Placing a question mark at the end of a phrase does not make it a sensible question. Try to use a whole sentence to describe what it is that you want answered.
Yes, that is correct.
Yes! Any multiple of 11 that is in the form of 11x10^n +209 will have an odd digit sum and will be divisible by 11. 209 is divisible by 11, and 11X10^n is too. Remember that n has to be over 1,000 for it too work. ex. 11x10^5+209=1100209. 1100209/11=100019
they are infinite.
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]:--------------------------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:20911209110209110020911000209110000209...----------------[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.-Sqrxz
The only stable pattern is that the difference between the sum of all the digits in odd locations and the sum of all digits in even locations is a multiple of 11.For example, 5678*11 = 62458Sum of odd locations = 6 + 4 + 8 = 18Sum of even locations = 2 + 5 = 7Difference = 18 - 7 = 11 which is a multiple of 11.
There is an infinite amount of common multiples starting with 77.
495
41
6 9 11 23
1+3+7=11
Lets find out.2 + 3 = 5 (Odd)4 + 7 = 11 (Odd)12 + 13 = 25 (Odd)So it appears that the sum of an even and odd number is an odd number.
27