Aptitude Number System Test Yourself

5358 x 51= 5358 x$\left(50 + 1\right)$= 5358 x 50 + 5358 x 1= 267900 + 5358= 273258.

Required sum = 2 + 3 + 5 + 7 + 11 = 28.

Note: 1 is not a prime number.

Definition: A prime number [or a prime] is a natural number that has exactly two distinct natural number divisors: 1 and itself.

Let the smaller number be $ x $. Then larger number = $x$ + 1365.

$\therefore x $ + 1365 = 6$ x $ + 15

$\Rightarrow$ 5$ x $ = 1350

$\Rightarrow x $ = 270

$\therefore$Smaller number = 270.

Given Exp. =$ \dfrac{(12)^3 \times 64}{432} = \dfrac{(12)^3 \times 64}{12 \times 6^2} $= (12)² x 6² = (72)² = 5184

72519 x 9999= 72519 x $\left(10000 - 1\right)$= 72519 x 10000 - 72519 x 1= 725190000 - 72519= 725117481.

Sum of digits = 5 + 1 + 7 + $ x $ + 3 + 2 + 4 = 22 + $ x $, which must be divisible by 3.

$\therefore$  $ x $ = 2.

Sum of digits = (4 + 8 + 1 + x + 6 + 7 + 3) = (29 + x), which must be divisible by 9.

$\therefore$   $ x $ = 7.

Local value of 7 - Face value of 7 = 70000 - 7= 69993

6$n$² + 6$n$ = 6$n$ $\left(n + 1\right)$, which is always divisible by 6 and 12 both, since $n$ $\left( n + 1\right)$ is always even.

Let the number be $ x $ and on dividing $ x $ by 5, we get $ k $ as quotient and 3 as remainder.

$\therefore$    $ x $ = 5k + 3

$\Rightarrow$    $ x $² = (5k + 3)²

   = (25 k ² + 30$ k $ + 9)

   = 5(5 k ² + 6$ k $ + 1) + 4

$\therefore$On dividing $ x $² by 5, we get 4 as remainder.

Required numbers are 102, 108, 114, ... , 996

This is an A.P. in which $ a $ = 102, $ d $ = 6 and $ l $ = 996

Let the number of terms be $ n $. Then,

$ a $ + $\left( n - 1\right)$ d = 996

  $\Rightarrow$ 102 + $\left( n - 1\right)$ x 6 = 996

  $\Rightarrow$ 6 x $\left( n - 1\right)$ = 894

  $\Rightarrow$ $\left( n - 1\right)$ = 149

  $\Rightarrow$ $ n $ = 150.

Let $ x $ be the number and $ y $ be the quotient. Then,

$ x $ = 357 $x y $ + 39

  = $\left(17 \times 21 xy\right )$ + $\left(17 \times 2\right)$ + 5

  = 17 x $\left(21y + 2\right) + 5$

$\therefore$Required remainder = 5.

Clearly, 4864 is divisible by 4.

So, 9P2 must be divisible by 3. So, 9 + P + 2 must be divisible by 3.

$\therefore$ P = 1.

6) 4456 (742

    42

    -------

     25

     24

    -------

      16

      12

    --------

       4

Therefore, Required number = (6 - 4) = 2.

6 = 3 x 2. Clearly, 5 * 2 is divisible by 2. Replace * by $ x $.

Then, 5 + $x$ + 2 must be divisible by 3. So, $ x $ = 2.

Given sum

= [1 + 1 + 1 + ... to n terms]-$(\dfrac{1}{n}+\dfrac{2}{n}+\dfrac{3}{n}+ ... $)to n terms

= $ n $ -$\dfrac{ n }{2}$$\left(\dfrac{1}{ n }\right)$+ 1    [ Ref: $ n $th terms = $\left( n / n\right )$ = 1]

= $ n $ -$\dfrac{ n + 1}{2} $

=$ \dfrac{1}{2} \left( n - 1\right)$

S_n = $\left(1 + 2 + 3 + ... + 50 + 51 + 52 + ... + 100\right)$ - $\left(1 + 2 + 3 + ... + 50\right)$

    =$ \dfrac{100}{2} $ x (1 + 100) -$ \dfrac{50}{2} $x (1 + 50)

    = $\left(50 \times 101\right)$ - $\left(25 \times 51\right)$

    = 5050 - 1275

    = 3775.

45 = 5 x 9, where 5 and 9 are co-primes.

Unit digit must be 0 or 5 and sum of digits must be divisible by 9.

Among given numbers, such number is 202860.

(3²⁵ + 3²⁶ + 3²⁷ + 3²⁸) = 3²⁵ x (1 + 3 + 3² + 3³) = 3²⁵ x 40

     = 3²⁴ x 3 x 4 x 10

     = (3²⁴ x 4 x 30), which is divisible by30.

Let the numbers be $ a $ and $ b $. Then, $ a $ + $ b $ = 12 and ab = 35.

$\therefore \dfrac{a + b}{ab} $=$ \dfrac{12}{35} $      $\Rightarrow$  $ \left(\dfrac{1}{b} +\dfrac{1}{a} \right) $=$ \dfrac{12}{35} $

$\therefore$ Sum of reciprocals of given numbers =$ \dfrac{12}{35} $