Find-the-number-of-4-digit-numbers-so-that-when-decomposed-into-prime-factors-have-the-sum-of-prime-factors-equal-to-the-sum-of-the-exponents-

Question Number 211250 by RojaTaniya last updated on 01/Sep/24

$${Find}\:{the}\:{number}\:{of}\:\mathrm{4}\:{digit}\:{numbers} \\ $$$$\:{so}\:{that}\:{when}\:{decomposed}\:{into}\:{prime} \\ $$$$\:{factors},\:{have}\:{the}\:{sum}\:{of}\:{prime}\:{factors} \\ $$$$\:{equal}\:{to}\:{the}\:{sum}\:{of}\:{the}\:{exponents}? \\ $$

Answered by mr W last updated on 03/Sep/24

say the number is N=p^a q^b r^c ... 1000≤N≤9999 p,q,r,... ∈ P and p<q<r<... a,b,c,... ∈ N p+q+r+...=a+b+c+... (p^a q^b r^c )_(min) =p^(p+q+r−2) q^1 r^1 (p^a q^b r^c )_(max) =p^1 q^1 r^(p+q+r−2) no solution if p^(p+q+r−2) q^1 r^1 >9999 or p^1 q^1 r^(p+q+r−2) <1000 case 1: N=p^a only one solution: p=a=5 since 3^3 =27<1000, 5^5 =3125, 7^7 =823543>9999 case 2: N=p^a q^b 2^1 3^4 =162<1000 ⇒no solution 2^6 5^1 =320 2^3 5^4 =5000, 2^4 5^3 =2000 2^8 7^1 =1792 2^7 7^2 =6272 2^(12) 11^1 =45056>9999 ⇒no solution 3^7 5^1 =10935>9999 ⇒no solution case 3: N=p^a q^b r^c 2^8 3^1 5^1 =3840 2^7 3^2 5^1 =5760, 2^7 3^1 5^2 =9600, 2^6 3^3 5^1 =8640 2^(10) 3^1 7^1 =21504>9999 ⇒no solution case 4: N=p^a q^b r^c s^d 2^(14) 3^1 5^1 7^1 =1720320>9999 ⇒no solutio summary: there are 9 such 4−digit numbers: 1792, 2000, 3125, 3840, 5000, 5760, 6272, 8640, 9600

