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What 3 digit odd number has the most multiples?
All integers have an infinite amount of multiples.
Is all multiples of 5 even numbers?
No as 5, 15 and an infinite amount of other multiples of 5 are odd
This was on my homework............. Are there an infinite amount of multiples of 11 that have an odd digit sum?
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
How many two digit numbers are not multiples of 2?
All of the odd numbers between 11 and 99 inclusive are two digit numbers that are not multiples of 2. There are 45 of them.
What are all prime numbers that are odd numbers?
There is an infinite amount of prime numbers all of which are odd numbers
How many odd factors are there?
Since any odd number can be a factor, there is an infinite amount of them.
What is the greatest odd number?
There are an infinite amount of numbers, so there can be no "greatest" odd number.
How many multiples of 9 are odd numbers?
9 multiplied by any other odd number will still be odd. 9 multiplied by any even number will be even - so the multiples of 9 between 1 and 120 that are odd would be:1x9 = 93x9 = 275x9 = 457x9 = 639x9 = 8111x9 = 9913x9 =117The next odd number you could multiply 9 by is 15 but 15x9 = 135, which is greater than 120 so does not fall within the constraints of the question.
What are numbers larger then 225 and 999 that are odd and multiples of three?
There are an infinite number of numbers that satisfy these criteria.
What are 2-digit multiples of 5?
-3
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
Why are all multiples of 3 odd and even?
The multiples of all odd numbers are odd and even. Odd x odd = odd. Odd x even = even. Since odd and even numbers alternate, the multiples will alternate as well.
