Question

if (a-7)(b-10)(c-12)=1000 and a,b,c belongs to N,then the least possible value of (a+b+c)is?

Answer 1

Sorry next time I will better perform______$$$$_______$$________&_____......

Answer 2

The least value is 59.

Step-by-step explanation:

Since we have given that

$(a-7)(b-10)(c-12)=1000$

We need to find the least value of (a+b+c).

so, it becomes,

$(a-7)(b-10)(c-12)=1000\\\\(a-7)(b-10)(c-12)=10^3\\\\(a-7)(b-10)(c-12)=10\times 10\times 10$

Now, comparing the each term on left hand side with corresponding term on right hand side.

So, it becomes

$a-7=10\\\\a=10+7\\\\a=17\\\\and\\\\b-10=10\\\\b=10+10\\\\b=20\\\\and\\\\c-12=10\\\\c=10+12\\\\c=22$

So, the least value would be

$a+b+c\\\\=17+20+22\\\\=59$

Hence, the least value is 59.

# learn more:

If (x - 5)(y + 6 )(z - 8)= 1331 then find the minimum value of x+y+z​

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