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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 is the smallest two-digit prime number?
11 is the smallest two-digit prime number.
When this 3 digit number is rounded to the nearest hundred it rounds to 300. The digit in the ones place is 4. The sum of the digits in the 3 - digit number is 11. What is the number?
254, 344
What is the smallest two digit prime number?
11
When the greatest four digit number is divided by the greatest two digit number what is the quotient?
11
Are there an infinite amount of multiples of 11 that have an odd digit sum?
Yes, that is correct.
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 the rules of multiples of 11?
The only multiples of 11 are 1 and itself, so it is a prime number. This is the answer to the question "what are the factors of 11". Multiples of 11 can be determined as follows. Find the sum, mod 11 of the alternate digits in the number. The difference will be zero. Examples: - 1,529. 1+2 = 3 [the thousands digit plus the tens digit], 5+9 = 14 which is 3(mod 11) [the hundreds digit plus the units digit]. Difference between the two sums is zero, therefore 1,529 is a multiple of 11 . 12,345. 1+3+5 = 9, 2+4 = 6. difference is 3, therefore 12,345 is not a multiple of 11.
You are a three-digit number and a common multiple of 27 and 11 The sum of your digits is 10 What number are you?
Common three-digit multiples of 27 and 11 are 297, 594 and 891. None of their digits add up to 10.
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
What is the sum of the first three two-digit prime numbers?
11 + 13 + 17 = 41
What are two digit multiples of 11?
22, 33, 44, 55, 66, 77, 88, 99
What is the digit sum of 7301?
11
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
The sum of the digit 65?
6 + 5 = 11
Why is any 2 digit number with 2 identical digits divisible by 11?
The 2-digit multiples of 11 all have identical digits.
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.
