But \(15.2^{2n}\) for odd numbers n will be congruent to 60 modulo 100 (that's not hard to see), so we have proven in inductively.
–
Bogdan Simeonov
·
3 years, 2 months ago

@Finn Hulse
–
I believe it should be pronounced like this: pronounce log, but with a B instead of an l, then say the name Dan :D
–
Bogdan Simeonov
·
3 years, 2 months ago

## Comments

Sort by:

TopNewestHint:Induction – Bogdan Simeonov · 3 years, 2 months ago

Log in to reply

– Finn Hulse · 3 years, 2 months ago

Okay... Can you explain?Log in to reply

Multiply by 16^2 and add \(15.2^{2n+4}\) and you will get the next such number.So the new number will be congruent to

\(256.28+15.2^{2n+4}\equiv 68+15.2^{2n+4}(mod100)\)

But \(15.2^{2n}\) for odd numbers n will be congruent to 60 modulo 100 (that's not hard to see), so we have proven in inductively. – Bogdan Simeonov · 3 years, 2 months ago

Log in to reply

– Finn Hulse · 3 years, 2 months ago

Oh, I guess so! Great job Bogdan. :DLog in to reply

– Bogdan Simeonov · 3 years, 2 months ago

You probably have no idea how to pronounce my name :DLog in to reply

– Finn Hulse · 3 years, 2 months ago

Probably. How? :DLog in to reply

– Bogdan Simeonov · 3 years, 2 months ago

Try using the speech feature on it.It sounds stupid :DLog in to reply

– Finn Hulse · 3 years, 2 months ago

HAHA yeah. But how is your name actually pronounced?Log in to reply

– Bogdan Simeonov · 3 years, 2 months ago

I believe it should be pronounced like this: pronounce log, but with a B instead of an l, then say the name Dan :DLog in to reply

– Finn Hulse · 3 years, 2 months ago

Oh, that's what I thought. :DLog in to reply