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What is a two digit multiple of 5 whose digits add up to 13?
The number is 85.
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.
What number is a factor of 36 and it is a multiple of 3 and as 2 digits and they add up to three?
12
What is special about the multiples of 9?
If I learned why this is true, then I must've forgotten. I did an Internet search on sum of digits and came up with a bunch of depreciation and loan pay-off formulas. So I sat down to try to figure it out.Let's take a 2-digit number N and the digits are ab, where a & b are digits 0-9. The value of N = 10*a + bNot knowing if it's a multiple of nine, set a + b = 9, and let's see what happens.Since a+b=9, then b=9-a, and substitute this into the first equation:N = 10*a + 9 - a = 9*a + 9 --> 9*(a + 1). Since a is a whole number then (a+1) is also a whole number, so N is a multiple of 9. So by setting the sum of the digits equal 9, the number is constrained to be a multiple of nine. I know this isn't a formal proof, but it does show how it works.When you get to 99, the sum is not nine but a multiple of nine. Any higher numbers, and you have 3 digits and more to work with. There should be some sort of general proof out there somewhere for any-digit numbers, but this is a start.--------------I found the following on The Math Forum @ Drexel:Sum of Digits of Multiples of NineDate: 08/12/2004 at 05:19:50 From: SabaSubject: number theory: multiples of 9Why is it that when you add the individual digits of any multiple ofnine until a single digit answer is reached the answer is always nine?Is it possible to prove this?For example, 99 => 9 + 9 = 18 => 1 + 8 = 9Why doesn't it work with other numbers between 1-9 either?Date: 08/12/2004 at 10:10:23From: Doctor LuisSubject: Re: number theory: multiples of 9Hi Saba,Good job finding that pattern! The reason is that the sum of thedigits of ANY multiple of 9 is also a multiple of 9. Since you keepadding the digits (each time getting a new multiple of 9, but asmaller multiple), eventually you'll end up with a single digitnumber. Eventually you'll get to the multiple 9 itself.Now, how do I know that the sum of the digits is always a multiple of9? Suppose that a number N has digits a,b,c,d,...(from right to left),N = a + 10b + 100c + 1000d + ...= a + (b + 9b) + (c + 99c) + (d + 999d) + ...= (a + b + c + d + ...) + (9b + 99c + 999d + ...)= (a + b + c + d + ...) + 9*(b + 11c + 111d + ...)N = (sum of digits of N) + 9 * (some number)Now, look at that equation carefully. It means that(sum of digits of N) = N - 9 * (some number)Since N is assumed to be a multiple of 9, we can write it in terms ofanother integer k, so that N = 9k(sum of digits of N) = 9 * k - 9 * (some number)= 9 * (k - (some number))= 9 * (some other number)Since we showed that the sum of the digits is 9 times some integer,then it is also a multiple of 9 itself.To summarize, starting from a multiple of 9, you keep adding thedigits, each time arriving to a multiple of 9. This establishes achain of decreasing multiples of 9, until eventually you reach 9 (froma two-digit multiple). Does that make sense?It doesn't work for other integers because the chain is broken. Forexample, multiples of 8 such as 56 don't add up to a multiple of 8.
How do you calculate multiples of 3?
134 x 3= 402.... If you divide the number by 3 and if you have no remainder it is a multiple of 3...
