Ugh. This is difficult.

Question

anonymous · Answer

http://data.artofproblemsolving.com/aops20/latex/images/aae92a678b59848a0eb062327f398b7149810abc.png

anonymous · Answer

Can anyone help?

abb0t · Answer

That really is tough. Lol

anonymous · Answer

I know right? I've been working on it for some time already. sigh.

jdoe0001 · Answer

I don't get it either way
it mentions one integer N
then goes on to say 3 numbers then
then the 3 numbers are divisible by 4 amounts

anonymous · Answer

im assuming they mean that one of the numbers is divisible by 2 of the amounts.

anonymous · Answer

the three numbers are n, n+1, and n+2

anonymous · Answer

how about 48 , 49 and 50 ? i mean n = 48

anonymous · Answer

:)

anonymous · Answer

simple right ! :)

anonymous · Answer

@orple8

anonymous · Answer

wait. inputing the answer into the computer...

anonymous · Answer

is it a test?

anonymous · Answer

hmm...it says it's wrong...

anonymous · Answer

no its a practice thing.

anonymous · Answer

@SandeepReddy None of those numbers are divisible by 9?

debbieg · Answer

@SandeepReddy which of this is divisible by 3^2? :)

debbieg · Answer

*those

anonymous · Answer

true...

anonymous · Answer

oops biggest mistake , i took just numbers not their squares lol :)

anonymous · Answer

wait

anonymous · Answer

I believe 98,99,100 work. But I can't prove they're the smallest. And that's hardly a satisfactory solution...

anonymous · Answer

i too believe the same

anonymous · Answer

yes that would be the minimum number i mean 98

anonymous · Answer

yay! it says it's right!  thank you so much erin!!!

debbieg · Answer

Any smaller set would have to include 49 and 50, and we've already shown that doesnt work....?

anonymous · Answer

hmm

anonymous · Answer

im reading the solution they give right now.

anonymous · Answer

In fact you can show it's the smallest... look around the multiples of 7^2=49. Around 49, you have 47,48,49, or 49,49,50 or 49,50,51. And clearly none of those work. Now look around 2*49=98... and 98,99,100 work.

I'd prefer a nicer solution though.

anonymous · Answer

lol they did almost the same thing we did http://data.artofproblemsolving.com/aops20/latex/images/a9acb5925f6469a46233e4fca1f988f0c88c3605.png

anonymous · Answer

Oh that's boring :P Not as nice as your last question, orple!

debbieg · Answer

"Start with guess k. If guess k doesn't work, move on to guess k+1. Continue as necessary." ;) lol

anonymous · Answer

lol i'll find some better ones eventually :)