Question Number 518 by Yugi last updated on 25/Jan/15
$${Give}\:{the}\:{result}\:{of}\:{the}\:{following}\:{computation}\:{as}\:{an}\:{integer}\:{in}\:{the}\:{usual}\:{decimal} \\ $$$${form}.\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{303},\mathrm{000},\mathrm{000},\mathrm{000},\mathrm{303}×\mathrm{3},\mathrm{300},\mathrm{000},\mathrm{033}}{\mathrm{1},\mathrm{000},\mathrm{100},\mathrm{010},\mathrm{001}} \\ $$
Answered by prakash jain last updated on 22/Jan/15
![((303(10^(12) +1)×33×(10^8 +1))/((10^(12) +10^8 +10^4 +1))) =((303×33×(10^(12) +1)(10^8 +1))/([10^8 (10^4 +1)+(10^4 +1)])) =((303×33×[(10^4 )^3 +1][10^8 +1])/((10^8 +1)(10^4 +1))) =((303×33(10^4 +1)(10^8 −10^4 +1))/(10^4 +1)) =303×33×(10^8 −10^4 +1) =9999×(99990000+1) =9999×9999×10000+9999 =(10000−1)^2 ×10000+9999 =(100000000−20000+1)×10000+9999 =(99980000+1)×10000+9999 =999800010000+9999 =999800019999](Q523.png)
$$\frac{\mathrm{303}\left(\mathrm{10}^{\mathrm{12}} +\mathrm{1}\right)×\mathrm{33}×\left(\mathrm{10}^{\mathrm{8}} +\mathrm{1}\right)}{\left(\mathrm{10}^{\mathrm{12}} +\mathrm{10}^{\mathrm{8}} +\mathrm{10}^{\mathrm{4}} +\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{303}×\mathrm{33}×\left(\mathrm{10}^{\mathrm{12}} +\mathrm{1}\right)\left(\mathrm{10}^{\mathrm{8}} +\mathrm{1}\right)}{\left[\mathrm{10}^{\mathrm{8}} \left(\mathrm{10}^{\mathrm{4}} +\mathrm{1}\right)+\left(\mathrm{10}^{\mathrm{4}} +\mathrm{1}\right)\right]} \\ $$$$=\frac{\mathrm{303}×\mathrm{33}×\left[\left(\mathrm{10}^{\mathrm{4}} \right)^{\mathrm{3}} +\mathrm{1}\right]\left[\mathrm{10}^{\mathrm{8}} +\mathrm{1}\right]}{\left(\mathrm{10}^{\mathrm{8}} +\mathrm{1}\right)\left(\mathrm{10}^{\mathrm{4}} +\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{303}×\mathrm{33}\left(\mathrm{10}^{\mathrm{4}} +\mathrm{1}\right)\left(\mathrm{10}^{\mathrm{8}} −\mathrm{10}^{\mathrm{4}} +\mathrm{1}\right)}{\mathrm{10}^{\mathrm{4}} +\mathrm{1}} \\ $$$$=\mathrm{303}×\mathrm{33}×\left(\mathrm{10}^{\mathrm{8}} −\mathrm{10}^{\mathrm{4}} +\mathrm{1}\right) \\ $$$$=\mathrm{9999}×\left(\mathrm{99990000}+\mathrm{1}\right) \\ $$$$=\mathrm{9999}×\mathrm{9999}×\mathrm{10000}+\mathrm{9999} \\ $$$$=\left(\mathrm{10000}−\mathrm{1}\right)^{\mathrm{2}} ×\mathrm{10000}+\mathrm{9999} \\ $$$$=\left(\mathrm{100000000}−\mathrm{20000}+\mathrm{1}\right)×\mathrm{10000}+\mathrm{9999} \\ $$$$=\left(\mathrm{99980000}+\mathrm{1}\right)×\mathrm{10000}+\mathrm{9999} \\ $$$$=\mathrm{999800010000}+\mathrm{9999} \\ $$$$=\mathrm{999800019999} \\ $$