Question 192950
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:


{{{3+1+z+5=9k}}} where "k" is an integer (ie whole number)


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So if {{{k=1}}} (the smallest possible k value where we get a solution), then... 


{{{3+1+z+5=9}}}


Solve for "z" to get: {{{z+9=9}}} ---> {{{z=0}}}



So when {{{k=1}}}, then {{{z=0}}} giving the number 3105



So 3105 is divisible by 9. Check: {{{3105/9=345}}} remainder 0.




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If {{{k=2}}}, then... 


{{{3+1+z+5=18}}}


Solve for "z" to get: {{{z+9=18}}} ---> {{{z=9}}}



So when {{{k=1}}}, then {{{z=9}}} giving the number 3195



So 3195 is divisible by 9. Check: {{{3195/9=395}}} remainder 0.





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If {{{k=3}}}, then...


{{{3+1+z+5=27}}}


Solve for "z" to get: {{{z+9=27}}} ---> {{{z=18}}}



But since "z" is a single digit, this means that {{{0<=z<=9}}}. 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".



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Answer:


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