Student A, Student B, and Student C have been hired to help scientists develop a new flavour of juice. There are 4200 samples to test. Each sample either contains blueberry or does not. Each student is asked to taste each sample and report whether or not they think it contains blueberry. Student A reports correctly on exactly $90 \%$ of the samples containing blueberry and reports correctly on exactly $88 \%$ of the samples that do not contain blueberry. The results for all three students are shown below.

|  | Student A | Student B | Student C |
| :---: | :---: | :---: | :---: |
| Percentage correct on samples  containing blueberry | $90 \%$ | $98 \%$ | $(2 m) \%$ |
| Percentage correct on samples  not containing blueberry | $88 \%$ | $86 \%$ | $(4 m) \%$ |

Student B reports 315 more samples as containing blueberry than Student A. For some positive integers $m$, the total number of samples that the three students report as containing blueberry is equal to a multiple of 5 between 8000 and 9000 . The sum of all such values of $m$ is
(A) 45
(B) 36
(C) 24
(D) 27
(E) 29
Of the 4200 samples to test, let the number of samples that contain blueberry be $x$.

Since each sample either contains blueberry or it does not, then there are $4200-x$ samples that do not contain blueberry.

Student A reports correctly on $90 \%$ of the $x$ samples that contain blueberry.

Thus, Student A reports $\frac{90}{100} x$ of these samples contain blueberry.

Student A reports correctly on $88 \%$ of the $4200-x$ samples not containing blueberry and thus incorrectly reports that $100 \%-88 \%=12 \%$ of these samples contain blueberry.

(That is, when a student is wrong when reporting "no blueberry", it means that there is blueberry.)

Therefore, Student A reports $\frac{12}{100}(4200-x)$ of these samples contain blueberry.

In total, Student A reports that $\frac{90}{100} x+\frac{12}{100}(4200-x)$ samples contain blueberry.

Similarly, Student B reports that $\frac{98}{100} x+\frac{14}{100}(4200-x)$ samples contain blueberry, and Student C reports that $\frac{2 m}{100} x+\left(\frac{100-4 m}{100}\right)(4200-x)$ samples contain blueberry.

Student B reports 315 more samples as containing blueberry than Student A, and so

$$
\left(\frac{98}{100} x+\frac{14}{100}(4200-x)\right)-\left(\frac{90}{100} x+\frac{12}{100}(4200-x)\right)=315
$$

Clearing fractions by multiplying by 100 , the equation becomes

$$
98 x+14(4200-x)-(90 x+12(4200-x))=31500
$$

Solving this equation, we get

$$
\begin{aligned}
98 x+14(4200-x)-(90 x+12(4200-x)) & =31500 \\
98 x+58800-14 x-(90 x+50400-12 x) & =31500 \\
98 x+58800-14 x-90 x-50400+12 x & =31500 \\
6 x+8400 & =31500 \\
6 x & =23100 \\
x & =3850
\end{aligned}
$$

Thus, there are 3850 samples that contain blueberry and $4200-3850=350$ samples that do not contain blueberry.

Together, the three students report

$$
\left(\frac{98}{100}(3850)+\frac{14}{100}(350)\right)+\left(\frac{90}{100}(3850)+\frac{12}{100}(350)\right)+\left(\frac{2 m}{100}(3850)+\left(\frac{100-4 m}{100}\right)(350)\right)
$$

samples as containing blueberry.

Simplifying this expression, we get

$$
\begin{aligned}
& \left(\frac{98}{100}(3850)+\frac{14}{100}(350)\right)+\left(\frac{90}{100}(3850)+\frac{12}{100}(350)\right)+\left(\frac{2 m}{100}(3850)+\left(\frac{100-4 m}{100}\right)(350)\right) \\
& =3773+49+3465+42+77 m+350-14 m \\
& =7679+63 m
\end{aligned}
$$

For some positive integers $m$, together the three students report $7679+63 m$ samples as containing blueberry.

If $7679+63 m$ is greater than 8000 , then $63 m$ is greater than $8000-7679=321$ and so $m$ is greater than $\frac{321}{63} \approx 5.09$.

Since $m$ is a positive integer, then $m$ is greater than or equal to 6 .

Similarly, if $7679+63 m$ is less than 9000 , then $63 m$ is less than $9000-7679=1321$ and so $m$ is less than $\frac{1321}{63} \approx 20.97$.

Since $m$ is a positive integer, then $m$ is less than or equal to 20 .

Thus, we want all integers $m$ from 6 to 20 inclusive for which $7679+63 m$ is equal to a multiple of 5 .

An integer is a multiple of 5 exactly when its units (ones) digit is 0 or 5 .

The units digit of the total number of samples, $7679+63 m$, is determined by the sum of the units digits of 7679 , which is 9 , and the units digit of the value of $63 m$.

Thus, $7679+63 \mathrm{~m}$ is a multiple of 5 exactly when the value of $63 \mathrm{~m}$ has a units digit of 1 or 6 (since $9+1$ has a units digit of 0 and $9+6$ has a units digit of 5 ).

The value of $63 m$ has a units digit of 1 exactly when $m$ has a units digit of 7 (since $3 \times 7$ has a units digit of 1 ).

The value of $63 m$ has a units digit of 6 exactly when $m$ has a units digit of 2 (since $3 \times 2$ has a units digit of 6 ).

The values of $m$ from 6 to 20 inclusive having a units digit of 7 or 2 are 7,12 and 17 .

We can confirm that when $m$ is equal to 7,12 and 17 , the values of $7679+63 m$ are 8120,8435 and 8750 respectively, as required.

The sum of all values of $m$ for which the total number of samples that the three students report as containing blueberry is a multiple of 5 between 8000 and 9000 is $7+12+17=36$.

ANSWER: (B)