Question

Determinați ultima cifră a numerului b=7+7²+7³+...+7⁴³
c=3^0+3¹+3²+3³+...+3⁴²
d=8+8²+8³+...+8^88​

Answer 1

Hei! :)

$b=7+7^{2} +7^{3} +....+7^{43}= 7*(1+7+7^{2} +7^{3}+.....+7^{42})\\b=7*[ (1 +7 +7^{2} +7^{3})+7^{4}(1 +7 +7^{2} +7^{3})+...+7^{36}*(1 +7 +7^{2} +7^{3})+7^{40} (1 +7 +7^{2} +7^{3})]\\b=7*400*(1+7^{4} +..+7^{36} )+7*7^{40} *57\\=> uc(b)=uc(0+19)=9$

$c=3^{0} +3^{1} +3^{2} +3^{3} +....+3^{42} \\c=(1 + 3^{2} ) + 3(1 + 3^{2} ) + .....+ 3^{39}*(1 + 3^{2} )+ 3^{42}\\c=10(1 + 3 + ........+3^{39}) +3^{42}\\uc(c)=uc(3^{42})=9$

$d=8+8^{2} +8^{3} +.....+8^{88} \\d=(8+8^{2} +8^{3}+8^{4} )+ 8^{4} (8+8^{2} +8^{3}+8^{4} )+...+8^{84} (8+8^{2} +8^{3}+8^{4})\\d= 4680(1 +8^{4}+ 8^{84}) \\=> uc(d)=0$

• unde uc- ultima cifra a numarului respectiv