101²+2A1¹⁶+21¹⁰
101 = 1∙22+0∙21+1∙20 = 4+0+1 = 5 (10)
2А1 = 2∙162+10∙161+1∙160 = 512+160+1 = 673 (10)
5 + 673 + 21 = 699
Ответ: 699
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101²+2A1¹⁶+21¹⁰
101 = 1∙22+0∙21+1∙20 = 4+0+1 = 5 (10)
2А1 = 2∙162+10∙161+1∙160 = 512+160+1 = 673 (10)
5 + 673 + 21 = 699
Ответ: 699