Question 75098
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Simplify:

{{{2^31*4^31*5^62*7^31*8^93=y}}}

Notice that all the exponents are multiples
of 31.

Write the {{{62}}} power as {{{2*31}}}

and write the {{{93}}} power as {{{3*31}}}

{{{2^31*4^31*5^(2*31)*7^31*8^(3*31)=y}}}

Now write the {{{5^(2*31)}}} as {{{(5^2)^31}}}

and write the {{{8^(3*31)}}} as {{{(8^3)^31}}}

{{{ 2^31*4^31*(5^2)^31*7^31*(8^3)^31=y }}}

Now write the {{{5^2}}} as {{{25}}}
and write the {{{8^3}}} as {{{512}}}

{{{ 2^31*4^31*25^31*7^31*512^31=y }}}

Now since all factors are raised to the 31 power,

we can write the above as:

{{{ (2*4*25*7*512)^31=y }}}

Then finally multiply the numbers in the
parentheses and you end up with

{{{ (102400)^31=y }}}

or

{{{102400^31=y}}}

and you would just leave the answer like
that because that number written out
would have hundreds of digits.

Edwin</pre>