2017 AMC 10 B

Let her digit number be . Multiplying by makes it a multiple of , meaning that the sum of its digits is divisible by . Adding on increases the sum of the digits by , and reversing the digits keeps the sum of the digits the same; this means that the resulting number must be more than a multiple of . There are two such numbers between and : and .Now that we have narrowed down the choices, we can simply test the answers to see which one will provide a twodigit number when the steps are reversed:For ,we reverse the digits, resulting in Subtracting , we get .We can already see that dividing this by will not be a two-digit number, so does not meet our requirements.Therefore, the answer must be the reversed steps applied to .We have the following: Therefore, our answer is .

Working backwards, we reverse the digits of each number from and subtract from each, so we have ,,,,The only numbers from this list that are divisible by are and . We divide both by , yielding and . Since is not a twodigit number, the answer is .

You can just plug in the numbers to see which one works. When you get to , you multiply by and add to get . When you reverse the digits of , you get , which is within the given range. Thus, the answer is .

If Sofia ran the first meters of each lap at meters per second and the remaining meters of each lap at meters per second, then she took seconds for each lap. Because she ran laps, she took a total of seconds, or minutes and seconds.

The answer is .

Notice that must be positive because . Therefore the answer is .

The other choices:

As grows closer to , decreases and thus becomes less than .

can be as small as possible , so grows close to as approaches .

  For all , , and thus it is always negative.

The same logic as above, but when this time.

Rearranging, we find , or . Substituting, we can convert the second equation into .

Substituting each and with , we see that the given equation holds true, as . Thus,.

Let . The first equation converts into , which simplifies to . After a bit of algebra we found out , which means that . Substituting into the second equation it becomes .

Denote the number of blueberry and cherry jelly beans as and respectively. Then and . Substituting, we have , so , .

From the problem, we see that less than one of the answer choices must be a multiple of and positive. The only answer choice satisfying this is . We can check that blueberry and cherry jelly beans indeed does work.

We find that the volume of the larger block is , and the volume of the smaller block is . Dividing the two, we see that only a maximum of four by by blocks can fit inside the by by block. Drawing it out, we see that such a configuration is indeed possible. Therefore, the answer is .

Let's call the distance that Samia had to travel in total as , so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both , or .She bikes at a rate of kph, so she travels the distance she bikes in hours. She walks at a rate of kph, so she travels the distance she walks in hours.The total time is . This is equal to of an hour. Solving for , we have:,,,,,Since is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about .

Notice that Samia walks times slower than she bikes, and that she walks and bikes the same distance. Thus, the fraction of the total time that Samia will be walking is ,Then, multiply this by the time minutes , minutes is a little greater than of an hour so Samia traveled kilometers,The answer choice a little greater than is . (Note that we could've multiplied by and gotten the exact answer as well)

Since the distance that both rates travel are equivalent, we can find the harmonic mean of the two rates to find the average speed. Therefore, let and ,,Now, we can find the overall distance by multiply the average speed by the time(hours), minutes.Let ,,Because the distance Samia walks is half the total distance, the distance she walks is .With some knowledge of sixths, one finds that Samia walked, in kilometers,.

Since , then is isosceles, so . Therefore, the coordinates of are .

Calculating the equation of the line running between points and , . The only coordinate of that is also on this line is .

There are two ways the contestant can win.

Case : The contestant guesses all three right. This can only happen of the time.

Case : The contestant guesses only two right. We pick one of the questions to get wrong, , and this can happen of the time. Thus, .So, in total the two cases combined equals .

Complementary counting is good for solving the problem and checking work if you solved it using the method above.

There are two ways the contestant can lose.

Case : The contestant guesses zero questions correctly.

The probability of guessing incorrectly for each question is . Thus, the probability of guessing all questions incorrectly is .

Case : The contestant guesses one question correctly. There are ways the contestant can guess one question correctly since there are questions. The probability of guessing correctly is .so the probability of guessing one correctly and two incorrectly is .

The sum of the two cases is . This is the complement of what we want to the answer is .

