Number System - Aptitude MCQ Questions and Solutions with Explanations

In each set of 100 numbers, there are 10 numbers whose tens digit and unit digit are same. Again in the same set there are 10 numbers whose hundreds and tens digits are same. But one number in each set of 100 numbers whose Hundreds, Tens and Unit digit are same as 111, 222, 333, 444 etc
Hence, there are exactly (10 + 10 - 1) = 19 numbers in each set of 100 numbers. Further there are 9 such sets of numbers
Therefore such total numbers = 19 × 9 = 171

Alternatively,
9 × 10 × 10 - 9 × 9 × 9 = 900 - 729 = 171

Since, the magazine has 84 pages it means that there are 21 sheets of paper, which are folded in the middle. The pattern of pages will be this way,

Left	Right
1, 2	83, 84
3, 4	81, 82
- - - -	- - - -
29, 30	55, 56

$$\eqalign{ & {\text{Since}}, \cr & \frac{6}{{10}} \to \,{\text{remainder}}\,{\text{is}}\,{\text{6}}{\text{}} \cr & \frac{{{6^6}}}{{10}} \to \,{\text{remainder}}\,{\text{is}}\,{\text{6}}{\text{}} \cr & \frac{{{6^{{6^6}}}}}{{10}} \to \,{\text{remainder}}\,{\text{is}}\,{\text{6}}{\text{}} \cr & \frac{{{6^{{6^{{6^6}}}}}}}{{10}} \to \,{\text{remainder}}\,{\text{is}}\,{\text{6}}{\text{}} \cr & \frac{{{6^{{6^{{6^{{6^6}}}}}}}}}{{10}} \to \,{\text{remainder}}\,{\text{is}}\,{\text{6}}\,{\text{and}}\,{\text{so}}\,{\text{on}}. \cr & \cr & {\text{Thus,}}\,{\text{in}}\,{\text{all}}\,{\text{the}}\,{\text{such}}\,{\text{cases}}\, \cr & {\text{The}}\,{\text{remainder}}\,{\text{will}}\,{\text{always}}\,{\text{be}}\,6 \cr} $$

$$\eqalign{ & \frac{{\left[ {\left( {888 \times 888 \times 888} \right) - \left( {222 \times 222 \times 222} \right)} \right]}}{{\left[ {\left( {888 \times 888} \right) + \left( {888 \times 222} \right) + \left( {222 \times 222} \right)} \right]}} \cr & = \frac{{\left( {{{888}^3} - {{222}^3}} \right)}}{{\left[ {{{888}^2} + \left( {888 \times 222} \right) + {{222}^2}} \right]}} \cr & = 888 - 222 \cr & = 666 \cr} $$

$$\eqalign{ & {32^{{{32}^{32}}}}\,{\text{means}}\,{32^{\left( {32.32.32.32.......32\,\text{times}} \right)}} \cr & {\text{and}}\,{\text{the}}\,{\text{remainder}}\,{\text{of}}\,\frac{{32}}{7}\,{\text{is}}\,4 \cr & {\text{So}}, \cr & \frac{{{4^{\left( {32.32.32.32.......32\,\text{times}} \right)}}}}{7} \cr & \frac{{{4^{\left( {2.2.2.2.2........32\,\text{times}} \right)}}}}{7} \cr & {\text{Remainder}} = 4 \cr & {\text{Since}},\,\frac{4}{7} \to \,{\text{Remainder}}\,4 \cr & \frac{{{4^2}}}{7} \to \,{\text{Remainder}}\,2 \cr & \frac{{{4^3}}}{7} \to \,{\text{Remainder}}\,1 \cr & \frac{{{4^4}}}{7} \to \,{\text{Remainder}}\,4 \cr} $$

Remainder of $$\frac{{ {{{888}^{222}} + {{222}^{888}}} }}{3}$$     = 0
Since, 888 and 222 both (bases) are divisible by 3

The remainder when 10 is divided by 6 is 4.
The remainder when 10² is divided by 6 is 4.
The remainder when 10³ is divided by 6 is 4.
The remainder when 10⁴ is divided by 6 is 4.
Thus, remainder is always 4.
So, the required remainder,
$$\eqalign{ & = \frac{{4 + 4 + 4 + 4 + \,.....\,99{\text{ times}}}}{6} \cr & = \frac{{396}}{6} \cr} $$
Thus, Required remainder is 0.

Since, $${3^4} = 81$$  gives remainder 1 on divided by 80
So, $$\frac{{{3^{4n}}}}{{80}}$$  gives remainder 1
Thus, $$\frac{{{3^{67!}}}}{{80}}$$  will also give the remainder as 1
Since, 67! = 4n for a positive integer n

Since, $$\frac{{{{\text{a}}^{\text{n}}}}}{{{\text{a}} + 1}}$$   gives remainder 1 when 'n' is even.
Now since, 97! is an even number so remainder will be 1.
Note:-Factorial of any no. is even.

$$\eqalign{ & {2^{59}}\,{\text{can}}\,{\text{be}}\,{\text{expressed}}\,{\text{as}}\,{2^{8n}} \cr & {\text{So}}, \cr & {\text{Remainder}}\,\frac{{{2^{59}}}}{{255}} \cr & = {\text{Remainder}}\,\frac{{{2^{8n}}}}{{255}} \cr & = {\text{Remainder}}\,\frac{{{{256}^n}}}{{255}} = 1 \cr & {\text{Required}}\,{\text{remainder}} = 1 \cr} $$