Question 1100247: When 64329 is divided by the number x, the successive remainders are 175,114 and 213. The sum of digit of x is?
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! 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  was to be divided by  ,
dividing successive quotients by  ,
with  ,  , and  being the first successive remainders in that order,
calculating the sum the digits of ,
and repeating the adding of digits until a 1-digit sum was obtained,
the final sum of digits would be  .
Why?
Adding digits of a number and the successive sums of digits,
the final sum is the remainder of dividing that number by  .
The final sums for
, , and
are multiples of  ,
 ,  and 
meaning that , , and are multiples of .
That would make the successive quotients multiples of .
With and the fist quotient  being multiples of ,
 would be a multiple of  .
That means that the remainder from dividing by
is the same as the remainder from dividing by ,which is the final sum of digits for , :
WHY THE PROBLEM AS POSTED DI NOT MAKE SENSE:
"When 64329 is divided by the number x" means 64329 ÷ x .
That number is the divisor.
The phrase "successive remainders" suggests that
is divided by to get a quotient and a remainder,
then that quotient is divided by to get a second quotient and a second remainder,
and that second quotient is divided by to get a third quotient and a third remainder.
That cannot happen with the remainders listed.
If in a first division, we divide by ,
getting quotient and remainder ,
we know that
 .
So,  .
As 32077 is a prime number, that would mean  .
The phrase "successive remainders" suggests that
is divided by to get a quotient and as a remainder;
the quotient obtained in the first division is divided by again to get a second quotient  and as a remainder,
and that second quotient is divided by getting a third quotient  and as a remainder.
In the second division,  would be divided by  ,
getting  as the quotient and as a remainder.
Clearly, if dividing by gives for a remainder, would be , the quotient would be ,
and we would not get as a reminder when dividing by  .
If a number  that is not
is divided by a divisor  (that may noy be the the problem is about)
to get a quotient and as a remainder;
the quotient obtained in the first division is divided by again to get a second quotient and as a remainder,
and that second quotient is divided by getting a third quotient and as a remainder,
 ,  , and 
Then,
 
As the divisor must be greater than all remainders,
it must be a whole number that satisfies  .
That makes for a pretty big number to be divided.
The smallest such number would be  ,
what we get with  and  .
|
|
|
