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How many two-digit numbers has a digit sum equal to 5?

There are five such numbers.


How many 5 digits numbers are there whose digits sum to 22?

There are 5460 five digit numbers with a digit sum of 22.


How many one-digit numbers are twice the size the sum of their digits?

None. The sum of one digit can't be twice the size of the digit.


How many 3 digit numbers have a digit sum of 27?

Only one . . . 999 .


How many distinct 4-digit numbers have a digit sum of 7?

-3


What is the least possible sum of two 4-digit numbers?

what is the least possible sum of two 4-digit numbers?what is the least possible sum of two 4-digit numbers?


How many three digit numbers have the sum of their numbers equal to 5?

30,25,15,20, or 10.


How many numbers are equal to the sum of two odd one digit numbers?

2,4,6,8,10,12,14,16,18


How many even two digit numbers are there where the sum of the digits is 5?

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.


What is the greatest sum of two 2 digit numbers?

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 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


How many numbers between 100 and 300 have a digit sum of 4?

Seven of them.