Number System - Aptitude MCQ Questions and Solutions with Explanations

HCF of 2⁴, 8², 16², 20³ = 2⁴
LCM of 2⁴, 8², 16², 20³ = 2⁴ = 2⁸ × 125 = 32000

The least possible number = (LCM of 4, 5, 6 and 7) + 3
= 420 + 3
= 423
The next higher number is,
(420m + 3), now we put a least value of m such that
(420m + 3) ≥ 1000
At m = 3,
value = 420 × 3 + 3 = 1263

Let one part of string is x.
Now,
x + $$\frac{{9{\text{x}}}}{4}$$ = 221 m
x = 68 meter
and, $$\frac{{9{\text{x}}}}{4}$$ = 153 m
Difference between two parts = 153 - 68 = 85m.

1 - 3 + 5 - 7 + 9 - 11 .......... + 197 - 199
= (- 2) + (-2) + (-2) + .......... + (-2) (50 times)
= 50 × (-2) = -100

$$\eqalign{ & \frac{{ {{6^{36}}} }}{{215}},\,{\text{can}}\,{\text{be}}\,{\text{written}}\,{\text{as}} \cr & \frac{{{{\left( {{6^3}} \right)}^{12}}}}{{215}} \cr & Or,\,\frac{{{{216}^{12}}}}{{215}},\,[216\,{\text{on}}\,{\text{divided}}\,{\text{by}}\,215,\,{\text{gives}}\,{\text{remainder}}\,1] \cr & \frac{{{1^{12}}}}{{215}} \cr & {\text{The}}\,{\text{remainder}}\,{\text{will}}\,{\text{be}}\,1 \cr} $$

$$\eqalign{ & \frac{{{{12}^{13}}}}{{11}}\,{\text{gives}}\,{\text{Remainder}}\,1 \cr & {\text{It}}\,{\text{can}}\,{\text{be}}\,{\text{written}}\,{\text{as}}\, \cr & \frac{{\left( {12 \times 12 \times 12 \times 12\,........\,13\,\text{times}} \right)}}{{11}} \cr & {\text{On}}\,{\text{dividing}}\,{\text{it}}\,{\text{gives}}\,{\text{remainder}}\,{\text{1,}}\,{\text{each}}\,{\text{time}}. \cr & 1 \times 1 \times 1 \times 1\,.........\,13\,{\text{times}} \cr & {\text{So,}}\,{\text{final}}\,{\text{remainder}}\,{\text{will}}\,{\text{be}}\,1 \cr & \frac{{{{23}^{13}}}}{{11}} \Rightarrow \,{\text{Remainder}}\,1 \cr & {\text{Thus}}, \cr & {\text{The}}\,{\text{remainder}}\,{\text{of}}\,\frac{{\left( {{{12}^{13}} + {{23}^{13}}} \right)}}{{11}} \cr & = \left( {1 + 1} \right) \cr & = 2 \cr} $$

$$\frac{{{{75}^{{{75}^{75}}}}}}{{37}}$$
When 75 is divided by 37, it leaves remainder of 1.
So, the expression will become,
$$\frac{{{1^{{{75}^{75}}}}}}{{37}}$$
Now, for any power of 1, we always get 1 and expression becomes $$\frac{1}{{37}}$$ that leaves remainder 1

The only possible number which is both perfect square and perfect cube is 729 which is cube of 9 and square of 27
So, 7 × 2 × 9 = 126

Going through options:
LCM of 204 and 170 = 1020.

$$\eqalign{ & {\text{Total}}\,{\text{Marks}} = 90 \cr & {\text{Marks}}\,{\text{in}}\,{\text{QA}} = \frac{7}{9}\,\text{of}\,\,90 \cr & = \frac{{\left( {7 \times 90} \right)}}{9} \cr & = 7 \times 10 \cr & = 70 \cr} $$