$${say}\:{the}\:{number}\:{is}\:\boldsymbol{{N}}=\boldsymbol{{p}}^{\boldsymbol{{a}}} \boldsymbol{{q}}^{\boldsymbol{{b}}} \boldsymbol{{r}}^{\boldsymbol{{c}}} … \\ $$$$\mathrm{1000}\leqslant{N}\leqslant\mathrm{9999} \\ $$$${p},{q},{r},…\:\in\:\mathbb{P}\:{and}\:{p}<{q}<{r}<… \\ $$$${a},{b},{c},…\:\in\:\mathbb{N} \\ $$$$\boldsymbol{{p}}+\boldsymbol{{q}}+\boldsymbol{{r}}+…=\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}+… \\ $$$$ \\ $$$$\left(\boldsymbol{{p}}^{\boldsymbol{{a}}} \boldsymbol{{q}}^{\boldsymbol{{b}}} \boldsymbol{{r}}^{\boldsymbol{{c}}} \right)_{{min}} =\boldsymbol{{p}}^{{p}+{q}+{r}−\mathrm{2}} \boldsymbol{{q}}^{\mathrm{1}} \boldsymbol{{r}}^{\mathrm{1}} \\ $$$$\left(\boldsymbol{{p}}^{\boldsymbol{{a}}} \boldsymbol{{q}}^{\boldsymbol{{b}}} \boldsymbol{{r}}^{\boldsymbol{{c}}} \right)_{{max}} =\boldsymbol{{p}}^{\mathrm{1}} \boldsymbol{{q}}^{\mathrm{1}} \boldsymbol{{r}}^{{p}+{q}+{r}−\mathrm{2}} \\ $$$${no}\:{solution}\:{if}\: \\ $$$$\boldsymbol{{p}}^{{p}+{q}+{r}−\mathrm{2}} \boldsymbol{{q}}^{\mathrm{1}} \boldsymbol{{r}}^{\mathrm{1}} >\mathrm{9999}\:{or} \\ $$$$\boldsymbol{{p}}^{\mathrm{1}} \boldsymbol{{q}}^{\mathrm{1}} \boldsymbol{{r}}^{{p}+{q}+{r}−\mathrm{2}} <\mathrm{1000} \\ $$$$ \\ $$$$\underline{{case}\:\mathrm{1}:\:{N}={p}^{{a}} } \\ $$$${only}\:{one}\:{solution}:\:{p}={a}=\mathrm{5} \\ $$$${since}\:\mathrm{3}^{\mathrm{3}} =\mathrm{27}<\mathrm{1000},\:\mathrm{5}^{\mathrm{5}} =\mathrm{3125},\:\mathrm{7}^{\mathrm{7}} =\mathrm{823543}>\mathrm{9999} \\ $$$$\underline{{case}\:\mathrm{2}:\:{N}={p}^{{a}} {q}^{{b}} } \\ $$$$\mathrm{2}^{\mathrm{1}} \mathrm{3}^{\mathrm{4}} =\mathrm{162}<\mathrm{1000}\:\Rightarrow{no}\:{solution} \\ $$$$\mathrm{2}^{\mathrm{6}} \mathrm{5}^{\mathrm{1}} =\mathrm{320} \\ $$$$\:\:\:\:\mathrm{2}^{\mathrm{3}} \mathrm{5}^{\mathrm{4}} =\mathrm{5000},\:\mathrm{2}^{\mathrm{4}} \mathrm{5}^{\mathrm{3}} =\mathrm{2000} \\ $$$$\mathrm{2}^{\mathrm{8}} \mathrm{7}^{\mathrm{1}} =\mathrm{1792} \\ $$$$\:\:\:\:\mathrm{2}^{\mathrm{7}} \mathrm{7}^{\mathrm{2}} =\mathrm{6272} \\ $$$$\mathrm{2}^{\mathrm{12}} \mathrm{11}^{\mathrm{1}} =\mathrm{45056}>\mathrm{9999}\:\Rightarrow{no}\:{solution} \\ $$$$\mathrm{3}^{\mathrm{7}} \mathrm{5}^{\mathrm{1}} =\mathrm{10935}>\mathrm{9999}\:\Rightarrow{no}\:{solution} \\ $$$$\underline{{case}\:\mathrm{3}:\:{N}={p}^{{a}} {q}^{{b}} {r}^{{c}} } \\ $$$$\mathrm{2}^{\mathrm{8}} \mathrm{3}^{\mathrm{1}} \mathrm{5}^{\mathrm{1}} =\mathrm{3840} \\ $$$$\:\:\:\mathrm{2}^{\mathrm{7}} \mathrm{3}^{\mathrm{2}} \mathrm{5}^{\mathrm{1}} =\mathrm{5760},\:\mathrm{2}^{\mathrm{7}} \mathrm{3}^{\mathrm{1}} \mathrm{5}^{\mathrm{2}} =\mathrm{9600},\:\mathrm{2}^{\mathrm{6}} \mathrm{3}^{\mathrm{3}} \mathrm{5}^{\mathrm{1}} =\mathrm{8640} \\ $$$$\mathrm{2}^{\mathrm{10}} \mathrm{3}^{\mathrm{1}} \mathrm{7}^{\mathrm{1}} =\mathrm{21504}>\mathrm{9999}\:\Rightarrow{no}\:{solution} \\ $$$$\underline{{case}\:\mathrm{4}:\:{N}={p}^{{a}} {q}^{{b}} {r}^{{c}} {s}^{{d}} } \\ $$$$\mathrm{2}^{\mathrm{14}} \mathrm{3}^{\mathrm{1}} \mathrm{5}^{\mathrm{1}} \mathrm{7}^{\mathrm{1}} =\mathrm{1720320}>\mathrm{9999}\:\Rightarrow{no}\:{solutio} \\ $$$$ \\ $$$${summary}: \\ $$$${there}\:{are}\:\mathrm{9}\:{such}\:\mathrm{4}−{digit}\:{numbers}: \\ $$$$\mathrm{1792},\:\mathrm{2000},\:\mathrm{3125},\:\mathrm{3840},\:\mathrm{5000},\: \\ $$$$\mathrm{5760},\:\mathrm{6272},\:\mathrm{8640},\:\mathrm{9600} \\ $$

Commented by Rasheed.Sindhi last updated on 03/Sep/24