Consider all integer pairs (
a,
b) such that (b-a) = sod(a) + sod(b) (with a < b)
Now consider Set A, the set of all possible values for "a", the smaller member of each pair.
Part 1:
Prove that every "a" is a multiple of 9, or find a counterexample.
Part 2: Set M contains every element of Set A divided by 9.
But not every natural number appears in M.
Can you find a pattern to predict which positive integers do not appear in M?
Inspired by a problem on the YouTube channel Dr Peyam Show which asked "Are you smarter than a first grader?" which was itself inspired by Nob Yoshigahara's masterpiece
*** "sod" denotes "sum of digits"
for any positive integer n, n (mod 9) = sod(n) (mod 9)
Therefore, if (b-a) = sod(a) + sod(b),
Then mod 9
b-a = b+a
so 0 = 2a (mod 9)
but 2 is relatively prime to 9, so it follows that
0 = a (mod 9)
so a is a multiple of 9.
THIS IS ONLY TRUE FOR NON-NEGATIVE INTEGERS, as sod is not defined for negative integers.
for instance, let (a,b) = (-2,0).
b - a = 2
sod(b) = 0
sod(a) = sod(-2). If you defined sod(-2) as 2,
then set A contains -2, which would be a counterexample.
Fortunately, sod(-2) is undefined
Edited on January 28, 2022, 8:30 pm
part 1 (spoiler)
