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How to find out the last two digits of 14^40

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by Anurag@Gurome » Wed Dec 05, 2012 3:39 am
This problem can be easily solved by remembering the blue colored fact.

14^40 = (2*7)^40 = (2^40)*(7^40)

Now, 7^4 = 2401. Hence, 7^4 raised to any power will end in ...01.
Hence, last two digits of 7^40 = (7^4)^10 is 01.

And, 2^10 = 1024
24 raised to any odd power ends with 24 and 24 raised to any even power ends with 76. This is because 24^2 ends with 76 and 76 raised to any power always ends 76.
Hence, 2^40 = (2^10)^4 ends with 76.

Hence, last two digits of 14640 is last two digits of 76*01 = 76
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by maximiliano » Sun Jun 23, 2013 12:00 pm
Yes, I understand the process. But when I go and check in the calculator, the result is:14^40=7.0003769659107e+45 (this notation in the calculator means 7.0003769659107 + 10^45 = 7000376965910700000000000000000000000000000000)

Then, the last digit is zero? I don´t understand.
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by Atekihcan » Mon Jun 24, 2013 1:14 am
maximiliano wrote:Yes, I understand the process. But when I go and check in the calculator, the result is:14^40=7.0003769659107e+45 (this notation in the calculator means 7.0003769659107 + 10^45 = 7000376965910700000000000000000000000000000000)

Then, the last digit is zero? I don´t understand.
I guess you meant to write 'x'
Anyway, standard calculators cannot show large numbers properly as the display filed is limited to 10 or 12 or 15... digits. So calculator approximates the number and show it in short form.

For example, a calculator with 10 display digits will show the number 12345678900012 as 1234567890 x 10^4

Hope that helps.
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