Gas

[Please note this solution was written when the category of the problem was Number Theory.]

To begin with, we note that any price with a 9 in the thousands place can be thought of as 0.001 less than a "normal" dollar amount with the usual two decimal places; e.g. $3.429 = $3.43 - $0.001, $3.799 = $3.80 - $0.001, etc. To simplify our notation, we will convert the prices to cents, so our two previous examples would now be $3.429 = 343¢ - 0.1¢ and $3.799 = 380¢ - 0.1¢. (For the remainder of this solution prices will be in cents unless accompanied by the $ sign.)

If we're purchasing $G$ gallons, and the per gallon price is $P = n - 0.1$ , for $G,n \in \mathbb{N}, 301 \le n \le 400$ ; then the total cost $GP = Gn - 0.1G$ and we want this to round to a multiple of 100.

Suppose $6 \le G \le 15$ ; then $0.6 \le 0.1G \le 1.5$ . $Gn$ is a whole number, so if we want $GP$ to round to a multiple of 100, then we need $Gn$ to end in $01$ (or more technically, to be congruent to $1 \pmod{100}$ .) This eliminates more than half the possible values of $G$ , as no even number can have a multiple ending in $01$ , and neither can $15$ ; Thus the only possible values for $G$ in this range are $7, 9, 11$ and $13$ ; we find multiples of these numbers ending in $01$ . [This is not a random search; e.g. for $G = 7$ , we know that the ones digit of $n$ has to be a $3$ , then since $3 \times 7 = 21$ , the tens digit of $n$ multiplied by $7$ must end in an $8$ , so the tens digit of $n$ must be a $4$ , which means $n = 343$ . The other values of $n$ are found similarly.]

$\begin{array}{rlrclcl} \bullet & G = \ \ 7: & n = 343 & \Rightarrow & Gn = 2401 & \longrightarrow & \ \ 7 \times \$3.429 = \$24.003 \approx \$24.00 \\ \bullet & G = \ \ 9: & n = 389 & \Rightarrow & Gn = 3501 & \longrightarrow & \ \ 9 \times \$3.889 = \$35.001 \approx \$35.00 \\ \bullet & G = 11: & n = 391 & \Rightarrow & Gn = 4301 & \longrightarrow & 11 \times \$3.909 = \$42.999 \approx \$43.00 \\ \bullet & G = 13: & n = 377 & \Rightarrow & Gn = 4901 & \longrightarrow & 13 \times \$3.769 = \$48.997 \approx \$49.00 \end{array}$

Note that for each value of $G$ , there was only one possible value of $n$ ; this is because all the values of $G$ were relatively prime to $100$ , so any other value of $n$ would have to be at least $100$ higher and thus outside our range.

Now suppose $16 \le G \le 25$ ; then $1.6 \le 0.1G \le 2.5$ , which means we now need $Gn$ to end in $02$ . This doesn't eliminate as many possible values of $G$ as before, but we can still clearly eliminate $20$ and $25$ , and also $16$ and $24$ as no multiple of $4$ can end in $02$ . This leaves us six possible values for $G$ , namely $17, 18, 19, 21, 22$ and $23$ . Note also that two of the values ( $18, 22$ ) have a common factor of $2$ with $100$ , so they will yield two possible values of $n$ , which will differ by $50$ .

$\begin{array}{rlrclcl} \bullet & G = 17: & n = 306 & \Rightarrow & Gn = 5202 & \longrightarrow & 17 \times \$3.059 = \$52.003 \approx \$52.00 \\ \bullet & G = 18: & n = 339 & \Rightarrow & Gn = 6102 & \longrightarrow & 18 \times \$3.389 = \$61.002 \approx \$61.00 \\ & & n = 389 & \Rightarrow & Gn = 7002 & \longrightarrow & 18 \times \$3.889 = \$70.002 \approx \$70.00 \\ \bullet & G = 19: & n = 358 & \Rightarrow & Gn = 6802 & \longrightarrow & 19 \times \$3.579 = \$68.001 \approx \$68.00 \\ \bullet & G = 21: & n = 362 & \Rightarrow & Gn = 7602 & \longrightarrow & 21 \times \$3.619 = \$75.999 \approx \$76.00 \\ \bullet & G = 22: & n = 341 & \Rightarrow & Gn = 7502 & \longrightarrow & 22 \times \$3.409 = \$74.998 \approx \$75.00 \\ & & n = 391 & \Rightarrow & Gn = 8602 & \longrightarrow & 22 \times \$3.909 = \$85.998 \approx \$86.00 \\ \bullet & G = 23: & n = 374 & \Rightarrow & Gn = 8602 & \longrightarrow & 23 \times \$3.739 = \$85.997 \approx \$86.00 \\ \end{array}$

