chimath35 said:
Also if you are bored, here is a number theory problem:
An integer is divisible by 9 if and only if the sum of its digits is divisible by 9
Proof by induction?
I am stuck ac=b 9c=b ex. 9c=81 so c is an int. thus divisible let b= d+dn so 9c=d+dn then 9c=d+dn+dn+9
jgens said:
Write n = ∑ak10k and notice that n is divisible by 9 if and only if n ≡ 0 (mod 9). So this means n is divisible by 9 if and only if ∑ak ≡ 0 (mod 9) and the result follows.
Hi jgens, I am pretty sure that your sketch of the proof is very elegant and efficient, but I feel "mod" is an unnecessary artifice for somebody trying to understand (and like) this type of pure math course.
I have something which is a little bit simpler to handle for a novice:
Suppose k is a positive integer. Then k = a
n10
n + a
n-110
n-1 + ... + a
310
3 + a
210
2 + a
110
1 + a
0, for some non-negative integers n, a
n, a
n-1, ..., a
3, a
2, a
1, a
0.
(Note that the ai's are the digits of k.)
Dividing the positive integer k by 9 yields the following equation:
k/9 = (a
n10
n + a
n-110
n-1 + ... + a
310
3 + a
210
2 + a
110
1 + a
0)/9.
The above equation can be re-written as follows:
k/9 = [a
n(10
n -1 +1) + a
n-1(10
n-1 -1 +1) + ... + a
3(10
3 -1 +1) + a
2(10
2 -1 +1) + a
1(10
1 -1 +1) + a
0]/9.
It follows that
k/9 = [a
n(10
n -1) + a
n-1(10
n-1 -1) + ... + a
3(10
3 -1) + a
2(10
2 -1) + a
1(10
1 -1)]/9 + [ a
n+ a
n-1 + ... + a
3 + a
2 + a
1 + a
0]/9.
Clearly, the first term on the right hand-side of the above equation is an integer since a
n(10
n -1) + a
n-1(10
n-1 -1) + ... + a
3(10
3 -1) + a
2(10
2 -1) + a
1(10
1 -1) is divisible by 9.
(Note how each (10i -1) contains only 9's as digits, so each (10i -1) is divisible by 9.)
Now suppose that k is not divisible by 9. Then k/9 is not an integer. It follows that [ a
n+ a
n-1 + ... + a
3 + a
2 + a
1]/9 is not an integer, so the sum of the digits of k is not divisible by 9. Thus, if the sum of the digits of k is divisible by 9, k is divisible by 9.
Next suppose the sum of the digits of k is not divisible by 9. Then [ a
n+ a
n-1 + ... + a
3 + a
2 + a
1]/9 is not an integer. It follows that k/9 is not an integer, and hence k is not divisible by 9. Thus, if k is divisible by 9, the sum of its digits is divisible by 9.
Therefore, we conclude that integer k is divisible by 9 if and only if the sum of its digits is divisible by 9. This completes the proof.