Nice number theory problem: Given that 14! = 87a78bc1200, find a, b, and c

[imath]14!=87178291200 = 2^11\times3^5\times5^2\times7^2\times11\times13[/imath] (22 prime factors, 6 distinct)
 
Use divisibility rules for 9, 11, 7 and 32. For 32 note that 32 = 2^5 | 2^5*5^5 = 10^5 (just like 8|1000)
Come on tutors--try to solve this!
 
I can't pass up a good challenge. But there's about 2 hours of my life I'm never going to get back!

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See below for rest of solution.

Just to make sure there is only one solution, I then checked the other options ie a=3, c=7; a=5, c=5; a=7, c=3; a=9, c=1 which were not divisible by 7. (Trust me, I tried them all!)

SO, a=1, b=2 and c=9.

Nice problem, Steven G. I didn't particularly like using trial and error in a few spots. I'd like to see a more efficient solution. Anyone? :unsure:

Edit: Are my photos legible?
 

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Harry_the_cat said:
I can't pass up a good challenge. But there's about 2 hours of my life I'm never going to get back!

View attachment 36582
See below for rest of solution.

Just to make sure there is only one solution, I then checked the other options ie a=3, c=7; a=5, c=5; a=7, c=3; a=9, c=1 which were not divisible by 7. (Trust me, I tried them all!)

SO, a=1, b=2 and c=9.

Nice problem, Steven G. I didn't particularly like using trial and error in a few spots. I'd like to see a more efficient solution. Anyone? :unsure:

Edit: Are my photos legible?
Click to expand...
Nicely done!
I do have two comments, it's 9|18, not 18|9 but no big deal.
You missed one detail and that was a+b+c and a-b+c must have the same parity! That would have help a bit.
In the end you did it and it only took about two hours.
Later tonight, I'll post my solution.
 
Please allow = to also mean congruent.

Clearly 9|14! ==> 9|87a78bc1200 iff 8+7+a+7+8+b+c+1+2 =0(mod9) iff 33+ a+b+c = 0(mod9) iff a+b+c = 3(mod9). a+b + c = 3, 12 or 21

Clearly 11|14!
11|87a78bc1200 iff 8-7+a-7+8-b+c-1+2 = 0(mod 11) iff 3+a-b+c = 0(mod 11) iff a-b+c = 8(mod 11)
Now a-b+c can only equal 8 (since a=9, c=9, b=0 yields a-b+c = 18<19)
a-b+c=8

NOTE: a+b+c and a-b+c has the same parity, which is even since a-b+c=8. As a result we have a+b+c=12

7|14!
7|87a78bc1200 iff 7| (200) - (bc1) + (a78) -(87) iff 7| 200 - 100b - 10c - 1 + 100a + 78 - 87 iff 7| 190 +100(a-b)-10c iff 7| 1 + 2(a-b) + 4c iff 1+ 2(a-b) + 4c = 0(mod7) iff 2(a-b) + 4c = 6(mod7) iff a-b+ 2c = 3(mod 7).
a-b+2c = 3, 10, or 17

32| 87a78bc1200 iff 32|c1200 iff 32| 10000c + 1200 iff 32|16c + 16 iff 16c + 16 =0(mod 32) iff 16c = 16(mod32) which implies that c is odd

Combining all bold:
a+b + c = 3, 12 or 21----not needed any more as we know that a+b+c=12
(1)a-b+c=8
(2)a+b+c=12
(3)a-b+2c = 3, 10, or 17
(4)c is odd

(2)-(1): 2b=4 or b=2
(3) & (1): a-b+2c = (a-b+c)+c =8 +c. Now since c is odd, we have 8+c is odd and 8<8+c<17 or 0<c<9==> c =9

If a+b + c =12, then a=1

In the end, a=1, b=2 and c=9.

...and this problem came from the very beginning of my studying number theory.
I never studied number theory before because I thought it wasn't going to be a branch of mathematics that I would like. However, I am enjoying it!
 
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Steven G said:
Please allow = to also mean congruent.

Clearly 9|14! ==> 9|87a78bc1200 iff 8+7+a+7+8+b+c+1+2 =0(mod9) iff 33+ a+b+c = 0(mod9) iff a+b+c = 3(mod9). a+b + c = 3, 12 or 21

Clearly 11|14!
11|87a78bc1200 iff 8-7+a-7+8-b+c-1+2 = 0(mod 11) iff 3+a-b+c = 0(mod 11) iff a-b+c = 8(mod 11)
Now a-b+c can only equal 8 (since a=9, c=9, b=0 yields a-b+c = 18<19)
a-b+c=8

NOTE: a+b+c and a-b+c has the same parity, which is even since a-b+c=8. As a result we have a+b+c=12

7|14!
7|87a78bc1200 iff 7| (200) - (bc1) + (a78) -(87) iff 7| 200 - 100b - 10c - 1 + 100a + 78 - 87 iff 7| 190 +100(a-b)-10c iff 7| 1 + 2(a-b) + 4c iff 1+ 2(a-b) + 4c = 0(mod7) iff 2(a-b) + 4c = 6(mod7) iff a-b+ 2c = 3(mod 7).
a-b+2c = 3, 10, or 17

32| 87a78bc1200 iff 32|c1200 iff 32| 10000c + 1200 iff 32|16c + 16 iff 16c + 16 =0(mod 32) iff 16c = 16(mod32) which implies that c is odd

Combining all bold:
a+b + c = 3, 12 or 21----not needed any more as we know that a+b+c=12
(1)a-b+c=8
(2)a+b+c=12
(3)a-b+2c = 3, 10, or 17
(4)c is odd

(2)-(1): 2b=4 or b=2
(3) & (1): a-b+2c = (a-b+c)+c =8 +c. Now since c is odd, we have 8+c is odd and 8<8+c<17 or 0<c<9==> c =9

If a+b + c =12, then a=1

In the end, a=1, b=2 and c=9.

...and this problem came from the very beginning of my studying number theory.
I never studied number theory before because I thought it wasn't going to be a branch of mathematics that I would like. However, I am enjoying it!
Click to expand...
"Now since c is odd, we have 8+c is odd and 8<8+c<17 or 0<c<9==> c =9"

0<c<9 We know this anyway because c is a single digit.

I get the first part of this, but how do you conclude that c=9 ?
 
Harry_the_cat said:
"Now since c is odd, we have 8+c is odd and 8<8+c<17 or 0<c<9==> c =9"

0<c<9 We know this anyway because c is a single digit.

I get the first part of this, but how do you conclude that c=9 ?
Click to expand...
From (3) a-b+2c = a-b+c+c = 8+c= 3, 10, or 17.
Since we know c is positive and odd, then it must be 8+c=17 => c=9.
 
An alternative approach to conclude c is odd.

[imath]\dfrac{14!}{100}=87a78bc12[/imath]

By the divisibility rule of 8, c12 must be divisible by 8. Since the last two digits are 12, the modulo pattern will alternate between 0 and 4. We can quickly check that all odd values of c in {0,1,...,9} give remainders of 0.
 
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