+238 a 5-digit integer is reversed the difference of the original and the reversed number is not zero the largest prime number that must be a factor of this number is ??
+129899 I'll give this one a shot.......
Let the larger number be :
a10^4 + b10^3 + c10^2 + d10 + e
And let the smaller number be :
e10^4 + d10^3 + c10^2 + b10 + a
And subtracting the second from the first, we have
(a - e)10^4 + (b - d)10^3 + (d - b)10 + (e - a) rewrite as
(a - e) 10^4 + (b -d)10^3 - (b - d)10 - ( a - e) =
(a - e)(10^4 - 1) + (b - d)(10^3 - 10) =
(a - e)(9999) + 990(b - d) =
(a - e)(3^2 x 11 x 101) + (b - d) (2 x 3^2 x 5 x 11) ]
11 [ (a - e)(3^2 x 101) + (b - d)(2 x 3^2 x 5) ]
And since a,b,c,d,e are random digts between 0 and 9 inclusive, neither (a - e) nor (b - d) can be any prime number greater than 7
So it appears that 11 is the largest prime number that must be a factor of this number
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Can some other mathematician see if this is correct???
+129899 I'll give this one a shot.......
Let the larger number be :
a10^4 + b10^3 + c10^2 + d10 + e
And let the smaller number be :
e10^4 + d10^3 + c10^2 + b10 + a
And subtracting the second from the first, we have
(a - e)10^4 + (b - d)10^3 + (d - b)10 + (e - a) rewrite as
(a - e) 10^4 + (b -d)10^3 - (b - d)10 - ( a - e) =
(a - e)(10^4 - 1) + (b - d)(10^3 - 10) =
(a - e)(9999) + 990(b - d) =
(a - e)(3^2 x 11 x 101) + (b - d) (2 x 3^2 x 5 x 11) ]
11 [ (a - e)(3^2 x 101) + (b - d)(2 x 3^2 x 5) ]
And since a,b,c,d,e are random digts between 0 and 9 inclusive, neither (a - e) nor (b - d) can be any prime number greater than 7
So it appears that 11 is the largest prime number that must be a factor of this number
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Can some other mathematician see if this is correct???
