pnyy gur obkrf 1 gb 5 gb fgneg jvgu sbe pbairavrapr.abj cynpr bar sebz obk bar, gjb sebz obk 2 naq 3 sebz obk 3 ba gur fpnyrf. gura sbe gur frpbaq gvzr 3 sebz obk 3, 4 sebz obk 4 naq 5 sebz obk 5gur gbgny jrvtug jvyy gura or 1k jrvtug bs bar onyy sebz obk 1, 2k jrvtug bs obk 2 rgp... tvivat lbh na rdhngvba. guvf tvirf 2 rdhngvbaf.yrg'f fnl obk 2 jnf urnil naq obk 3 jnf yvtug gur rdhngvbaf jbhyq ybbx yvxr guvf:1ko1 + 2ko2 + 3ko3= 1 + 2.2 + 2.7 = 5.9 (0.1 bhapr yvtugre guna vs nyy onyyf jrer 1 bhapr, bs pbhefr gur zvqqyr rdhngvba vf abg xabja whfg gur svefg naq ynfg)3k o3 + 4ko4 +5ko5 =2.7 4 +5 = 11.7 qbvat vg guvf jnl rafher gurer vf nyjnlf ng yrnfg bar qvfpercnapl sebz jung gur fpnyrf fubhyq fnl vs nyy onyyf jrvturq 1 bhapr.vg pna or frra gur svefg gvzr gurer vf bar zber 0.9 bhapr onyy guna 1.1 nf vg vf haqre jrvtug. be o1 vf 0.9 naq gur erfg ner abezny jvgu o4 naq o5 orvat gur qvssrerag barf. sebz gur frpbaq gurer ner 3 zber 0.9 gung 1.1. nf vg vf haqre jrvtug.fb vg vf frra gur bayl jnl sbe 3 zber 0.9f vf vs o3 vf gur yvtug jrvtug obk.xabjvat va gur svefg rdhngvba gurer vf bar zber 0.9 onyy naq gurer ner 3 bs gurz gurer zhfg or 2 1.1 onyyf naq gurer vf bayl bar onyy glcr jvgu 2, o2.gurer ner znal cbffvovyvgvrf bs pbhefr ohg gur zrgubq jbexf sbe nyy. jevgvat na rknhfgvir yvfg jbhyq or irel grqvbhf, v ubcr v qba'g unir gb.
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So I think @uber-dragon got there before me with a different method but fun problem never the less. I will have to be super quick next time!