Question

31m5 is a multiple of 9 and two values are obtain by m,why where m is adigit

Answer 1

Click here to see ALL problems on Divisibility and Prime Numbers
Question 192950: If 31z5 is a multiple of 9, where z is a digit, what is the value of z?
(Scroll Down for Answer!)
Answer by jim_thompson5910(34448) (Show Source):
You can put this solution on YOUR website!
If 31z5 is a multiple of 9 (where z is a single digit), then the digits MUST add to a number that is a multiple of 9 (hopefully, they will add to 9 itself).


So the digits must add to 0, 9, 18, 27, 36, 45, 54, 63, 72, etc...


So this gives us the equation:

where "k" is an integer (ie whole number)

-----------------------------------------------------

So if (the smallest possible k value where we get a solution), then...



Solve for "z" to get: --->


So when , then giving the number 3105


So 3105 is divisible by 9. Check: remainder 0.



-----------------------------------------------------

If , then...




Solve for "z" to get: --->


So when , then giving the number 3195


So 3195 is divisible by 9. Check: remainder 0.




-----------------------------------------------------

If , then...



Solve for "z" to get: --->


But since "z" is a single digit, this means that . So ANY value of "k" greater than 2 will give us a "z" value greater than 9. So this means that we're done looking for values of "z".


=========================================================

Answer:

So the two values of "z" are 0 and 9 produce the multiples of 9 3105 and 3195 respectively.