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by samirpandeyit62 » Thu Sep 06, 2007 9:22 am
7 ^ (4n+3) * 6^ n

the number 7 has a property that when it is raised to a power,

The last digit of the product will repeat after every 4 continous powers

i.e 7 ^ 1 =7
7 ^ 2 =49
7 ^ 3 = 343
7 ^ 4 = ...1
7 ^ 5 =....7 (repetition starts here)

Hence as we put values of n in 7^ (4n+3) like 1,2...

so n =1 will give 7 ^ 7 whose last digit would be same as 7^ 3 (7-3 =4)i.e 3 hence for all +ve values of n 7^ (4n+3) will have last digit as 3

similary 6 ^n will always give last digit 6 as 6 X 6 =36, 36 X 6 =..6 so on

hence upon multiplying 7^ (4n+3) with 6 ^ n

we are multiplying their last digits i.e 3 X 6 =18

hence the product will always end in 8

upon division with 10 it will always give a remainder of 8

hence ans should be E (J here) "8"
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by ri2007 » Thu Sep 06, 2007 10:03 am
Hi samirpandeyit62

that was a great explaination - thanks for posting the answer.

another request - how did u think of this solution? do you know of any good material I could review to learn the properties of the numbers or did u just figure the property of no 7 by solving it on you own?
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by samirpandeyit62 » Thu Sep 06, 2007 10:34 am
Thanks ri2007,
If this sort of a question u would get only one answer if there is a pattern, here 7 was raised to 4n+3, so u can make educated guess why n is made a multiple of 4 i.e 4n, a rational answer would be a pattern, so just try to solve upto 4+1 i.e. 5 powers of 7 considering only the last digit and see if ur correct, in most situations this has to work

as far as 6 is concerned it definately will have 6 as last digit for all powers.
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by ri2007 » Thu Sep 06, 2007 12:46 pm
Thanks a lot samirpandeyit62. That was just the kind of explaination I was hoping to get.
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