Question

The smallest number which when diminished by 4 is divisible by 18,27,36 and 45 is..
A) 112.B)544C)184D)364​

Answer 1

The smallest number which when diminished by 4 is divisible by 18 , 27 , 36 and 45 is 544

Given :

The numbers 18 , 27 , 36 and 45

To find :

The smallest number which when diminished by 4 is divisible by 18 , 27 , 36 and 45 is

A) 112

B) 544

C) 184

D) 364

Solution :

Step 1 of 3 :

Write down the given numbers

The given numbers are 18 , 27 , 36 and 45

Step 2 of 3 :

Find LCM of 18 , 27 , 36 and 45

We prime factorise the given numbers

18 = 2 × 3 × 3

27 = 3 × 3 × 3

36 = 2 × 2 × 3 × 3

45 = 3 × 3 × 5

LCM of 18 , 27 , 36 and 45

= 2 × 2 × 3 × 3 × 3 × 5

= 540

Step 3 of 3 :

Find the required number

Since the required number is the smallest number which when diminished by 4 is divisible by 18 , 27 , 36 and 45

So the required number is obtained by adding 4 with the LCM of the given numbers

So the required number

= 540 + 4

= 544

Hence the correct option is B) 544

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Answer 2

find smallest number when diminished by four is divisible by the given numbers

Explanation:  

1. let the required smallest number be x
2. given numbers are $18,\ 27,\ 36,\ and\ 45$  then we have, $(x-4)$ divisible by all of them
3. now let, $x=112\ \ \ \ \ ->x-4=108$                                                                                  $=>108=3x3x3x2x2\\ =>\ \ \ \ \ \ =18(3x2)\\=>\ \ \ \ \ \ =27(2x2)\\=>\ \ \ \ \ \ =36(3)$      -----> (A) not divisible by $45$      
4. now, let $x=544\ \ \ \ \ ->x-4=540$                                                                                     $=>540=5x3x3x3x2x2\\ =>\ \ \ \ \ \ =18(5x3x2)\\=>\ \ \ \ \ \ =27(5x2x2)\\=>\ \ \ \ \ \ =36(5x3)\\=>\ \ \ \ \ \ =45(3x2x2)$     ------> (B) divisible by all four numbers
5. now let $x=184\ \ \ \ \->x-4=180$                                                                            $=>180=5x3x3x2x2\\ =>\ \ \ \ \ \ =18(5x2)\\=>\ \ \ \ \ \ =36(5)\\=>\ \ \ \ \ \ =45(2x2)$         ------> (C) not divisible by $27$  
6. now let $x=364\ \ \ \ \ \->x-4=360$                                                                                           $=>360=5x3x3x2x2x2\\ =>\ \ \ \ \ \ =18(5x2x2)\\=>\ \ \ \ \ \ =36(5x2)\\=>\ \ \ \ \ \ =45(2x2x2)$  ----> (D) not divisible by $27$  
7. hence, the required number is (B)$544$