Writing each equation in slopeintercept form, we get and . We observe the slope of each equation is and , respectively. Because the slope of a line perpendicular to a line with slope is , we see that ,because it is given that the two lines are perpendicular. This equation simplifies to . Because is a solution of both equations, we deduce and . Because we know that , the equations reduce to and . Solving this system of equations, we get .

of the people that claim that they dislike dancing actually like it,and . Therefore, the answer is .

Suppose that his old car runs at km per liter. Then his new car runs at km per liter, or km per of a liter. Let the cost of the old car's fuel be , so the trip in the old car takes dollars, while the trip in the new car takes . He saves .

Because they do not give you a given amount of distance, we'll just make that distance miles.Then, we find that the new car will use . The old car will use . Thus the answer is .

You can find that the ratio of fuel used by the old car and the new car in a same amount of distance is , and the ratio between the fuel price of these two cars is . Therefore, by multiplying these two ratios, we get that the costs of using these two cars is ,So the percentage of money saved is .

By PIE (Property of Inclusion/Exclusion),the answer is .

By Fermat's Little Theorem, (mod ), when is relatively prime to . Hence, this happens with probability .

Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits . The pattern for is , no matter what power, so doesn't work. Likewise, the pattern for is always . Doing the same for the rest of the digits, we find that the units digits of ,,,,,, and all have the remainder of when divided by , so .

We can use modular arithmetic for each residue of (mod ).

If (mod ), then (mod ),

If (mod ), then (mod ),

If (mod ), then (mod ).

If (mod ), then (mod ).

If (mod ), then (mod ).

In out of the cases, the result was (mod ), and since each case occurs equally as (mod ), the answer is .

First, note that because is a right triangle. In addition, we have , so . Using similar triangles within , we get that and . Let be the foot of the perpendicular from to . Since and are parallel, is similar to . Therefore, we have . Since , . Note that is an altitude of from , which has length . Therefore, the area of is .

Alternatively, we can use coordinates. Denote as the origin. We find the equation for as , and as . Solving for yields . Our final answer then becomes .

We note that the area of must equal area of because they share the base and the height of both is the altitude of congruent triangles. Therefore, we find the area of to be .

We know all right triangles are , so the areas are proportional to the square of like sides. Area of is of . Using similar logic in Solution , Area of is the same as .

We can use complementary counting. There are positive integers in total to consider, and there are onedigit integers, two digit integers without a zero, three digit integers without a zero, and fourdigit integers starting with a without a zero. Therefore, the answer is .

Case : monotonous numbers with digits in ascending order.

There are ways to choose n digits from the digits to . For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. Also, (the empty set) isn't included because it doesn't generate a number. The sum is equivalent to .

Case : monotonous numbers with digits in descending order.

There are ways to choose n digits from the digits to . For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that is included since we are allowed to end numbers with zeros. However, (the empty set) still isn't included because it doesn't generate a number. The sum is equivalent to .

We discard the number since it is not positive. Thus there are here.

Since the digit numbers to satisfy both case and case , we have overcounted by . Thus there are monotonous numbers.

Like Solution , divide the problem into an increasing and decreasing case:

Case : Monotonous numbers with digits in ascending order.

Arrange the digits through in increasing order, and exclude because a positive integer cannot begin with .

To get a monotonous number, we can either include or exclude each of the remaining digits, and there are ways to do this. However, we cannot exclude every digit at once, so we subtract to get monotonous numbers for this case.

Case : Monotonous numbers with digits in descending order.

This time, we arrange all digits in decreasing order and repeat the process to find ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only , so we subtract to get monotonous numbers for this case. At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract from our total.

Thus our final answer is .

First we figure out the number of ways to put the blue disks. Denote the spots to put the disks as from left to right, top to bottom. The cases to put the blue disks are ,,,,,. For each of those cases we can easily figure out the number of ways for each case, so the total amount is .

Solution

Note that by symmetry, is also equilateral. Therefore, we only need to find one of the sides of to determine the area ratio. WLOG, let . Therefore,  and . Also, , so by the Law of Cosines, . Therefore, the answer is .

Solution

As mentioned in the first solution, is equilateral. WLOG, let . Let be on the line passing through such that is perpendicular to . Note that is a with right angle at . Since , and . So we know that . Note that is a right triangle with right angle at . So by the Pythagorean theorem, we find .Therefore, the answer is .