Number System - Aptitude MCQ Questions and Solutions with Explanations

$$59535 = 3 \times 5 \times \boxed{3 \times 3} \times \boxed{3 \times 3} \times 7 \times 7$$
⇒ To make a perfect square we should multiply by = 3 × 5 = 15
x = 15 [given]
⇒ Sum of digits of number = 1 + 5 = 6

211X
Go through option C
(46)² = 2116

6 < $$\root 3 \of {\text{N}} $$ < 7
On cubing,
216 < N < 343

Number of terms = (342 - 217) + 1 = 125 + 1 = 126

Sum of all prime numbers (x) = 2 + 3 + 5 + 7 + . . . . . . . .
Sum of all odd prime numbers (y) = 3 + 5 + 7 + . . . . . . . .
x - y = 2 Answer

$$\eqalign{ & \frac{{\sqrt {29.16} }}{{\sqrt {1.1664} }} + \frac{{\sqrt {0.2916} }}{{\sqrt {116.64} }} + \frac{{\sqrt {0.0036} }}{{\sqrt {0.36} }}\, \cr & = \frac{{5.4}}{{1.08}} + \frac{{0.54}}{{10.8}} + \frac{{0.06}}{{0.6}} \cr} $$

\[\left[ \begin{gathered} \because \sqrt {294} = 54 \hfill \\ \sqrt {11664} = 108 \hfill \\ \sqrt {36} = 6 \hfill \\ \end{gathered} \right.\]

$$\eqalign{ & = \frac{{540}}{{108}} + \frac{{54}}{{1080}} + \frac{6}{{60}} \cr & = 5 + \frac{1}{{20}} + \frac{1}{{10}} \cr & = \frac{{103}}{{20}} \cr} $$

$$\eqalign{ & 2000{\text{ - - - - - - }}7000 \cr & {\text{Perfect square}} = \sqrt {4096} = 64 \cr & {\text{Perfect cube}} = \sqrt {4096} = 16 \cr} $$

6 digit number ⇒ abc abc
= abc(1001)
So, number will be divisible by 1001.

$$\eqalign{ & \frac{7}{2} + \frac{{11}}{3} + \frac{7}{6} + \frac{{11}}{{15}} + \frac{7}{{12}} + \frac{{11}}{{35}} + \,........\, + \frac{7}{{156}} + \frac{{11}}{{575}} \cr & = \left( {\frac{7}{2} + \frac{7}{6} + \frac{7}{{12}} + \,........\, + \frac{7}{{156}}} \right) + \left( {\frac{{11}}{3} + \frac{{11}}{{15}} + \frac{{11}}{{35}} + \,........\, + \frac{{11}}{{575}}} \right) \cr & = 7\left( {\frac{1}{{1 \times 2}} + \frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} + \,........\, + \frac{1}{{12 \times 13}}} \right) + \frac{{11}}{2}\left( {\frac{2}{{1 \times 3}} + \frac{2}{{3 \times 5}} + \frac{2}{{5 \times 7}} + \,........\, + \frac{1}{{23 \times 25}}} \right) \cr & = 7\left( {1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \,........\, + \frac{1}{{12}} - \frac{1}{{13}}} \right) + \frac{{11}}{2}\left( {1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \,........\, + \frac{1}{{23}} - \frac{1}{{25}}} \right) \cr & = 7\left( {1 - \frac{1}{{13}}} \right) + \frac{{11}}{2}\left( {1 - \frac{1}{{25}}} \right) \cr & = \frac{{7 \times 12}}{{13}} + \frac{{11 \times 24}}{{2 \times 25}} \cr & = \frac{{84}}{{13}} + \frac{{132}}{{25}} \cr & = \frac{{2100 + 1716}}{{325}} \cr & = \frac{{3816}}{{325}} \cr} $$

3600 → 9 × 4 × 25 × 4 = 3² × 2⁴ × 5²
Factor: aⁿ → (n + 1)
Total number of factors = 3 × 5 × 3 = 45

m¹² - 1
= (m⁶)² - (1)²
= (m⁶ + 1)(m⁶ - 1)
= (m⁶ + 1){(m³)² - (1)²}
= (m⁶ + 1)(m³ - 1)(m³ + 1)
= (m⁶ + 1)(m³ - 1)(m + 1)(m² + m + 1)
Hence m + 1 is factor of m¹² - 1.
It will completely divide it and give remainder 0.