SOLUTION: When 64329 is divided by the number x, the successive remainders are 175,114 and 213. The sum of digit of x is?
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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 .
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