Remainder=?

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2 Answers
Mar 7, 2018

This can be calculated in a number of ways. One way using brute force is

#27^1/7# has a remainder #=6# .....(1)
#27^2/7=729/7# has a remainder #=1# .....(2)
#27^3/7=19683/7# has a remainder #=6# …….. (3)
#27^4/7=531441/7# has a remainder #=1# ….. (4)
#27^5/7=14348907/7# has a remainder #=6# …..(5)
#27^6/7=387420489/7# has remainder #=1# …. (6)

As as per emerging pattern we observe that the remainder is #=6# for an odd exponent and the remainder is #=1# for an even exponent.

Given exponent is #999-># odd number. Hence, remainder #=6.#

Mar 7, 2018

Alternate solution

Explanation:

Given number needs to be divided by #7#. Hence it can be written as

#(27)^999#
#=>(28-1)^999#

In the expansion of this series, all terms which have various powers of #28# as multiplicants will be divisible by #7#. Only one term which is #=(-1)^999# now needs to be tested.

We see that this term #(-1)^999=-1# is not divisible by #7# and therefore, we are left with remainder #=-1.#
Since remainder can not be #=-1#, we will have to stop division process for remaining terms of expansion when the last #7# remains.

This will leave remainder as #7+(-1)=6#