Question

Find the smallest number which when divided by 12,15,18&27 leaves as remainder 8,11,14,23 respectively.

Answer 1

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{536}}}}}$

Step-by-step explanation:

Solution :

Here, 12 - 8 = 4

15 - 11 = 4

18 - 14 = 4

27 - 23 = 4

Thus, every divisor is greater than its remainder by 4. So, the required smallest number is the difference of the L.C.M of the given number and 4

Finding the L.C.M

First of find the prime factors of each numbers

12 = 2 × 2 × 3

15 = 3 × 5

18 = 3 × 3 × 3

27 = 3 × 3 × 3

Take out the common prime factors : 3 , 3 and 3

Also take out the other remaining prime factors : 2 , 2 and 5

Now, Multiply those all prime factors and obtain L.C.M

L.C.M = Common factors × Remaining factors

= 3 × 3 × 3 × 2 × 2 × 5

= 540

L.C.M of 12 , 15 , 18 and 27 = 540

So, The required smallest number = 540 - 4

= 536

Hope I helped!

Best regards!!