# find( without using calculators or computer programs) the last two decimal digits of : (2007)^2007?

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• Anonymous

To find the last two digits of 2007^2007,

we will have to determine the pattern with the last digits of 2007^n......

2007^1 = 2007

Last 2 Digits Of 2007^1 --> 07

2007^2 = 4 028 049

Last 2 Digits Of 2007^2 --> 49

2007^3 = 8 084 294 343

Last 2 Digits Of 2007^3 --> 43

2007^4 = 16 225 178 746 401

Last 2 Digits Of 2007^4 --> 01

2007^5 = 32 563 933 744 026 807

Last 2 Digits Of 2007^5 --> 07

2007^6 = 65 355 815 024 261 801 649

Last 2 Digits Of 2007^6 --> 49

2007^7 = 131 169 120 753 693 435 909 543

Last 2 Digits Of 2007^7 --> 43

2007^8 = 263 256 425 352 662 725 870 452 801

Last 2 Digits Of 2007^8 --> 01

By observation,

when n is a multiple of 4, the last digits are 01.

when (n + 1) is a multiple of 4, the last digits are 43.

when (n + 2) is a multiple of 4, the last digits are 49.

when (n + 3) is a multiple of 4, the last digits are 07.

Therefore, the last 2 digits of 2007^2007 is 43 since

(2007 + 1) is divisible by 4.

Use cyclicity man its repeats at

7^1 units = 07

7^2 units = 49

7^3 units = 43

7^4 = 01

7^5 = 07

7^6 = 49

7^7 = 43

cyclicity =04

and since the 2007 ^2007 has zeroes the units and tens places shall be 43(343)