Proving 8 Digit Number Divisible by 137 by Induction

[haoku]

Homework Statement

Given a "8 digit number" abcdabcd.That means 1st digit =5th digit, 2nd digit =6th digit etc. Also, note that the smallest possible "8 digit number" is 10001(i.e.00010001). Prove this by induction that it is divisible by 137.

Homework Equations

The Attempt at a Solution

I can prove this actually but not able to find it by induction. I don't know how to present and start the assumption and k+1 phase.

Thanks for helping :)

 

[CompuChip]

The smallest one is given, so you start by showing that 10001 is divisible by 137.
Suppose now that for some four digit number (possibly padded by zeros from the left) A, the assumption is true, that is the number obtained by writing A twice in a row is divisible by 137. Can you write a formula for this number? Then the induction step consists of adding 1 to A and showing that the difference is divisible by 137.

 

[Architeuthis Dux]

Firstly, let's define the "8 digit number" as abcdabcd, where a, b, c, and d are the digits of the number. We can also write this number as 10000a + 1000b + 100c + 10d + a + b + c + d.

Now, let's assume that this number is divisible by 137, i.e. abcdabcd = 137k, where k is an integer.

For the base case, let's consider the smallest possible "8 digit number", 10001. This can be written as 137 x 73, where 73 is an integer.

Next, we need to prove that if abcdabcd is divisible by 137, then (a+1)(b+1)(c+1)(d+1)(a+1)(b+1)(c+1)(d+1) is also divisible by 137.

Expanding the expression, we get (a+1)(b+1)(c+1)(d+1)(a+1)(b+1)(c+1)(d+1) = (10000a + 1000b + 100c + 10d + a + b + c + d + a + 1)(a + 1)(b + 1)(c + 1)(d + 1)

= (10001a + 1000b + 100c + 10d + 1)(a + 1)(b + 1)(c + 1)(d + 1)

= (137k + 1)(a + 1)(b + 1)(c + 1)(d + 1) [substituting abcdabcd = 137k]

= 137k(a + 1)(b + 1)(c + 1)(d + 1) + (a + 1)(b + 1)(c + 1)(d + 1)

= 137k(a + 1)(b + 1)(c + 1)(d + 1) + (1000a + 100a + 10a + a + 100b + 10b + b + 100c + 10c + c + 100d + 10d + d + 1)

= 137k(a + 1)(