63
None. The sum of one digit can't be twice the size of the digit.
-3
30,25,15,20, or 10.
The greatest sum of two two-digit numbers occurs when you add the largest two-digit numbers together. The largest two-digit number is 99. Therefore, the greatest sum of two two-digit numbers is 99 + 99, which equals 198.
Seven of them.
There are five such numbers.
There are 5460 five digit numbers with a digit sum of 22.
None. The sum of one digit can't be twice the size of the digit.
Only one . . . 999 .
-3
what is the least possible sum of two 4-digit numbers?what is the least possible sum of two 4-digit numbers?
30,25,15,20, or 10.
2,4,6,8,10,12,14,16,18
To find the even two-digit numbers where the sum of the digits is 5, we need to consider the possible combinations of digits. The digits that sum up to 5 are (1,4) and (2,3). For the numbers to be even, the units digit must be 4, so the possible numbers are 14 and 34. Therefore, there are 2 even two-digit numbers where the sum of the digits is 5.
The greatest sum of two two-digit numbers occurs when you add the largest two-digit numbers together. The largest two-digit number is 99. Therefore, the greatest sum of two two-digit numbers is 99 + 99, which equals 198.
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
Seven of them.