Finding the last two digits of a number using modular artihmetic

Finding the lass two digits of 7^{7^{7^7}} using properties of congruences.
Example 5.5.6 in Logic and Discrete Mathematics, by W. Conradie and V. Goranko.

[@avinashdongare6410]

After searching 2 hours here and there ..I got perfect video ❤️

[@bohsu182]

Fermat's little theorem is another option to figure this out. 7^7^7^7=7^7^7mod(100)=7^7mod(100)=7^3mod(100)=43.

[@subramaniamdhakshinamurthy1449]

you can do this in a easier way in less time. as you got 7^4 = 1 mod 100. multiply by 85 (7^4)^85 = 1^85mod 100 now multiply by 7^3. 7 ^3 leaves a remainder 43 when divided by 100. so 7^3 * 7^340 = 43 * 1^85 mod 100. that is 7^343 = 43 mod 100.

[@user-gx9lr2mh2i]

thank you! great video.

[@zimaelohey1524]

Thank you!

[@2infinitybeyond655]

Nicely Explained

[@ishayadav6039]

Thanks a lot!

[@vidajohn2199]

thank you!

[@taylorhorne1046]

i swear the amount of times i try to understand this

[@sunchess5950]

Still I don't get anything . I can't get anything useful

[@mostafanaga5021]

i don't understand how (7^4)^z changed to 1

[@jacobolobo2106]

Gracias

[@wajeehakhan6985]

Will u plz solve me for this find the last two digits of 7^7^7upto so on 7 without congrence relation

[@paulvincentaspi7763]

0

[@jarunam321]

Everything was very well explained and I understood the thing completely.... what frustrates me is the length of these videos, try to speak less (sorry for being blunt).

[@surajmishra-td3uy]

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[@varunpradhan7289]

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[@sudhirsirohi7202]

Solution is too long.
This can be done within 10 seconds without congruence relation.

[@TheRooster1337]

Your accent gave me an intraventricular hemorrhage

[@abdelalsnayyan959]

Thank you!