Equation in factorials

Jan 2012
140
2
For some natural N, the number of positive integral 'x' satisfying the equation:

\(\displaystyle 1!+2!+3!+.....+(x!)=\left ( N \right )^{2}\)

Ans. is x=1 and x=3, but how can it be proved?

Thanks.
 
Feb 2013
281
0
A) N is even
  • N^2 is even. Therefore 1! + even = even^2. No solution for x.
B) N is odd
  • (N-1) and (N+1) both are even. N^2-1 = (N+1)*(N-1)
  • 2!+...+x!=N^2-1=4*M
  • x=1 is a solution
  • if x>1, 2*(1+3!/2 + 4!/2 + .... + x!/2)=4M --> x=3
 
Jul 2013
180
11
Croatia
csak said:
A) N is even
  • N^2 is even. Therefore 1! + even = even^2. No solution for x.
B) N is odd
  • (N-1) and (N+1) both are even. N^2-1 = (N+1)*(N-1)
  • 2!+...+x!=N^2-1=4*M
  • x=1 is a solution
  • if x>1, 2*(1+3!/2 + 4!/2 + .... + x!/2)=4M --> x=3
Can you explain how exactly did you get those solutions?
 
Jan 2012
140
2
csak said:
A) N is even
  • N^2 is even. Therefore 1! + even = even^2. No solution for x.
B) N is odd
  • (N-1) and (N+1) both are even. N^2-1 = (N+1)*(N-1)
  • 2!+...+x!=N^2-1=4*M
  • x=1 is a solution
  • if x>1, 2*(1+3!/2 + 4!/2 + .... + x!/2)=4M --> x=3
For N=even, okay. I am confused at N=odd. Clearly x=1 is the visible solution. But the last equation you wrote comes out to be:

(1+3!/2 + 4!/2 + .... + x!/2)=2M, for x>1, M being any positive integer (odd or even). But even if x=4 or 5, still the LHS side is even and right side also even...?

Thanks.
 

skipjack

Forum Staff
Dec 2006
21,478
2,470
It's simpler to use the original equation. For x = 2 or x > 3, the LHS ends in 3, so it cannot be N².
 
Jan 2012
140
2
Re:

skipjack said:
It's simpler to use the original equation. For x = 2 or x > 3, the LHS ends in 3, so it cannot be N².
Okay, this is fine. Also, can you please point me to the link which proves that if a natural number ends in 2, 3, 7 or 8 it can never be a perfect square, so that I can give that too to the students?

As a matter of interest, can the following be clarified..., the same thing:

(1+3!/2 + 4!/2 + .... + x!/2)=2M, for x>1 which how serves the same result (like previously depicted), which I was not able to get...

But anyways, it (the reason of '3' in the end!) works.

Thanks.
 
Jul 2013
180
11
Croatia
happy21 said:
skipjack said:
It's simpler to use the original equation. For x = 2 or x > 3, the LHS ends in 3, so it cannot be N².
Okay, this is fine. Also, can you please point me to the link which proves that if a natural number ends in 2, 3, 7 or 8 it can never be a perfect square, so that I can give that too to the students?

As a matter of interest, can the following be clarified..., the same thing:

(1+3!/2 + 4!/2 + .... + x!/2)=2M, for x>1 which how serves the same result (like previously depicted), which I was not able to get...

But anyways, it (the reason of '3' in the end!) works.

Thanks.
Let N be some integer, form of N=10A+n, A is an integer and n is last digit of N. (n can go from 0 to 9)
Then N² is :

N² =(10A+n)²= 100A²+20An+n²=10(10A²+2Aa)+n²

We can see that last digit of N² depends on last digit of n², where n?<0,9> and n is natural.
for n=1, n²=1 last digit of N is 1
for n=2 n²=4 last digit of N is 4
for n=3, n²=9 last digit of N is 9
for n=4, n²=16 last digit of N is 6
for n=5, n²=25 last digit of N is 5
for n=6, n²=36 last digit of N is 6
for n=7, n²=49 last digit of N is 9
for n=8, n²=64 last digit of N is 4
for n=9, n²=81 last digit of N is 1
for n=0, n²=0 last digit of N is 0
possible last digits OF N² are : 0,1,4,9,6 and 5.
 

skipjack

Forum Staff
Dec 2006
21,478
2,470
As N = 5K + r, where r is 0, 1, 2, 3 or 4, N² = 25K² + 10rK + r²,
and has a last digit that is the last digit of 25K² + r²,
but 25K² ends in 0 or 5 and r² ends in 0, 1, 4, 9, or 6, so their sum cannot end in 3.

Much the same original problem occurs here and Google finds an answer on the same site (albeit without the full detail relating to the final digit).
 
Jan 2012
140
2
crom said:
Let N be some integer, form of N=10A+n, K is an integer and n is last digit of N. (n can go from 0 to 9)
Then N² is :

N² =(10A+n)²= 100A²+20Kn+n²=10(10K²+2Ka)+n²=10(10K²+2Kn)+n²

We can see that last digit of N² depends on last digit of n², where n?<0,9> and n is natural.
for n=1, n²=1 last digit of N is 1
for n=2 n²=4 last digit of N is 4
for n=3, n²=9 last digit of N is 9
for n=4, n²=16 last digit of N is 6
for n=5, n²=25 last digit of N is 5
for n=6, n²=36 last digit of N is 6
for n=7, n²=49 last digit of N is 9
for n=8, n²=64 last digit of N is 4
for n=9, n²=81 last digit of N is 1
for n=0, n²=0 last digit of N is 0
possible last digits for N are : 0,1,4,9,6 and 5.
Oh I see. Thanks...it serves.......
 
Jan 2012
140
2
And the equation x=10K+n covers all the numbers from 10 to what one wishes. And the squares of first 9 numbers -- easily calculated.

Thanks.