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Are there infinitely many multiples of 11 with an odd digit sum?

Yes.

One proof [I'm not sure that this is the simplest proof for this, but it is a proof]:

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Note that 209 is divisible by 11 and has an odd digit sum (11). Now consider the number 11000*10i+209. Because 11000 is divisible by 11 and 209 is divisible by 11,

11000*10i+209 is divisible by 11 for all whole number values of i (of which there are infinitely many).

Further, the digit sum for 11000*10i+209 is odd for all whole number values of i because the hundreds, tens and ones places will always be 2, 0, and 9 respectively, and the other digits will be either all zeros (for i=0) or two ones followed by zeros, down to and including the thousands places. Thus, the digit sum of 11000*10i+209 is 11 for i=0 and 13 for all other whole numbers i.

Thus, we have have the following set of numbers (of which there are infinitely many) which are multiples of 11 and which have an odd digit sum:

209
11209
110209
1100209
11000209
110000209
.
.
.
----------------
[Note there are other multiples of 11 that have an odd digit sum (e.g., 319, 11319, 110319, ...).]
___________________________________________________________
Late addition:

Here is the simplest proof:
Prove that x+2=3 implies that x=1.

proof:
FIRST, assume the hypothesis, that x+2=3. What we try to do is reach the conclusion (x=1) using any means possible. I have some algebra skills, so I'll subtract 2 from both sides, which leads me to x = 1.
QED.

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