Problem Solving

When positive integer x is divided by positive integer y, the result is 59.32. What is the sum of all possible 2-digit remainders for xy?

560
616
672
728
784


Please also explain logical thinking

There is a typo in the problem above.
The problem should read as follows:

oquiella wrote:When positive integer x is divided by positive integer y, the result is 59.32. What is the sum of all possible 2-digit remainders for x/y?

560
616
672
728
784

When one positive integer is divided by another, we typically represent what's left over either as a REMAINDER or as a DECIMAL.
There is a relationship between the two representations:

Remainder/Divisor = Decimal.

When 5 is divided by 2:
Remainder representation: 5/2 = 2 R1.
Decimal representations: 5/2 = 2.5.
Remainder/Divisor = 1/2.
Decimal = .5.
Since the two values are equal:
Remainder/divisor = decimal.

It can be helpful to write the decimal representation AS A FRACTION IN ITS MOST REDUCED FORM.

In the problem above:
Remainder = R.
Divisor = y.
Decimal = 0.32 = 32/100 = 8/25.
Plugging these values into remainder/divisor = decimal, we get:
R/y = 8/25.

The resulting equation implies that R must be a MULTIPLE OF 8.
The prompt asks for the sum of all possible 2-DIGIT remainders.
2-digit multiples of 8:
16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96.

The values above constitute an EVENLY SPACED SET.
Sum of an evenly spaced set = (number of terms)(median).
Median of the 11 terms above = 56.
Thus:
Sum = (11)(56) = 616.

The correct answer is B.

Another approach:

x/y = 59.32, so

x = 59.32y, and

100x = 5932y, and

25x = 1483y

So our solutions must in this form, that is {x, y} = {1483, 25} or {2966, 50}, or {4449, 75}, etc.

Notice that the first remainder (1483/25) is 8, the second remainder (2966/50) is 16, etc.

Since our remainder can be ANY multiple of 8, we only care about the two digit multiples of 8, or

16 + 24 + 32 + ... + 96, which is

8 * (2 + 3 + ... + 12), or

8 * ((sum from 1 to 12) - 1), or

8 * (12*13/2 - 1), or

8 * (78 - 1), or 616

Mitch´s ("GMATGuruNY") solution shows a relationship between remainders and (what he calls) "decimals" through examples.
Scott´s ("Scott@TargetTestPrep") solution also deals with this relationship through mixed fractions.

Mathematically speaking, those arguments are immediate consequences of the Division Algorithm:
$$N,D\,\, \ge \,\,1\,\,{\rm{ints}}$$
$$N = QD + R\,\,\,\,\,\,\left\{ \matrix{
\,Q\,\,{\mathop{\rm int}} \hfill \cr
\,0 \le R \le D - 1 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\left( * \right)$$
\[\frac{N}{D}\,\,\mathop = \limits^{\left( * \right)} \,\,Q + \boxed{\,\frac{R}{D}\,} = Q\,\, + \,\,\boxed{\,{\text{decimal}}\,}\]

This comment follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.