Question 176346
Break down each factor to its prime factors.
{{{6=2*3=2^1*3^1}}}
{{{8=2*2*2=2^3}}}
{{{12=2*2*3=2^2*3^1}}}
You need to keep the max exponent for each prime factor.
The lowest common factor would be 
{{{2^3*3^1=8*3=24}}}
We need to find the common factor greater than 99.
Multiplying by 4 would only reach 96. 
Multiply by 5.
{{{24*5=120}}}
"which is the smallest 3 digit number which is exactly divisible by 6,8, 12"
120.
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{{{8=2*2*2=2^3}}}
{{{10=2*5=2^1*5^1}}}
{{{12=2*2*3=2^2*3^1}}}
The lowest common factor would be 
{{{2^3*3^1*5^1=8*3*5=120}}}
Its also 3 digits.
"which is the samllest 3 digit number which is exactly divisible by 8,10,12"
120.
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.
.

Again starting with 24 (6,8,12), the largest multiplier than keeps the product under 1000 is 41.
"which is the greatest 3 digit number which is exactly divisible by 6,8, 12"
{{{41*24=984}}}
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.
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Again starting with 120 (8,10,12), the largest multiplier than keeps the product under 1000 is 8.
"which is the greatest 3 digit number which is exactly divisible by 6,8, 12"
{{{120*8=960}}}