what multiples do 7,5 and 9 have that are the same
Both digits are successive counting numbers.
The first 5 multiples of 5 are: 5, 10, 15, 20, 25 The first 5 multiples of 9 are: 9, 18, 27, 36, 45 The first 5 multiples of both 5 and 9 are: 45, 90, 135, 180, 225
The first 5 common multiples are the first 5 multiples of their lowest common multiple (LCM) LCM(9, 10) = 90 → first 5 common multiples are 90, 180, 270, 360, 450.
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.
If they are multiples of 9, the digits add up to a multiple of 9.
Because they're in multiples of three - a quick way to tell if a number is a multiple of three is to add up the digits and see if the digits add up to a multiple of three e.g 576, 5+7+6=18, 1+8=9, 9 is a multiple of three
That the sum of its digits add up to 9 as for example 9*9 = 81 and 8+1 = 9
the only multiples of 5 are 30.35.40.45.50, and 55 the only sum to 9 is 45
117. All multiples of 9 have digits that add up to something divisible by 9. Examples, 117, 36, 45, 54 all have digits that add up to 9.
If you add up the two digits and they make either 3, 6, or 9
The sum of the digits is also a multiple of 9. And, if the sum of the digits is too large, you can sum those digits and keep going.
So there is a rule about multiples of 9 in general: something is a multiple of 9 if and only if its digits add up to a multiple of 9. 6+4 = 10, so it is not a multiple of 9. There's also a neat trick to remembering the multiples of 9 below 99 with your fingers, but you can look this up if you're interested.
The digits of 498 do not add to a multiple of 9 so it isn't.
There are 720 of them. The three digit counting numbers are 100-999. All multiples of 5 have their last digit as 0 or 5. There are 9 possible numbers {1-9} for the first digit, There are 10 possible numbers {0-9} for each of the first digits, There are 8 possible numbers {1-4, 6-9} for each of the first two digits, Making 9 x 10 x 8 = 720 possible 3 digit counting numbers not multiples of 5.
The total sum of the digits has to equal nine.
You can experiment a bit for this one. Or you can apply some reasoning:* Multiples of 45 end with 5 or 0. To avoid a 5, you would have to multiply by an even number; that is, you are looking for multiples of 90. * Since multiples of 90 are also multiples of 9, the sum of the digits must needs be 9. This means that you must provide 9 ones (or a multiple thereof), and the last digit must be 0. The smallest number that satisfies these criteria is 1111111110.