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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?

Sep 15, 2018
edited by yasbib555  Sep 15, 2018

#1
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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.

Sep 15, 2018
edited by Guest  Sep 15, 2018
edited by Guest  Sep 15, 2018
#2
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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

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

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Sep 15, 2018