II gimnazjum

Matematyka wokół nas 2

215+216+217+218=215⋅1+215⋅2+215⋅22+215⋅23=215⋅(1+2+22+23)=215(1+2+4+8)=215⋅15=214⋅2⋅15=214⋅30 <section class="answer"><h4 class="answer-heading">Odpowiedź:</h4> A, D</section>

Rozwiązanie 1

Zapisujemy ostatnie cyfry kolejnych potęg 3 (starając się zauważyć regułę, według której te cyfry się pojawiają): 31−&gt;3

−2−2=(−1)⋅2−2=(−1)⋅(21​)2=−41​

920⋅2−919⋅3+918⋅5=918⋅92⋅2−918⋅9⋅3+918⋅5=918⋅(92⋅2−9⋅3+5)=918⋅(162−27+5)=918⋅140=918⋅2⋅70=918⋅4⋅35=918⋅7⋅20=918⋅5⋅28

2,5⋅106cm3=2,5⋅106⋅0,001dm3=2,5⋅103dm3=2,5⋅103l=2,5⋅103⋅100cm⋅100cm⋅100cm=2,5⋅103⋅106mm3=2,5⋅109mm3

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