Question on numbers (equation using variables)?

The initial number is 3163.

Let y be the number formed by the 2 leftmost digits and let x be the number formed by the 2 rightmost digits of the initial number ==>

The initial number is 100y+x, while after the exchange of the digits the new number obtained will be 100x+y ==>

100x + y = 2(100y+x) + 5

98x = 199y + 5

The above is a linear Diophantine equation and there is a pattern for solving it:

x = (199y + 5)/98 = 2y + (3y+5)/98 ..... (1)

x is an integer in case (3y+5)/98=k is an integer ==>

3y+5 = 98k ==> y=(98k-5)/3=32k -1+(2k-2)/3 ..... (2)

y is an integer when (2k-2)/3=m is an integer ==>

2k-2=3m ==> k=(3m+2)/2=m+1+m/2

k is an integer when m=2n ==>

k=2n+1+2n/2=3n+1

Substitute k=3n+1 in (2) ==>

y=32(3n+1) -1+[2(3n+1)-2]/3=96n+31+2n=98n+31

Substitute y=98n+31 in (1) ==>

x=2(98n+31)+[3(98n+31)+5]/98

x=196n+62+3n+1=199n+63

Since x and y are 2 digit numbers, they are obtained for n=0 ==>

x=63 and y=31, so the initial number is 3163

Check:

2*3163+5=6331

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