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restul impartirii nr natural 1+2+...+100 la 15 este egal cu:2 4 6 8 10
valoarea nr 2 la puterea 1x 2 la puterea a 2-a...2 la puterea a 10-a


Răspuns :

S=1+2+....+100
S=100*101:2
S=50*101
S=5050
5050:15=336 rest 10

2^1 *2^2 *......*2^10=
=2^(1+2+....+10)=
=2^(10*11:2)=
=2^(5*11)=2^55
1+2+3+++100=
100x100:2=5050
5050:15=336rest 10



2¹x2².....x2¹°
=2⁽₁₊₂₊₃₊₊₊₁₀⁾
=2⁽¹°ˣ¹¹:²⁾
=2⁽ˣ¹¹⁾=2⁵⁵