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Mathematics 7 Online
OpenStudy (anonymous):

simplify i^13 i^53 i^5

OpenStudy (anonymous):

divide the exponent by 4 and take the integer remainder

OpenStudy (anonymous):

so first one is \[i^{13}=i^1=i\]

myininaya (myininaya):

\[i^p=i^{4 \cdot q+r}\] just divide p each time by 4 and the remainder will be number that you really want to look at

OpenStudy (anonymous):

second one is \[i^{53}=i^1=i\]

OpenStudy (anonymous):

i i i

OpenStudy (anonymous):

and the last one is... \[i^5=i^1=i\] the i's have it

myininaya (myininaya):

since \[i^p=i^{4 \cdot q+r}=(i^4)^qi^r=1^qi^r=1i^r=i^r\]

OpenStudy (anonymous):

ooh fancy division algoriddim!

OpenStudy (anonymous):

i i i

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