Tiny Math Challenge

7 & 11&5&3 are also factors of 1155 ... //0,2,4,8,6 comes as last digit 2^n , on substracting 1... we get 9,1,3,7,5 as last digit , still some possibility of getting divisible bcz of 5 as last digit also possible & now what ? //there are numbers possible which have last digit as 5 & div. by 7,11&3 ... //i am also confused 😅

take log ☺👍 use log2=.3010 & log as .4771 log y = 2^2007+ 3^2008 -2009 . log2 //will see 1 later y=10^((2^2007+3^2008-2009).log2) =10^(10^(2007.log2)+10^(2008.log3)-2009) . log2 then check whether y-1 is div by 1155 or not

It's much better, Gaurav👍👍👍 continue, please, extend and COMPLETE your answer

Without logarithms...please

Look at the last Gaurav's answer, Jay. He began correctly...

Hint: Answer - yes. Why?..

@Jacob, it's not about code, it's challenge (aka, question) because I can't make pretty formating presentation of it here without html + css😎

Nope(not very correct) @Gaurav, don't calculate, please🙏 Multiplicity and divisibility will help you, dear☺

@jay, just try to effort to thinking😁 It's not so hard, believe me, mate👍

@LunarCoffee Nope(not very correct) @Gaurav, don't calculate, please🙏 Multiplicity and divisibility will help you, dear☺

@Learnsolo, no, without calculator, please!☺

Count me out, math hates me :)

It's 12:17am... I math better in the morning. I will check it out/try though :)

Can you use a calculator? 😅 I need calculator to complete 😂

I'm in Grade 8 (Canada) and have no idea how to solve this without just... calculating it.

I just found this challenge and its pretty late here I'll try it tomorrow. Even though its already answered it will be a good exercise.