Question

4) A number whens divided by 36, 24 and 16, leaves the remainder 11 in each case.
Find the smallest value of this number?​

Answer 1

Answer:

hope it helps..........

Answer 2

We know that,

$The \: least \:number \:which$

$when \:divided \:by \:x,y \:and \:z$

$leaves \:the \:same \: remainder \:'r'$

$each \:case$

$= (LCM \:of \:x,y,z )+r$

*****

$\underline{\blue{ Finding \:LCM \: of \: 36,24 \:and \:16 }}$

2| 36,24,16

__________

2| 18,12,8

__________

2| 9,6,4

__________

3| 9,3,2

__________

** 3 , 1, 2

$LCM \:of \: 36,24\:and \:16$

$= 2\times2\times 2 \times 3 \times 3 \times 2$

$= 144$

$The \: smallest \: number$

$whens\: divided \:by\: 36, 24 and 16,$

$leaves\: the\: remainder\: 11\: in\: each\: case$

$= (LCM\:of\:36,24,16) + remainder$

$= 144 + 11$

$= 155$

Therefore ,

$\red{ Required \:number }\green { = 155}$

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