Aptitude Discussion

Q. |
The sum of three digit number is subtracted from the number. The resulting number is always: |

✖ A. |
divisible by 6 |

✖ B. |
not divisible by 6 |

✔ C. |
divisible by 9 |

✖ D. |
not divisible by 9 |

**Solution:**

Option(**C**) is correct

Let the three digit number be $439$

Sum of digits $=16$

Difference $= 439-16 = 423$ which is **divisible by 9**.

**Arun**

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The question reads 'always'!

**DUDE**

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This is the property which will be true for all. Doesn't matter how may digits you take it will be always divisible by 9.

ie.2,3,4,5,6...and so on

**Riya**

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It holds for 153 too..

sum of digits $=9.$

subtracting 9 from 153 you get 144 which is divisible by 9

it holds for 6 also which is also given in the options

**Shreya**

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what if we take the example 153?

**Deepak**

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Let the $3$ digit number be: $XYZ$

As per the question we have to find the divisibility of number $P$ (let),

$\Rightarrow P=(100X+10Y+Z)-(X+Y+Z)$

$\Rightarrow P=(99X+9Y)$

$\Rightarrow P=9 \times (11X+Y)$

which is divisible by 9.

So if the sum of three digit number is subtracted from the number. The resulting number is **always divisible by 9**.

ur answer also not divisible by 6 at all.

if we take no as 999 then it get divisible by both 6 and 9.