Mixing
Finding last two digits of $14^{ 14^{large 14}}!$ without mod method
$begingroup$
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?
elementary-number-theory alternative-proof
edited Feb 17 at 2:24
Bill Dubuque
213k29196654
asked Jan 30 at 4:27
mavericmaveric
89412
$endgroup$
add a comment |
$begingroup$
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?
elementary-number-theory alternative-proof
edited Feb 17 at 2:24
Bill Dubuque
213k29196654
asked Jan 30 at 4:27
mavericmaveric
89412
$endgroup$
$begingroup$
Prior question on this.
$endgroup$
– Bill Dubuque
Feb 9 at 1:31
$begingroup$
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?
$endgroup$
– fleablood
Feb 9 at 1:41
add a comment |
$begingroup$
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?
elementary-number-theory alternative-proof
edited Feb 17 at 2:24
Bill Dubuque
213k29196654
asked Jan 30 at 4:27
mavericmaveric
89412
$endgroup$
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?
elementary-number-theory alternative-proof
elementary-number-theory alternative-proof
edited Feb 17 at 2:24
Bill Dubuque
213k29196654
asked Jan 30 at 4:27
mavericmaveric
89412
edited Feb 17 at 2:24
Bill Dubuque
213k29196654
asked Jan 30 at 4:27
mavericmaveric
89412
edited Feb 17 at 2:24
Bill Dubuque
213k29196654
edited Feb 17 at 2:24
Bill Dubuque
213k29196654
edited Feb 17 at 2:24
Bill Dubuque
213k29196654
213k29196654
asked Jan 30 at 4:27
mavericmaveric
89412
asked Jan 30 at 4:27
mavericmaveric
89412
asked Jan 30 at 4:27
mavericmaveric
89412
89412
$begingroup$
Prior question on this.
$endgroup$
– Bill Dubuque
Feb 9 at 1:31
$begingroup$
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?
$endgroup$
– fleablood
Feb 9 at 1:41
add a comment |
$begingroup$
Prior question on this.
$endgroup$
– Bill Dubuque
Feb 9 at 1:31
$begingroup$
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?
$endgroup$
– fleablood
Feb 9 at 1:41
$begingroup$
Prior question on this.
$endgroup$
– Bill Dubuque
Feb 9 at 1:31
$begingroup$
Prior question on this.
$endgroup$
– Bill Dubuque
Feb 9 at 1:31
$begingroup$
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?
$endgroup$
– fleablood
Feb 9 at 1:41
$begingroup$
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?
$endgroup$
– fleablood
Feb 9 at 1:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $u$
$15cdot14^{14}=15(1+5u)=15+25(3u)$
So, $a=1-15-25(3u)=11+25(-3u-1)$
and $a$ is divisible by $4$
Hence $a=25+11$ when divided by $100$
answered Jan 30 at 5:06
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
1
$begingroup$
I would claim "finding remainders" is exactly the same thing as the "mod method" which is strictly forbidden.
$endgroup$
– fleablood
Feb 9 at 1:45
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 $u$
$15cdot14^{14}=15(1+5u)=15+25(3u)$
So, $a=1-15-25(3u)=11+25(-3u-1)$
and $a$ is divisible by $4$
Hence $a=25+11$ when divided by $100$
answered Jan 30 at 5:06
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
1
$begingroup$
I would claim "finding remainders" is exactly the same thing as the "mod method" which is strictly forbidden.
$endgroup$
– fleablood
Feb 9 at 1:45
add a comment |
$begingroup$
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 $u$
$15cdot14^{14}=15(1+5u)=15+25(3u)$
So, $a=1-15-25(3u)=11+25(-3u-1)$
and $a$ is divisible by $4$
Hence $a=25+11$ when divided by $100$
answered Jan 30 at 5:06
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
1
$begingroup$
I would claim "finding remainders" is exactly the same thing as the "mod method" which is strictly forbidden.
$endgroup$
– fleablood
Feb 9 at 1:45
add a comment |
$begingroup$
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 $u$
$15cdot14^{14}=15(1+5u)=15+25(3u)$
So, $a=1-15-25(3u)=11+25(-3u-1)$
and $a$ is divisible by $4$
Hence $a=25+11$ when divided by $100$
answered Jan 30 at 5:06
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
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 $u$
$15cdot14^{14}=15(1+5u)=15+25(3u)$
So, $a=1-15-25(3u)=11+25(-3u-1)$
and $a$ is divisible by $4$
Hence $a=25+11$ when divided by $100$
answered Jan 30 at 5:06
lab bhattacharjeelab bhattacharjee
228k15158279
answered Jan 30 at 5:06
lab bhattacharjeelab bhattacharjee
228k15158279
answered Jan 30 at 5:06
lab bhattacharjeelab bhattacharjee
228k15158279
answered Jan 30 at 5:06
lab bhattacharjeelab bhattacharjee
228k15158279
228k15158279
1
$begingroup$
I would claim "finding remainders" is exactly the same thing as the "mod method" which is strictly forbidden.
$endgroup$
– fleablood
Feb 9 at 1:45
add a comment |
1
$begingroup$
I would claim "finding remainders" is exactly the same thing as the "mod method" which is strictly forbidden.
$endgroup$
– fleablood
Feb 9 at 1:45
1
1
$begingroup$
I would claim "finding remainders" is exactly the same thing as the "mod method" which is strictly forbidden.
$endgroup$
– fleablood
Feb 9 at 1:45
$begingroup$
I would claim "finding remainders" is exactly the same thing as the "mod method" which is strictly forbidden.
$endgroup$
– fleablood
Feb 9 at 1:45
add a comment |
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$begingroup$
Prior question on this.
$endgroup$
– Bill Dubuque
Feb 9 at 1:31
$begingroup$
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?
$endgroup$
– fleablood
Feb 9 at 1:41