General aptitude

Question

What would be the smallest natural number which when divided either by 20 or by 42 or

by 76 leaves a remainder of 7 in each case?

(A) 3047

(B) 6047

(C) 7987

(D) 63847

Answer 1

$LCM(20,42,76)$  $=$ $7980$

What is the meaning of LCM (a,b,c) ? - It is the least common number which is divisible by all a, b and c.

$\therefore 7980 \text{ is the least number which gives remainder 0 when we divide with any of 20,42 or 76 }$

 

If we require remainder 1 in each case when divide with 20, 42 or 76, then the least number is 7980+1.

In general, we can write a number in the form:

a (least number which gives reminder 'r' in each case when divides with a, b and c)= LCM(a,b,c)+r for remainder 'r'.

Hence the number is 7980+7.

Verification:

$7987$ =$399*20+ 7$

$7987$ =$190*42+ 7$

$7987$ =$105*76+ 7$

So $C$ must be the ans

Answer 2

its just asking the smallest natural number which is divisible by all 3 given divisors and leaves a remainder 7 so just divide each option with that 3 divisors and see which option divides all three divisors and gives a remainder 7 for your information answer is C.

Comments

This approach looks easy and fast, but this depends upon the sequence of options. If answer is at option D, then we need to try all other options. On each occassion, we may need to check with all three. Instead of that, the approach mentioned in the best answer can be followed.

 

However, even I don't know the usage of LCM, still this method works.

---

yeah liitle bit time consuming $12$ times division will perform if option is $D$ :)