Number System - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{fraction}}\,{\text{be}}\,\frac{{3k}}{{4k}} \cr & \therefore \frac{{ {3k - 9} }}{{4k}} = {\frac{2}{3}} \cr & \therefore 9k - 27 = 8k \cr & \therefore k = 27 \cr } $$

Given that a^b + a^c + a^d = 0
Hence, out of these three terms, some are positive and some are negative. Since all of a, b, c, and d are negative integers, only that term which has an even power can be positive. Hence, the absolute value of at least one of a, b, c or d is even. Hence, a × b × c × d is also even

91 is divisible by 7. So, it is not a prime number.

$$\eqalign{ & {112 \times {5^4}} \cr & = 112 \times {\left( {\frac{{10}}{2}} \right)^4} \cr & = \frac{{112 \times {{10}^4}}}{{{2^4}}} \cr & = \frac{{1120000}}{{16}} \cr & = 70000 \cr} $$

Let 2³² = x.
Then,
(2³² + 1) = (x + 1)
Let (x + 1) be completely divisible by the natural number N.
Then,
(2⁹⁶ + 1)
= [(2³²)³ + 1]
= (x³ + 1)
= (x + 1)(x² - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.

$$\eqalign{ & 23)\,\,\,1056\,\,\,\,(45 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,92 \cr & \,\,\,\,\,\,\,\,\,\,\, - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,136 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,115 \cr & \,\,\,\,\,\,\,\,\,\,\, - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,21 \cr & \,\,\,\,\,\,\,\,\,\,\, - - - \cr & {\text{Required}}\,{\text{number}} \cr & = \left( {23 - 21} \right) \cr & = 2 \cr} $$

$$\eqalign{ & 1397 \times 1397 \cr & = {\left( {1397} \right)^2} \cr & = {\left( {1400 - 3} \right)^2} \cr & = {\left( {1400} \right)^2} + {\left( 3 \right)^2} - \left( {2 \times 1400 \times 3} \right) \cr & = 1960000 + 9 - 8400 \cr & = 1960009 - 8400 \cr & = 1951609 \cr} $$

132 = 4 x 3 x 11
So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.
264 ⇒ 11, 3, 4 (/)
396 ⇒ 11, 3, 4 (/)
462 ⇒ 11, 3 (X)
792 ⇒ 11, 3, 4 (/)
968 ⇒ 11, 4 (X)
2178 ⇒ 11, 3 (X)
5184 ⇒ 3, 4 (X)
6336 ⇒ 11, 3, 4 (/)
Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.
Required number of number = 4.

$$\eqalign{ & 935421 \times 625 \cr & = 935421 \times {5^4} \cr & = 935421 \times {\left( {\frac{{10}}{2}} \right)^4} \cr & = \frac{{935421 \times {{10}^4}}}{{{2^4}}} \cr & = \frac{{9354210000}}{{16}} \cr & = 584638125 \cr} $$

$$\eqalign{ & {\text{Largest}}\,{\text{4 - digit}}\,{\text{number}} = 9999 \cr & 88)\,\,\,\,9999\,\,\,\,(113 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,88 \cr & \,\,\,\,\,\,\,\,\,\,\, - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,119 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,88 \cr & \,\,\,\,\,\,\,\,\,\,\, - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,319 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,264 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,55 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - - - \cr & \text{Required number} \cr & = \left( {9999 - 55} \right) \cr & = 9944 \cr} $$