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There are 209/10 = 20.9 of them.
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How many tens all in the number 787?
There are 78.7 tens in the number 787
An example of a proof?
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.
What is the next number in the number sequence 298 209 129 58 -4?
-57 (each number that is subtracted from the first is 9 different from the previous (i.e. 298-89=209-80=129-71=58-62=-4-53=-57). As you can see, the amounts that are substracted are 89, 80, 71, 62, 53 (all 9 different from the one before it.)
How many integers have a two in the ones and tens place between 0 and 60?
Only one number between 0 and 60 has a two in the ones and tens place, and it's 22. However, if you are looking for all the numbers between 0 and 60 that have a two in EITHER the ones OR tens place, they are 2, 12, 21, 22, 23, 24, 25, 26, 27, 28, 29, 32, 42, and 52.
How do you write a number that has a 3 in the hundreds place a 6 in the thousands place and a 5 in the hundreds thousands place?
Remember you have Tens of Thousands, Tens, and Units all as 0's the number you describe is "Five Hundred and Six Thousand, Three Hundred" 506,300 Biggest numbers come first:-
How many tens all in the number 787?
There are 78.7 tens in the number 787
How many significant digits are in this number 546?
How many significant digest are in the number 0.0025
How many games did john worsfold play?
209 - all for West Coast.
How do you round 208930?
It all depends where you are rounding the number to. (ones, tens, hundreds etc.)
It is greater than 47 It has a 4 in the tens place It is an even number?
The only number that satisfies all three of these conditions is 48.
Are there an infinite amount of multiples of 11 with an odd digit sum?
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
An example of a proof?
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.
How many pages does All God's Children Need Traveling Shoes have?
All God's Children Need Traveling Shoes has 209 pages.
How many neutrons and protons in polonium 209?
84 protons, 84 electrons and 136 neutrons.
How many people use guns to commit a crime using a gun?
Between half and three fourths of people in the US. It is different in other countries.
How many Tae Kwon Do club it have in the world?
There are far too many Taekwondo dojang (clubs/schools) in the world to number them all. Many are with different organizations, and some are independent of any established or recognized associations, so there is no universal count. It would be safe to say they number well into the tens of thousands.
What is the average of 209 317 243 288 190?
To find the mean of the numbers, add all the numbers and divide by how many numbers are in the list: 190 + 209 +243 + 288 + 317 = 1247 1247 divided by 5 = 249.4 To find the median, list the numbers from smallest to largest. The median is the number in the middle: 243 The mode is the number that is repeated more than any other. This list of numbers has no mode.