Thus we find that there are a total of $\boxed{12}$ pairs of $(G,P)$ which will make the display a whole number of dollars.

Since the whole dollar part of $P$ doesn't matter, Here is $G*P=$ THIS SALE $ (rounded to the nearest cent)

$7.00*0.429=3.00$

$9.00*0.889=8.00$

$11.00*0.909=10.00$

$13.00*0.769=10.00$

$17.00*0.059=1.00$

$18.00*0.389=7.00$

$18.00*0.889=16.00$

$19.00*0.579=11.00$

$21.00*0.619=13.00$

$22.00*0.409=9.00$

$22.00*0.909=20.00$

$23.00*0.739=17.00$

I just made a spreadsheet and searched. These are all from fractions that have a number close to 9 in the thousandths place of the decimal, but since it's hard to say how close it needs to be, it was hard to search that way. For example, 3/7=6/14=.42857 is close enough for 7 gallons but not for 14. 0.429=7=3.003 rounds down, 0.429*14=6.006 rounds up.

I made the following assumptions: the pump delivers gasoline in pulses of one-thousandth of a gasoline and therefore the gasoline amount is exact, the computations within the pump are done to the mill ($0.001) and that 5/4 rounding occurs in the displayed price amount. The last assumption is reasonable as the example in the picture is for 24003 mills.

$\text{Length}[\text{Select}[\text{Flatten}[\text{Table}[((g\ p+5) \bmod 1000),\{g,6,25\},\{p,3009,3999,10\}]],0\leq \text{\$\#\$1}\leq 9\&]] \Longrightarrow 12$

There would be 201 combinations if the final digit of the gasoline were all possible digits. I am old enough to remember gasoline prices that were so.

A brief description of the computation: compute fractional dollar amount of each combination with the addition of 5 mills to cause the cents portion to be correctly 5/4 rounded, structure the result as a single list (Flatten), select those elements of the list that are 0 to 9 mills and count them.

Since this problem is marked as Computer Science , my solution uses some Python. The program runs for just 0.1s!

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pairs = set()

# For all possible G and P
for G in range(6, 26):
    for P in [3.009 + n / 100 for n in range(0, 100)]:  # 3.009 - 3.999
        # If it's a whole number, add it to the list
        if round(G * P, 2) == round(G * P, 0):
            pairs.add((G, P, round(G * P, 2)))

print(len(pairs))
print(pairs)

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# Output:
12
{(11, 3.909, 43.0), (7, 3.429, 24.0), (19, 3.579, 68.0), (13, 3.769, 49.0), 
(18, 3.389, 61.0), (22, 3.409, 75.0), (21, 3.619, 76.0), (22, 3.909, 86.0), 
(9, 3.889, 35.0), (17, 3.059, 52.0), (18, 3.889, 70.0), (23, 3.739, 86.0)}

I also used an Excel spreadsheet to solve it. It may be more appropriate to change it to Computer Science problem instead of Number Theory.

• The spreadsheet range is A1:U101 (cell A1: 12 and cell A101: 3.999).
• Formulas in the range B2:U101 are similar to that in cell B2: =IF(B$1 * $A2-INT(B$1 * $A2)<0.01, 1, 0) . This means that if the fractional part of product $GP$ is less than 0.01 assign 1 as the output of the cell otherwise assign 0.
• Then in cell A1: =SUM(B2:U101) . This sum up all the cases and the result is $\boxed{12}$ .

You should notice that the 1s are highlighted in red. I used the conditional formatting ( Conditional Formatting; Highlight Cells Rules; Equal To...; 1 ) so that it highlighted all cells contain 1 as output.