Aptitude Number System Practice QA

Let S_n =1 + 2 + 3 + ... + 45. This is an A.P. in which a =1, d =1, n = 45.

S_n =$ \dfrac{n}{2} $[2$ a $ + $\left( n - 1\right)$ d ]=$ \dfrac{45}{2} $ x [2 x 1 + (45 - 1) x 1]=$ \dfrac{45}{2} $x 46= 45 x 23

= 45 x (20 + 3)

= 45 x 20 + 45 x 3

= 900 + 135

= 1035.

Shorcut Method:

S_n =$ \dfrac{n(n + 1)}{2} $=$ \dfrac{45(45 + 1)}{2} $= 1035.

24 = 3 x8, where 3 and 8 co-prime.

Clearly, 35718 is not divisible by 8, as 718 is not divisible by 8.

Similarly, 63810 is not divisible by 8 and 537804 is not divisible by 8.

Consider option (D),

Sum of digits = (3 + 1 + 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 3.

Also, 736 is divisible by 8.

$\therefore$ 3125736 is divisible by 3 x 8, i.e., 24.

  $\Rightarrow$   20 - 20$ x $² = 9$ x $

  $\Rightarrow$   20$ x $² + 9$ x $ - 20 = 0

  $\Rightarrow$   20$ x $² + 25$ x $ - 16$ x $ - 20 = 0

  $\Rightarrow$   5 x $\left(4x+ 5\right)$ - 4$\left(4x + 5\right)$ = 0

  $\Rightarrow$   $\left(4 x + 5\right)$ x $\left(5 x - 4\right)$ = 0

$x$ =$ \dfrac{4}{5} $

Formula: [Divisor*Quotient] + Remainder = Dividend.

Soln:

[56*Q]+29 = D -------(1)

D%8 = R -------------(2)

From equation 2,

[56*Q+29]%8 = R.

=> Assume Q = 1.

=> [56+29]%8 = R.

=> 85%8 = R

=> 5 = R.

( x ⁿ + 1) will be divisible by ( x + 1) only when $ n $ is odd.

$\therefore$(67⁶⁷ + 1) will be divisible by (67 + 1)

$\therefore$(67⁶⁷ + 1) + 66, when divided by 68 will give 66 as remainder.

3-digit number divisible by 6 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 $ t $_n = 996.

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

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

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

$\Rightarrow$ ( n - 1) = 149

$\Rightarrow n $ = 150

$\therefore$ Number of terms = 150.

Required numbers are 24, 30, 36, 42, ..., 96

This is an A.P. in which $ a $ = 24, $ d $ = 6 and $ l $ = 96

Let the number of terms in it be $ n $.

Then t_n = 96  $\Rightarrow$  $ a $ + $\left(n - 1\right)$d = 96

$\Rightarrow$ 24 + $\left( n - 1\right)$ x 6 = 96

$\Rightarrow$ $\left( n - 1\right)$ x 6 = 72

$\Rightarrow$ $\left( n - 1\right)$ = 12

$\Rightarrow n $ = 13

Required number of numbers = 13.

4 a 3 |

9 8 4 } ==> a + 8 = b ==> b - a = 8

13 b 7 |

Also, 13b7 is divisible by 11   $\Rightarrow$   (7 + 3) - (b + 1) = (9 - b)

$\Rightarrow$   (9 - b) = 0

$\Rightarrow$   b = 9

$\therefore$ b= 9 and a= 1    $\Rightarrow$ (a + b) = 10.

Let $ x $ = 296$ q $ + 75

   = $\left(37 \times 8q + 37 \times 2\right)$ + 1

   = 37 $\left(8 q + 2\right)$ + 1

Thus, when the number is divided by 37, the remainder is 1.

5 | $x$         z = 13 x 1 + 12 = 25

--------------

9 | y - 4        y = 9 x z + 8 = 9 x 25 + 8 = 233

--------------

13| z - 8        $x$ = 5 x y + 4 = 5 x 233 + 4 = 1169

--------------

| 1 -12

585) 1169 (1

        585

       ----------

        584

       ----------

Therefore, on dividing the number by 585, remainder = 584.

87) 13601 (156

      87

      -------

        490

        435

      -------

         551

         522

        -------

          29

        -------

Therefore, the required number = 29.

Let the given number be 476 $x$$y$ 0.

Then $\left(4 + 7 + 6 + x + y + 0\right)$ = $\left(17 + x + y \right)$ must be divisible by 3.

And, $\left(0 + x + 7\right)$ - $\left( y + 6 + 4\right)$ = $\left( x - y -3\right)$ must be either 0 or 11.

$ x $ - $ y $ - 3 = 0   $\Rightarrow y $ = $ x $ - 3

17 + $x$ + $y$ = $\left(17 + x + x - 3\right)$ = 2$ x $ + 14

$\Rightarrow x $= 2 or $ x $ = 8.

$\therefore x $ = 8 and $ y $ = 5.

Given number = 97215 x 6

$\left(6 + 5 + 2 + 9\right)$ - $\left( x + 1 + 7\right)$ = $14 - x$, which must be divisible by 11.

$\therefore$  $ x $ = 3

(11² + 12² + 13² + ... + 20²) = (1² + 2² + 3² + ... + 20²) - (1² + 2² + 3² + ... + 10²)

Ref: (1² + 2² + 3² + ... + n²) =$\dfrac{1}{6} n(n + 1)(2n + 1)$    

= (2870 - 385)

= 2485.

Clearly, (2272 - 875) = 1397, is exactly divisible by N.

Now, 1397 = 11 x 127

$\therefore$ The required 3-digit number is 127, the sum of whose digits is 10.

987 = 3 x 7 x 47

So, the required number must be divisible by each one of 3, 7, 47

553681 $\rightarrow$ Sum of digits = 28, not divisible by 3

555181 $\rightarrow$ Sum of digits = 25, not divisible by 3

555681 is divisible by 3, 7, 47.

$ x $ = 13$ p $ + 11 and $ x $ = 17$ q $ + 9

$\therefore$ 13$ p $ + 11 = 17$ q $ + 9

$\Rightarrow$ 17$ q $ - 13$ p $ = 2

$\Rightarrow$ q=$ \dfrac{2 + 13p}{17} $

The least value of $ p $ for which q =$\dfrac{2 + 13 p }{17}$ is a whole number is $ p $ = 26

$\therefore x $ = $ \left(13 \times 26 + 11\right)$

   = 338 + 11

   = 349

72 = 9 x8, where 9 and 8 are co-prime.

The minimum value of x for which 73$ x $ for which 73$ x $ is divisible by 8 is, $ x $ = 6.

Sum of digits in 425736 = (4 + 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 9.

$\therefore$Required value of * is 6.

Let $ n $ = 4$ q $ + 3. Then 2$ n $ = 8$ q $ + 6   = 4 $\left(2 q + 1\right ) $ + 2.

Thus, when 2$ n $ is divided by 4, the remainder is 2.

Let the two consecutive even integers be 2$n $ and $\left(2 n + 2\right) $. Then,

$\left(2n + 2\right)$² = $\left(2 n + 2 + 2 n\right )$$\left(2 n + 2 - 2 n \right)$

     = 2$\left(4 n + 2\right) $

     = 4$\left(2 n + 1\right)$, which is divisible by 4.