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Question Number 16053 by vpawarksp@gmail.com last updated on 17/Jun/17

number of positive integers a and b and c satisfying  a^b^c  b^c^a  c^a^b  =5abc

$${number}\:{of}\:{positive}\:{integers}\:\boldsymbol{{a}}\:\boldsymbol{{and}}\:\boldsymbol{{b}}\:\boldsymbol{{and}}\:\boldsymbol{{c}}\:\boldsymbol{{satisfying}} \\ $$$$\boldsymbol{{a}}^{\boldsymbol{{b}}^{\boldsymbol{{c}}} } \boldsymbol{{b}}^{\boldsymbol{{c}}^{\boldsymbol{{a}}} } \boldsymbol{{c}}^{\boldsymbol{{a}}^{\boldsymbol{{b}}} } =\mathrm{5}\boldsymbol{{abc}} \\ $$

Commented by prakash jain last updated on 17/Jun/17

Given a,b,c are integers  at least one of a,b,c is a multiple of  5.  Let a=5  If a=5, RHS=25bc  case 1:  ⇒b^c =2 or b=2,c=1  5^2 2^1 1^(25) =50  case 2:  if a=5 and RHS is a muliple of 25  say b=5  a^b^c   increases at much faster rate  with c than 125c.  so no solutions witha=5,b=5  −−−−−  similarly for b=5 and c=5  solutions  a=5,b=2,c=1  b=5,c=2,a=1  c=5,a=2,b=1

$$\mathrm{Given}\:{a},{b},{c}\:\mathrm{are}\:\mathrm{integers} \\ $$$$\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{of}\:{a},{b},{c}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of} \\ $$$$\mathrm{5}. \\ $$$$\mathrm{Let}\:{a}=\mathrm{5} \\ $$$$\mathrm{If}\:{a}=\mathrm{5},\:\mathrm{RHS}=\mathrm{25}{bc} \\ $$$${case}\:\mathrm{1}: \\ $$$$\Rightarrow{b}^{{c}} =\mathrm{2}\:{or}\:{b}=\mathrm{2},{c}=\mathrm{1} \\ $$$$\mathrm{5}^{\mathrm{2}} \mathrm{2}^{\mathrm{1}} \mathrm{1}^{\mathrm{25}} =\mathrm{50} \\ $$$${case}\:\mathrm{2}: \\ $$$$\mathrm{if}\:{a}=\mathrm{5}\:\mathrm{and}\:\mathrm{RHS}\:\mathrm{is}\:\mathrm{a}\:\mathrm{muliple}\:\mathrm{of}\:\mathrm{25} \\ $$$$\mathrm{say}\:{b}=\mathrm{5} \\ $$$${a}^{{b}^{{c}} } \:\mathrm{increases}\:\mathrm{at}\:\mathrm{much}\:\mathrm{faster}\:\mathrm{rate} \\ $$$$\mathrm{with}\:{c}\:\mathrm{than}\:\mathrm{125}{c}. \\ $$$${so}\:{no}\:{solutions}\:{witha}=\mathrm{5},{b}=\mathrm{5} \\ $$$$−−−−− \\ $$$${similarly}\:{for}\:{b}=\mathrm{5}\:{and}\:{c}=\mathrm{5} \\ $$$${solutions} \\ $$$${a}=\mathrm{5},{b}=\mathrm{2},{c}=\mathrm{1} \\ $$$${b}=\mathrm{5},{c}=\mathrm{2},{a}=\mathrm{1} \\ $$$${c}=\mathrm{5},{a}=\mathrm{2},{b}=\mathrm{1} \\ $$

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