Question

which of the numbers 2, 3, 4, 5, 6, 8, 9 & 10 must divide 102, 103, X50 when the digit X is a multiple of 3?

Answer 1

is that 3 separate numbers or 1 number: 102,103,X50?

Answer 2

3 different numbers

Answer 3

just factor 102, 103, and 50 into a product of prime numbers. or if it is just one big number, same thing. If you want to know which one of the numbers in that list 1-10 will divide each one, easy, just match up the prime components of each. If you want to know which of the numbers in the list will simultaneously be divisors of 102, 103, and 50x at the same time use the chinese remainder theorem

Answer 4

Thanks

Answer 5

like for example, 50x, where x is of the form 3n, its decomposition is this: 3*5*5*2. So it is only divisible by 3,5, and 2, and multiples there of. rinse and repeat

Answer 6

@kbomeisl but it isn't 50x, it's x50...

Remember the divisibility test for 3 is that the sum of the digits is also divisible by 3. If x is a multiple of 3, can that be true of x50?

Answer 7

decomposition of 50: 5*5*2, then we multiply that by our mystery x, but mystery x is of the form 3n, that means it is really just n threes, so 50x is really 5*5*2 multiplied by a bunch of threes. n threes. 5*5*2(3+3+3+3...), use the distributive property here: 5*5*2(3)+5*5*2(3)+... that number definitely has to be divisible by three and five and 2. QED, if you like.

Answer 8

350/3 =
650/3 =
950/3 =

Discuss.

Answer 9

having the lead digit be a multiple of 3 means that the number is 50+3n(100), with \(n = 1,2,3)

None of those are divisible by 3, because 50 is not divisible by 3.

Answer 10

50x is not of that form, do you mean 50 multiplied by x of a number with x as the first digit and 50 as the last two?
If you want to be serious about number theory you have to just start seeing, or writing numbers in their prime decompositions, then you can study their form and all their fundamental properites pop out from there. GCD, LCM, all that stuff is easy to see if prime decomposition form.

Answer 11

If you mean x as the first digit and 50 as the last two digits then you have a number of this form: x(100) + 5(10)+ (0)1 = x(5*2*5*2)+ 5(5*2). This number is obviously only divisible by 5 and 2, regardless of what x is, x could be any number between 1-9 and this number would still only be divisible by 5 and 2. Again, if you want to be serious about number theory or anything in general, you need to ditch the formulas and heuristics and see numbers in their pure form. Think of every number as a product of primes, every polynomial as a product of prime polynomials, every material as a composition of atoms, etc, and all the right properties pop out naturally.

Answer 12

Oh and I should explain, that last decomposition is not just a prime decomposition, it is a decimal system decomposition. Any number can be represented like this:
1x+10y+100z+1000a... so you have your ones digit=x, tens digit=y, hundreds digit=z, etc. Any important properties of numbers are easy to test in this form.

Answer 13

I believe the interpretation of this problem as 3 separate numbers is incorrect. For starters, 103 is prime, so none of those numbers will divide it.

Actual problem should be which of {2,3,4,5,6,8,9,10} will divide 102103x50 where x is one of 3,6,9.

For that problem, the answer is that {2,3,5,6,10} will divide them.

"So it is only divisible by 3,5, and 2, and multiples there of." is incorrect, because it would have you say that 9 (a multiple of 3) divides all 3 numbers, and it does not.

@kbomeisl if you want to be serious about math problems, you need to read the problems carefully!

Answer 14

Read my replies more carefully, the first reply is if it is 50 multiplied by x, the second is if x is the hundreds digit in a number x50. I'm not sure what you are getting at...@whpalmer4

Answer 15

this is a really elementary number theory problem, and I am a doctoral candidate in mathematics...

Answer 16

Yeah, and your answer was still incorrect. 9 is a multiple of 3. 9 does not divide 102103x50 where x is one of 3,6,9. Maybe you can ask an undergrad for assistance :-)