There a trick you can use:
last 2 and 3 digits of fibonacci repeats each 1500 terms
last 4 digits repeats each 15,000
last 5 digits repeats each 150,000
last 6 digits repeats each 1,500,000
last 7 digits repeats each 15,000,000
x = 92.740
y = 4.562.603
z = 280.555.264
Nice idea about finding "the loop length" after which "endings" repeat :) Perhaps some other puzzle could be invented from it.
However my approach was simpler - just taking modulo by 10eX where X is amount of digits in the number we seek. Then numbers won't grow longer than this.
I suspect your and mine approaches are really two versions of the same fact, however
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