if p = (x!)^(y!)^(z!) & p is a single digit number, then find the possible value of p

(Q). If p = ${\left( {x!} \right)^{{{\left( {y!} \right)}^{\left( {z!} \right)}}}}$ & p is a single digit number then find the possible value of p.
Solution:-

➠ if x = 0 or 1 ⇒ x! = 1 then y & z can take any value & value of p will be = 1.

➠ if x = 2 ⇒ x! = 2
if y = 0 or 1 ⇒ y! = 1 then z can take any value & value of p will be 2.
if y = 2 ⇒ y! = 2 then z may take value 0 or 1 then p = 4

➠ if x = 3 ⇒ x! = 6 then y can be 0 or 1 & z can take any value & value of p will be = 6

Hence Possible values of p = 1,2,4,6 Answer

if p = (x!)^(y!)^(z!) & p is a single digit number, then find the possible value of p

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