费费第七部分第29题

tiger1234

2^(8n + 3) => 2^8n * 2^3 => 2^(4*2n) * 2^3, if the cycle repeats every 4 numbers, then 2^(4*2n) would have the same last digit as 2^4(which is 6), and 2^(4*2n) * 2^3 would have the 3rd number down the pattern since it's 2^3, which is 8;

tiger1234

so since u get the point 2^8n would have the same last digit as 2^4(which is 6), then 2^4 * 2 woule have 2 as the last digit, => 2^4 * 2^2 would have 4 => 2^4 * 2^3 would have 8, since 2^8n has the same last digit, then 2^8n * 2^3 would have 8 as the last digit as well

better?

tiger1234

we know the last digit goes in a cyle like: 2, 4, 8, 6, 2, 4, 8, 6... so on and so forth, so 2^1 => 2, 2^2 => 4, 2^3 = 8, 2^4 => 6, 2^5 => 2, 2^6 => 4, 2^7 => 8, 2^8 => 6...

so for every 2^4 we will get 6 as the last digit, and since 2^8n => 2^(4 * 2n) => 2^4 * 2^2n => we have 2^2n many of 2^4, and since every 2^4 has the same last digit 6, the product of 2^2n many of 2^4 would have 6 as its as digit as well;

now we know 2^3 = 8, and the product of 2^3 and 2^8n would have last digit 8 (8 * 6 = 48)

if it takes me three tries to explain, it's probably my bad, i hope this helps

[此贴子已经被作者于2004-8-19 8:38:44编辑过]