Last 2 digits of 13^11^7 , I'm stuck
>>374322
37, according to Wolfram
>>374325
anon I need to know the steps, because I am practicing for a math exam, thanks for taking the time though <3
>>374326
Raise 11 to the 7th power, then raise 13 to whatever the 11 to the 7th power is. It's a massive number, but if you have some sort of grapher at your disposal, you should be okay.
>>374327
would work but I need to do this by hand! thanks tho <3
13^2 -> 13 * 13 -> 69
13^4 -> 69 * 69 -> 61
13^8 -> 61 * 61 -> 21
13^16 -> 21 * 21 -> 41
13^20 -> 41 * 61 -> 01
13 has a 20 number cycle mod100, so 13^77 is congruent to 13^17
13^17 -> 41 * 13 -> 33
The answer is 33
To check:
13^19 -> 33 * 69 -> 77
13^38 -> 77 * 77 -> 29
13^76 -> 29 * 29 -> 41
13^77 -> 41 * 13
Sorry, just realized I may have misinterpreted how the powers are applied. If it's 13^(11^7), then you just need the value of 11^7(mod20) because of what's shown above.
Since 11^2(mod20) is congruent to 1, 11^7 -> 11, so in that case you just solve for 13^11(mod100), which is 37.