Show that there exists a positive integer $n$ such that the decimal representations of $3^n$ and $7^n$ both start with the digits $10$.
Problem
Source: Mongolia 1999 Grade 9 P6
Tags: number theory
BIBLEMATHS2015
05.05.2021 01:23
n=568 3^568 = 10112829396888047924753374725828018849670732041733382078406532751795368835633886194018580711258850594426594578884811997367862433155143665474101586569571936837621876662560878968224373935869601674808413962526575348665624380096619983277114735762894338280073902249156047441761 7^568 = 1036780279899191612409487659138867650016552658444828065406527086457850722699608648373713324736594370669631413307972068278953935967536739089834126843159833867283931042951617666016623324248459316932851220836724751941400399107654370327887843419681882597209751300447012581614306856245961519965296016404794270821269514176718134537783569528633424553812860707509780937293687364020193665186991470856988223780796936079398082806189512008824885990433106891869513205812053149318296561983700801 god is great - tony BARKER
jasperE3
05.05.2021 01:35
How to do this without a calculator? Also rare triple post glitch:
BIBLEMATHS2015
05.05.2021 02:10
since log(7/3,10) irrational, there exists integer k s.t. (7/3)^k is "arbitrarily close" to 10^l for some l. Thus: 3^k and 7^k start with same digits
Let x=3^(k/N), N very large s.t. x is very close to 1. There must exist M s.t. |x^M-10| is approx. 0. (if not close enough to 10, make N larger) Then 3^(kM) prefix is "10", and 7^(kM)/3^(kM) is close to 10^(lM), so 7^(kM) prefix is also "10"...
therefore n=kM
it can be more rigorous -tony BARKER God is great