As with solutions I have received recently for several of our problems; many of your solutions to "Back to Basics" were in the form of algebraic justifications. Whilst this offers a good view of the problem it does not entirely illustrate what is happening. At the end of the algebraic solution I have added a commentary to help you think beyond the algebra and back to the context. What is going on here and why?

Partial solutions were recieved from Mary Fortune of Birchwood Community HIgh School; Shu Cao and Andrei Lazanu, whose work is given below. Thanks for your contributions.

123₂₀ is 1 ×20² + 2 ×20 + 3 = 443₁₀

123₂₁ is 1 ×21² + 2 ×21 + 3 = 486₁₀

531 - 486 = 45
486 - 443 = 43

Investigate these differences when 123_b is converted to base 10 (for different values of b).

Try to explain what is happening.

123 in base b can be written as b²+2b+3=c ,
123 in base b+1 can be written as (b+1)²+2(b+1)+3=d ,

d−c = (b+1)2+2(b+1)+3−(b2+2b+3) = (b+1)2−b2+2(b+1)−2b = (b+1−b)(b+1+b)+2 = 2b+3

so when b=20, the difference is 43, when b=21, the difference is 45.

Commentary

The difference between the terms is a linear one of the form 2 times the base plus 3.

Is this because the tens digit of the original number is a 2 and the units digit is a three?

In other words, if the original number had been 142 would the difference be a linear one of the form 4b + 2?

Following from this - what would happen if the 100's digit had not been a 1?