+0  
 
0
41
6
avatar+52 

I got this question in a textbook and it really confused me. Can anyone help?

Suppose that A is a base 10 digit and the base 63 number A7894321(63) would end with a zero if it were written in base 10. Find all possible values of A.

 

(A7894321 is not A*7894321. A is the first digit of the number.)

 Aug 31, 2023
 #1
avatar
+1

If I understand your question, then:


When A==1 - THE DECIMAL NO =  35896572549181
When A==2 - THE DECIMAL NO =  8323708075012
When A==3 - THE DECIMAL NO =  12262688714179
When A==4 - THE DECIMAL NO =  16201669353346
When A==5 - THE DECIMAL NO =  20140649992513
When A==6 - THE DECIMAL NO =  24079630631680
When A==7 - THE DECIMAL NO =  28018611270847
When A==8 - THE DECIMAL NO =  31957591910014
When A==9 - THE DECIMAL NO =  35896572549181


As you can see, only when A==6, the decimal equivalent ends in a zero "0".

 Aug 31, 2023
 #4
avatar+52 
+1

Thank you so much!

Your explanation helped a lot.

choutowne  Sep 2, 2023
 #2
avatar
+1

We can write out the base 63 number A7894321(63) in base 10. This gives us the following number:

A * 63^6 + 7 * 63^5 + 8 * 63^4 + 9 * 63^3 + 4 * 63^2 + 3 * 63 + 2 * 63 + 1

We want this number to end with a zero. This means that the units digit of the number must be 0. The units digit of the number is equal to 1 * 1 + 0 = 1. Therefore, A must be a digit that, when multiplied by 1, gives a units digit of 0. The only digits that satisfy this condition are 1, 3, 5, 7, and 9.

Therefore, the possible values of A are 1, 3, 5, 7, and 9

 Aug 31, 2023
 #3
avatar+177 
+1

I read the other answers already provided for this particular problem, and I could not make sense of them, so I will post my own approach to this problem. Before we start with the nitty-gritty details of this problem, we should first determine what possible values of A exist in base 63. In the base 63 system, the smallest possible digit is the digit representing 0 in base 10, and the largest possible digit represents 62 in base 10. Therefore, we conclude that \(0 \leq A \leq 62\). This will become important when we generate values of A.

 

First, I will expand the base 63 number into its more familiar base 10 form by expansion. However, since we only care about the unit's digit of this particular number, I will simplify as much as possible to make the computation easier.

 

\(A7894321_{63} = 1 \times 63^0 + 2 \times 63^1 + 3 \times 63^2 + 4 \times 63^3 + 9 \times 63^4 + 8 \times 63^5 + 7 \times 63^6 + A \times 63^7\)

 

The first observation I made is that the unit's digit of \(63^p\) and \(3^p\) for nonnegative integer powers p have the same unit's digit. This occurs because \(63^p \pmod{10} \equiv 3^p \pmod{10}\). This means I can simplify the nasty expansion to the following. In order to ensure that I am writing mathematically, I will define a new function \(U(x)\) which returns the unit's digit of an integer x.

 

\(U(A7894321_{63}) = 1 \times 3^0 + 2 \times 3^1 + 3 \times 3^2 + 4 \times 3^3 + 9 \times 3^4 + 8 \times 3^5 + 7 \times 3^6 + A \times 3^7\)

 

I began exploring the powers of 3, and I immediately noticed a pattern among the powers of 3 and their units digit.

p 0 1 2 3 4 5 6 ...
3p 1 3 9 27 81 243 729 ...
U(3p) 1 3 9 7 1 3 9 ...

 

Observing the first few values of \(3^p\) should convince yourself that a never-ending pattern exists where 3 raised to the power of 0, 1, 2, and 3 have unique unit's digits, but the pattern repeats thereafter. This essentially means that we can simplify the powers further by applying the rule that \(U\left(3^p\right) = U\left(3^{p \pmod{4}}\right)\)

 

\(U(A7894321) = 1 \times 3^0 + 2 \times 3^1 + 3 \times 3^2 + 4 \times 3^3 + 9 \times 3^0 + 8 \times 3^1 + 7 \times 3^2 + A \times 3^3\)

 

Given our observations from the previous investigation of the powers of 3, we can reference that table to determine the unit's digit of the powers of 3.

 

\(U(A7894321) = 1 \times 1 + 2 \times 3 + 3 \times 9 + 4 \times 7 + 9 \times 1 + 8 \times 3 + 7 \times 9 + A \times 7\)

 

We can do these multiplications without a calculator unlike before where the calculation would have been unfeasible.

 

\(U(A7894321) = 1 + 6 +27 + 28 + 9 + 24 + 63 + 7A\)

 

We could do the addition now, but we can reduce the two-digit numbers to its final digit and do the addition because we only care about the unit's digit.

 

\(\begin{align*} U(A7894321) &= 1 + 6 + 7 + 8 + 9 + 4 + 3 + 7A \\ &= 38 + 7A \\ &= 8 + 7A \end{align*} \)

 

We finally have a nice and compact representation of the unit's digit. Of course, we want the unit's digit of this base 63 number to be zero. In order to make \(U(8 + 7A) = 0\). The first part of the addition has a unit's digit of 8. In order to make the sum have a final digit of 0, it would be required that \(U(7A) = 2\).

 

I will once again try to generate a table of values and observe a pattern

 

A 0 1 2 3 4 5 6 7 8 9 10 11 12 ...
7A 0 7 14 21 28 35 42 49 56 63 70 77 84 ...
U(7A) 0 7 4 1 8 5 2 9 6 3 0 7 4 ...

 

Based on these observations, \(U(7A) = 2\) when \(U(A) = 6\). In other words, the unit's digit of A needs to be 6. Now, all we need to do is list the values of A with a unit's digit of 6. Also, recall from the beginning that we determined that \(0 \leq A \leq 62\).

 

This means that \(A = \{6, 16, 26, 36, 46, 56\}\). All values in this set are possible values for A.

 Sep 1, 2023
 #5
avatar+52 
+1

Thank you very much!! 

Your detailed explanation made it much easier to understand the answer.

And sorry I didn't clarify, A must be a single-digit positive integer, so the only value possible in that set is 6, which is the answer :)

choutowne  Sep 2, 2023

8 Online Users

avatar
avatar
avatar
avatar
avatar
avatar
avatar