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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?

1. $3047$
2. $6047$
3. $7987$
4. $63847$

Simply use the virtual calculator provided.

I tried using virtual calculator, it says or not and so any number giving 7 as remainder would do right divided either by 20 or by 42
or by 76 leaves a remainder of 7 in each case?

3047 mod20=7 such type of mistakes I can do in the exam so please tell me how to interpret these type of questions.

Let $n$ be the smallest number which is divisible by $x, y$ and $z$ and leaves remainder $r$ in each case.

So$, n = LCM ( x , y , z ) + r$

According to Question

$x = 20 , y = 42 , z = 76$  and $r = 7$

Smallest number, $n = LCM (20, 42, 76 ) + 7$

$\implies n = 7980 + 7 = 7987$

So, option (C) is the right choice

Case 3: The smallest number which is divided by x,y,z and leaves remainder a,b,c respectively ⟹LCM(x,y,z)−k, Where k=(xa)=(yb)=(zc)

Not able to understand.Can anybody give any example??

Find the least number which when divided by $36, 48,$ and $64$ leaves the remaining $25, 37,$ and $53,$ respectively?

Since $(36 – 25) = (48 – 37) = (64 – 53) = 11,$ therefore the required smallest number (LCM of $36, 48$ and 6$4) –11 = 576 – 11 = 565.$

Got it thanks @LakshmanPatelRJIT

LCM$(20,42,76)$ = $7980$

Remainder is given $7$

NUMBER=$7980$+$7$=$7987$

Option - C $7987$

LCM(20 , 42 , 76 ) + 7

= 7980 + 7

= 7987

option C is correct

$x=20a+7$

$x=42b+7$

$x=76c+7$

$LCM(20,42,76)=7980$

$x-7=7980$

$x=7987$

Correct Answer$: C$