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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 ??

 Aug 24, 2015

Best Answer 

 #1
avatar+129849 
+5

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???

 

 

 Aug 25, 2015
 #1
avatar+129849 
+5
Best Answer

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

-------------------------------------------------------------------------------------------------------------------

Can some other mathematician see if this is correct???

 

 

CPhill Aug 25, 2015

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