Question Number 6788 by Tawakalitu. last updated on 26/Jul/16
Commented by Yozzii last updated on 27/Jul/16
$$\mathrm{60}=\mathrm{6}×\mathrm{10}=\left(\mathrm{3}×\mathrm{2}\right)\left(\mathrm{7}+\mathrm{3}\right) \\ $$$$\mathrm{17},\mathrm{23},\mathrm{32} \\ $$$${Add}\:{the}\:{two}\:{largest}\:{unit}\:{digit}\:{values}\:{among} \\ $$$${the}\:{three}\:{numbers}\:\left(\mathrm{3}\:\&\:\mathrm{7}\right)\:{and}\:{multiply}\:{this} \\ $$$${answer}\:{with}\:{the}\:{two}\:{largest}\:{digit}\:{values}\:{in} \\ $$$${tens}\:{columns}\:\left(\mathrm{2\&3}\right)\Rightarrow\left(\mathrm{7}+\mathrm{3}\right)×\mathrm{2}×\mathrm{3}=\mathrm{60}. \\ $$$$−−−−−−−−−−−−−−−−−−−− \\ $$$$\mathrm{13},\mathrm{23},\:?\:\rightarrow\mathrm{42} \\ $$$$\mathrm{42}=\mathrm{2}×\mathrm{3}×\mathrm{7}=\mathrm{2}×\mathrm{3}×\left(\mathrm{3}+\mathrm{4}\right) \\ $$$$\mathrm{13},\mathrm{23},\mathrm{34}\Rightarrow\mathrm{2}×\mathrm{3}×\left(\mathrm{3}+\mathrm{4}\right)=\mathrm{42} \\ $$$$?=\mathrm{34}\:\:{possibly} \\ $$
Commented by Tawakalitu. last updated on 27/Jul/16
$${I}\:{really}\:{appreciate}\:{your}\:{effort}\: \\ $$