Menu Close

Let-p-P-and-m-Z-Find-p-m-such-that-p-m-1-p-1-146410-




Question Number 4199 by Yozzii last updated on 01/Jan/16
Let p∈P and m∈Z^+ .  Find (p,m) such that p^(m−1) (p−1)=146410.
$${Let}\:{p}\in\mathbb{P}\:{and}\:{m}\in\mathbb{Z}^{+} . \\ $$$${Find}\:\left({p},{m}\right)\:{such}\:{that}\:{p}^{{m}−\mathrm{1}} \left({p}−\mathrm{1}\right)=\mathrm{146410}. \\ $$
Answered by Rasheed Soomro last updated on 01/Jan/16
p^(m−1) (p−1)=146410  146410=2×5×11^4    p∈P⇒p=2 ∣ p  is odd  For p=2,  146410 should be of 2^n  type but it  isn′t.  ∴ p is odd  ∴ p−1 is even  even factors of 146410 are 2,10,22  p−1=2,10,22 ⇒ p=3,11,23  ∵ 3 and 23 are not factors  ∴ p=11  11^(m−1) (11−1)=146410  11^(m−1) =11^4   m−1=4  m=5  (p,m)=(11,5)
$${p}^{{m}−\mathrm{1}} \left({p}−\mathrm{1}\right)=\mathrm{146410} \\ $$$$\mathrm{146410}=\mathrm{2}×\mathrm{5}×\mathrm{11}^{\mathrm{4}} \\ $$$$\:{p}\in\mathbb{P}\Rightarrow{p}=\mathrm{2}\:\mid\:{p}\:\:{is}\:{odd} \\ $$$${For}\:{p}=\mathrm{2},\:\:\mathrm{146410}\:{should}\:{be}\:{of}\:\mathrm{2}^{{n}} \:{type}\:{but}\:{it} \\ $$$${isn}'{t}. \\ $$$$\therefore\:{p}\:{is}\:{odd} \\ $$$$\therefore\:{p}−\mathrm{1}\:{is}\:{even} \\ $$$${even}\:{factors}\:{of}\:\mathrm{146410}\:{are}\:\mathrm{2},\mathrm{10},\mathrm{22} \\ $$$${p}−\mathrm{1}=\mathrm{2},\mathrm{10},\mathrm{22}\:\Rightarrow\:{p}=\mathrm{3},\mathrm{11},\mathrm{23} \\ $$$$\because\:\mathrm{3}\:{and}\:\mathrm{23}\:{are}\:{not}\:{factors} \\ $$$$\therefore\:{p}=\mathrm{11} \\ $$$$\mathrm{11}^{{m}−\mathrm{1}} \left(\mathrm{11}−\mathrm{1}\right)=\mathrm{146410} \\ $$$$\mathrm{11}^{{m}−\mathrm{1}} =\mathrm{11}^{\mathrm{4}} \\ $$$${m}−\mathrm{1}=\mathrm{4} \\ $$$${m}=\mathrm{5} \\ $$$$\left({p},{m}\right)=\left(\mathrm{11},\mathrm{5}\right) \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *