {{\left( {777} \right)}^{999}}} }}{{10}} \cr & {\text{Remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}}, \cr & \frac{{ {{3^{45}} \times {2^{77}} \times {7^{999}}} }}{{10}} \cr & Or,\,\frac{{ {3 \times 2 \times {7^3}} }}{{10}} \cr & Or,\,\frac{{ {6 \times 343} }}{{10}} \cr & {\text{Remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}}\,\frac{{ {6 \times 3} }}{{10}} \cr }$$
Thus, Required remainder and unit digit will be 8.

In such cases remainder will always be 1.

$$\eqalign{ & {\text{Required}}\,{\text{answer}}, \cr & = {\frac{{146}}{5}} + {\frac{{146}}{{{5^2}}}} + {\frac{{146}}{{{5^3}}}} \cr & = 29 + 5 + 1 \cr & = 35 \cr} $$
Note:
We have taken integral value only, not the fractional.
For example $$\frac{{146}}{5}$$ = 29.2 but we have taken 29 and so on.

$$\eqalign{ & {\text{Number of mint before Eswari has taken}}, \cr & = \left( {x - {\frac{x}{2}} } \right) + 2 = 17 \cr & Or,\,x = 30 \cr & {\text{Number of mint before Fatima has taken}}, \cr & = \left( {x - {\frac{x}{4}} } \right) + 3 = 30 \cr & Or,\,x = 36 \cr & {\text{Number of mint before Sita has taken}}, \cr & = \left( {x - {\frac{x}{3}} } \right) + 4 = 36 \cr & Or,\,x = 48 \cr & {\text{Hence,}}\,{\text{there}}\,{\text{were}}\,{\text{48}}\,{\text{mints}}\,{\text{originally}}{\text{.}} \cr} $$

When the birds sat one on a branch, there was one extra bird.
When they sat Two to a branch one branch was extra.
Now, we go through the options;

Checking option (A);
If there were 3 branches, there would be 4 birds.
(This would leave one bird without branch as per the question.)
When 4 birds would sit Two to a branch there would be one branch free
(as per the question.).
Hence, the option (A) is correct.

We go through the options,
Checking option (C)
$$\frac{{71}}{{17}}$$ = 4 quotient and remainder 3.
$$\frac{{71}}{8}$$ = 8 quotient and remainder 7 as remainder.

If we multiply the last three digits of each terms we get the last three digits.
345 × 321 = 110745
So, the last three digits are 745