Algebra book, chapter 5 problem set, page 124, question 17

[DanC]

I understand the rationale behind the explanation given for this question, but I don't understand the answer. The equation is 2n-1. Plugging in 26 and 34 does not yield a units digit of 3. How would you solve that problem with actual algebra, not just using the notion that they are 8 units apart?

[xerocoool]

I think the equation is An = 2^n - 1 for all n >=1

2^26 - 1 and 2^34 - 1, would be very difficult to compute. Instead we can try smaller values

n = 6

2^n - 1 => 2^6 - 1 => 64-1 = 63

Col A : Units digit of 63^26 => 9
Col B : Units digit of 63^34 => 9

n = 4

2^n - 1 => 2^4 -1 = 15

Col A : Units digit of 15^26 = 5
Col B : Units digit of 15^34 = 5

n = 26

2^n -1 => 2^26 -1 => 67,108,864 - 1 => 67,108,863

Just keep a track of the unit's digit ; ignore the rest (in the above case the unit's digit is 3 again, so the pattern will be same, both quantities will have unit's digit of 9)

Hope this helps

[DanC]

I guess it was a misprint in the book.

[tommywallach]

Nope. He's wrong. It's not about the units digit of the power, so you can't replace 26 with 6 or 34 with 4.

It's a pattern:

2^1 = 2
2^2 = 4
2^ 3 = 8
2 ^ 4 = 16
2 ^ 5 = 32
2 ^ 6 = 64
etc.

See the pattern? It repeats every 4.

2^1 ends in a 2, but so does 2^5, as will 2 ^ 9. Let's keep looking like this:

2^1 ends in 2
2^5 ends in 2
2^9
2^13
2^17
2^21
2^25
2^29
2^33

We need to know 2^26 and 2^34

2^26 is one further step in the pattern than 2^25, so it will end in 4.

2^34 is ALSO one bigger step in the pattern than one that ends in 2, so it will also end in 4.

Subtracting one from both of these will get us 3 as the units digit.

-t

P.S. Just to be clear, the equation is 2^n, not 2n. Perhaps that was your mistake? Either way, don't assume it's a misprint in the book!

[xerocoool]

Ah ! should have been more clear in the explanation.

Tommy is right :)

[DanC]

Yeah, once I discovered the equation in the fourth edition was wrong, it was easy. Thanks for the explanation though.

Nope. He's wrong. It's not about the units digit of the power, so you can't replace 26 with 6 or 34 with 4.

It's a pattern:

2^1 = 2
2^2 = 4
2^ 3 = 8
2 ^ 4 = 16
2 ^ 5 = 32
2 ^ 6 = 64
etc.

See the pattern? It repeats every 4.

2^1 ends in a 2, but so does 2^5, as will 2 ^ 9. Let's keep looking like this:

2^1 ends in 2
2^5 ends in 2
2^9
2^13
2^17
2^21
2^25
2^29
2^33

We need to know 2^26 and 2^34

2^26 is one further step in the pattern than 2^25, so it will end in 4.

2^34 is ALSO one bigger step in the pattern than one that ends in 2, so it will also end in 4.

Subtracting one from both of these will get us 3 as the units digit.

-t

P.S. Just to be clear, the equation is 2^n, not 2n. Perhaps that was your mistake? Either way, don't assume it's a misprint in the book!

[tommywallach]

Glad to help. You've been great at catching typos though, DanC!

-t