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Question Number 57985 by mr W last updated on 15/Apr/19
$${Find}\:{the}\:{number}\:{of}\:{integer}\:{solutions} \\ $$$${for}\:{a}×{b}×{c}×{d}=\mathrm{18900} \\ $$$${with}\:{a},{b},{c},{d}\geqslant\mathrm{1}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Apr/19
$$\mathrm{18900}=\mathrm{3}×\mathrm{3}×\mathrm{3}×\mathrm{7}×\mathrm{2}×\mathrm{2}×\mathrm{5}×\mathrm{5} \\ $$$$\mathrm{18900}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{3}} ×\mathrm{5}^{\mathrm{2}} ×\mathrm{7}^{\mathrm{1}} \\ $$$${number}\:{of}\:{divisor}\:{of}\:\mathrm{18900}\:{including}\:\mathrm{1}\:{and}\:\mathrm{18900} \\ $$$${are}\:\left(\mathrm{2}+\mathrm{1}\right)\left(\mathrm{3}+\mathrm{1}\right)\left(\mathrm{2}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{1}\right)=\mathrm{72} \\ $$$${now}\:{excluding}\:\mathrm{1}\:{and}\:\mathrm{18900}\rightarrow\mathrm{72}−\mathrm{2}=\mathrm{70} \\ $$$${for}\:{example}... \\ $$$${a}=\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$${b}=\mathrm{3}^{\mathrm{3}} =\mathrm{27} \\ $$$${c}=\mathrm{5}^{\mathrm{2}} =\mathrm{25} \\ $$$${d}=\mathrm{7}^{\mathrm{1}} =\mathrm{7} \\ $$$$\mathrm{4}×\mathrm{27}×\mathrm{25}×\mathrm{7}=\mathrm{18900} \\ $$$$ \\ $$$${now}\:{if}\:{a}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}=\mathrm{2}×\mathrm{3}^{\mathrm{3}} =\mathrm{54} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}=\mathrm{5}^{\mathrm{2}} =\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}=\mathrm{7} \\ $$$$\mathrm{2}×\mathrm{54}×\mathrm{25}×\mathrm{7}=\mathrm{18900} \\ $$$${in}\:{this}\:{way}\:{total}\:{integer}\:{solution} \\ $$$${are}=\mathrm{70} \\ $$$$\boldsymbol{{sir}}\:\boldsymbol{{pls}}\:\boldsymbol{{check}}\:\boldsymbol{{am}}\:\boldsymbol{{i}}\:\boldsymbol{{correct}}... \\ $$
Commented by mr W last updated on 15/Apr/19
$${i}\:{don}'{t}\:{have}\:{an}\:{answer}\:{yet}\:{sir}.\:{but} \\ $$$${i}\:{think}\:{there}\:{must}\:{be}\:{much}\:{more} \\ $$$${solutions}.\:{e}.{g}. \\ $$$$\mathrm{1}×\mathrm{1}×\mathrm{1}×\mathrm{18900} \\ $$$$\mathrm{1}×\mathrm{1}×\mathrm{18900}×\mathrm{1} \\ $$$$\mathrm{1}×\mathrm{18900}×\mathrm{1}×\mathrm{1} \\ $$$$\mathrm{18900}×\mathrm{1}×\mathrm{1}×\mathrm{1} \\ $$$${are}\:\mathrm{4}\:{different}\:{valid}\:{solutions}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Apr/19
$${yes}\:{sir}...{i}\:{have}\:{excluded}\:\mathrm{1}{in}\:{my}\:{solution}... \\ $$
Answered by MJS last updated on 17/Apr/19
$$\mathrm{there}\:\mathrm{are}\:\mathrm{382}\:\mathrm{solutions}\:\mathrm{with}\:{a}\leqslant{b}\leqslant{c}\leqslant{d}\:\mathrm{but} \\ $$$$\mathrm{without}\:\mathrm{this}\:\mathrm{restriction}\:\mathrm{there}\:\mathrm{are}\:\mathrm{exactly} \\ $$$$\mathrm{8000}\:\left(\mathrm{I}\:\mathrm{hope}\:\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{count}\:\mathrm{wrong}...\right) \\ $$