1. What is the remainder when \(1^3+2^3+3^3+...+100^3 \) is divided by \(6\)?

2. When a positive integer is expressed in base 7, it is \(AB_7\), and when it is expressed in base 5, it is \(B A_5 \).What is the positive integer in decimal?

yasbib555
Sep 15, 2018

#1**0 **

1)

∑[n^3, n, 1, 100] =25,502,500

**25,502,500 mod 6 =4**

**You can also use this formula to sum up your sequence: (n(n+1)/2)^2 =(100*(100+1)/2)^2 =25,502,500**

**2) 17 in base 10 =23 in base 7 =32 in base 5.**

Guest Sep 15, 2018

edited by
Guest
Sep 15, 2018

edited by Guest Sep 15, 2018

edited by Guest Sep 15, 2018

#2**+1 **

the above answer sort of misses the point I think.

We can notice that \(k^3 \pmod{6} = k,~k=1,2,\dots 5 \\ \\ \text{and obviously }6^3 \pmod{6} = 0\)

so we have a repeating sequence in the sum \( 1+2+3+4+5+0 + 1+2+\dots\)

there are 16 of these groups of 6 and then 4 additional elements of the sum

so we have

\(\sum \limits_{k=1}^{100}~k^3 \pmod{6} = 3\cdot 16 + 1+2+3+4 = 58 \\ \text{and} \\ \\ 58 \pmod 6 = 4 \\ \\ \text{so }\sum \limits_{k=1}^{100}~k^3 \pmod{6} = 4\)

Rom
Sep 15, 2018