Quantitative Ability Tech Test Divisibility Test 2

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.

5 | $x$         z = 13 x 1 + 12 = 25

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

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

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

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| 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

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

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.