Question 1100247
There is a typo, or something was lost in translation here,
because the wording does not make sense.
 
SUGGESTED PROBLEM AND ITS SOLUTION:
If a number {{{x}}} was to be divided by {{{64329=3*41*523}}} ,
dividing successive quotients by {{{64329}}} ,
with {{{175}}} , {{{114}}} , and {{{213}}}  being the first successive remainders in that order,
calculating the sum the digits of {{{x}}} ,
and repeating the adding of digits until a 1-digit sum was obtained,
the final sum of digits would be {{{4}}} .
 
Why?
Adding digits of a number and the successive sums of digits,
the final sum is the remainder of dividing that number by {{{9}}} .
The final sums for
{{{64329}}} , {{{114}}} , and {{{213}}}
are multiples of {{{3}}} ,
{{{system(6+4+3+2+9=24,2+4=6)}}} , {{{2+1+3=6}}} and {{{1+1+4=6}}}
meaning that {{{64329}}} , {{{114}}} , and {{{213}}} are multiples of {{{3}}} .
That would make the successive quotients multiples of {{{3}}} .
With {{{64329}}} and the fist quotient {{{q[1]}}} being multiples of {{{3}}} ,
{{{x-175=64329*q[1]}}} would be a multiple of {{{3*3=9}}} .
That means that the remainder from dividing {{{x}}} by {{{9}}}
is the same as the remainder from dividing {{{175}}} by {{{9}}} ,which is the final sum of digits for {{{175}}} , {{{4}}} :
{{{system(1+7+5=13,1+3=4)}}}
 
WHY THE PROBLEM AS POSTED DI NOT MAKE SENSE:
"When 64329 is divided by the number x" means 64329 ÷ x .
That number {{{x}}} is the divisor.
The phrase "successive remainders" suggests that
{{{64329}}} is divided by {{{x}}} to get a quotient and a remainder,
then that quotient is divided by {{{x}}} to get a second quotient and a second remainder,
and that second quotient is divided by {{{x}}} to get a third quotient and a third remainder.
That cannot happen with the remainders listed.
  
If in a first division, we divide {{{64329}}} by {{{x}}} ,
getting quotient {{{q[1]}}} and remainder {{{175}}} ,
we know that
{{{64329 = x * q[1] + 175}}} .
So, {{{x * q[1] = 64329 - 175 = 64154=2*32077}}} .
As 32077 is a prime number, that would mean {{{system(q[1]=2,x=32077)}}} .

The phrase "successive remainders" suggests that
{{{64329}}} is divided by {{{x}}} to get a quotient {{{q[1]}}} and {{{175}}} as a remainder; 
the quotient {{{q[1]}}} obtained in the first division is divided by {{{x}}} again to get a second quotient {{{q[2]}}} and {{{114}}} as a remainder,
and that second quotient {{{q[1]}}} is divided by {{{x}}} getting a third quotient {{{q[3]}}} and {{{213}}} as a remainder.

In the second division, {{{2}}} would be divided by {{{32077}}} ,
getting {{{0}}} as the quotient and {{{2}}} as a remainder.
 
Clearly, if dividing {{{64329}}} by {{{x}}} gives {{{175}}} for a remainder, {{{x}}} would be {{{32077}}} , the quotient would be {{{2}}} ,
and we would not get {{{114}}} as a reminder when dividing {{{2}}} by {{{x=32077}}} .
 
If a number {{{N}}} that is not {{{64329}}} 
is divided by a divisor {{{D}}} (that may noy be the {{{x}}} the problem is about)
to get a quotient {{{q[1]}}} and {{{175}}} as a remainder; 
the quotient {{{q[1]}}} obtained in the first division is divided by {{{D}}} again to get a second quotient {{{q[2]}}} and {{{114}}} as a remainder,
and that second quotient {{{q[1]}}} is divided by {{{D}}} getting a third quotient {{{q[3]}}} and {{{213}}} as a remainder,
{{{q[2]=D*q[3]+213}}} , {{{q[1]=D*q[2]+114}}} , and {{{N=D*q[1]+175}}} 
 Then,
{{{N=D*(D*(D*q[3]+213)+114)+175}}}
{{{"="}}}{{{D*(D^2*q[3]+D*213+114)+175}}}
{{{"="}}}{{{D^3*q[3]+D^2*213+D*114+175}}}
 
As the divisor must be greater than all remainders,
it must be a whole number that satisfies {{{D>=214}}} .
That makes for a pretty big number {{{N}}} to be divided.
The smallest such number would be {{{9779119}}} ,
what we get with {{{q[3]=0}}} and {{{D=214}}} .