For $ 14^{\large 14^{\Large 14}}\!$ using mod method I know the solution to find the last two digits. Even on this site I have found the solution but with mod method. But without using mod method can we solve this?
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Prior question on this. – Bill Dubuque Feb 09 '19 at 01:31
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Why don't you want to use the $\mod$ method? I can't think of any way to do this without the $\mod$ method but with the $\mod$ method is so easy why not use it? – fleablood Feb 09 '19 at 01:41
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Hint:
The number is clearly divisible by $4$
So, let us find the reminder when divided by $25$
$a=14^{14^{14}}=(1-15)^{14^{14}}=1-\binom{14^{14}}115+$ terms divisible by $25$
Now as $5$ divides $15,$ let us the reminder of $14^{14}$ when divided by $5$
$14^{14}=(1-15)^{14}=1+$ terms divisible by $5$
Let $14^{14}=1+5u$ for some integer $u$
$15\cdot14^{14}=15(1+5u)=15+25(3u)$
So, $a=1-15-25(3u)=11+25(-3u-1)$
and $a$ is divisible by $4$
Hence $a=25+11$ when divided by $100$
lab bhattacharjee
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1I would claim "finding remainders" is exactly the same thing as the "mod method" which is strictly forbidden. – fleablood Feb 09 '19 at 01:45