Questions and Discussion on OSN Mathematics 2023 at city/district level, junior high school level, Mathematics Doctor Hello friends, back again on the mathematics doctor channel, for those of you who are just visit our channel the mathematics doctor for the first time, let me introduce my name is Dr Jeff and on this channel I will discuss information about OSN (National Science Olympiad in Indonesia) or discussion of Olympic questions, well in this video we will continue the discussion of OSN at district level for junior high school level in 2023. For those of you who want to know the discussion of part one, you can check the video here, okay, without further delay, let's go straight into the discussion. the problem is. Well, we are going into the discussion of question number 16, so here we are given a set consisting of the numbers 1 2 3 4 to the number 9 and we are asked to look for a subset consisting of three members, well the three members here must be written as two odd numbers if there are two odd numbers here then the remaining one must be an even number so we need to pay attention to the possible odd numbers which are 1 3 5 7 9 Meanwhile for the even numbers they are 2 4 6 and 8 Well maybe the even ones are quite easy Yes, friends, from among the four numbers we choose one, there are four ways. Nam for odd numbers here there are five and we choose two. Maybe friends who have studied combinatorics here there is a term called Combination so we can use five combinations two of the five possible five numbers we choose du 5 C2 or 10 10 possibilities Well maybe for friends who don't understand the concept of combinations we can assign two places to the numbers we take now in front of this we can have five possibilities because there are 1 3 5 79 well then behind that there are only four possibilities left because the first number has already been taken for first place but don't forget because we are talking about subsets of the order ABC ACB BAC BCA Cab and CBA they are all counted as the same number so don't forget to We divide 5 * 4 by 2 because we calculate AB and ba twice so the total remains the same, without using the formula the number is 10. So here for the odd number 10 the even number is 4 so the number of subsets is 4 in* 10 means The total answer is 40. The answer is B. Let's move on to question number 17. OK, let's go into the discussion of question number 17. So here we are given a number theory question where given there are five positive integer numbers abcde with a less than equal to b less than the same as cd and e and it satisfies that a + b + c + e turns out to be the same as the multiplication. Well here maybe there are too many variables 5 and e the equation given is just one, well for trick problems like this usually you have to use the idea of limiting limiting here, especially because it says that a is less than B, less than C and less than e and so on. Well, here the method is like this: A + B + C + D + E turns out to be definitely less than equal to e + e + e + e + e why is that because his a is less than e his b is also less than e and also finally his e is the same as e so in his total there is no way he will exceed the number 5E e like that Well a + b + c + d + e itself turns out to be a multiplication of those numbers, so we change ab cde to multiplication, because above the writing is the same as OK. So here we can cross out e with e because E and e are positive integers so we can immediately cross out, we find that the product of the smallest number, AB CDE , is less than 5, well, in my opinion, this is quite a restrictive limit, right? So it can really limit the possibilities, friends, you need to explore further, here AB cd e- is less than 5 then the only possibility is that the multiplication is equal to 1 2 3 4 5 so we have to count case by case so for the first case if abcd is equal to 1 then automatically a is 1 b is 1 c is 1 d the number is 1 because there is this sequence, if a is equal to 1, b is equal to 1, c is equal to 1, d is equal to 1, then the equation will be the top one. Yes friends, I mean a + b + c + d + e ama the multiplication is 1 + 1 + 1 + 1 + E = 1 * 1 * 1 1, b, 1, c, 1, and d are 2, because there is an ordering property, 1 + 1 + 1 + 2 + E = 1 * 1 * 1 * 2 * e means 2E, then we will get our e cross out the left side and the right side then we get e equals 5 now this is a pretty good candidate answer okay let's move on to the third case if abcd is 3 then only 1 * 1 * 1 is possible, namely abc and d equals 3, okay, we plug it into the equation equals 3E, like that, finally we get that 6 = 2E, so e is equal to 3, well, this is one of the candidate answers for the fourth case, namely the abcd is equal to 4, then there are two possibilities, the first possibility is If a is equal to 1 and b is equal to 1, c is equal to 1, d is equal to 4 or the second possibility is that a is equal to 1, b is equal to 1, c is equal to 2, d is equal to 2, this is also possible OK, for the above case 1 + 1 + 1 + 4 we enter Dit e = 4E then 7 = 3E e is not an integer so there is no solution for the case e 4A for case 4b here 1 + 1 + 2 + 2 + e = 4E like that here it can be that 6 = 3E e which satisfies ama 2 this is also a candidate answer whose last abcd is equal to 5 then a is 1 b is 1 c is 1 and d is equal 5 gets 1 + 1 + 1 + 5 + e = 5E so the result is 8 = 4E and the e is 2 so we have several answers here but what is asked in the question is what is the largest value for E and we can get it in the case of when it is equal to 2, that is in the second case, so for the answer to question number 17, the value is c or 5. OK, let's move on to question number 18. Okay, let's go to question number 18. So here a billionaire wants to build a hotel so the rooms are The numbers from the room will be calculated by how many digits. So actually this question is a bit long, friends, you are required to have the ability to read it, your literacy skills to better understand the question. But the point is just how many digits are there in the hotel rooms. Well, because the total cost here is The amount used is 33,416,000 while each digit costs IDR 8,000 like this , so the number of digits that appear in each room in all the hotel rooms is 33,416 Dib 8 and the result is 4177 so the total digits for all the hotel rooms is 4177, well This is a question that comes up quite often in junior high school Olympiads, so counting the number of digits from 1 to n, the number can be thousands, so this is interesting, so if we look at the numbers 1 to 9, there is only one digit per number, so the total digit used is only 1 * 9 but for digits 10 to numbers 10 to 99, each digit of each number contains 2 digits so 2 is multiplied by because 10 to 9 there are 90 numbers, then 2 * 90 is 9, the bottom is 180. Then for the numbers 100 to 999 there are 3 the digits become 3 multiplied by 900, the total is 2700 2700. It turns out that if we monitor it up to here, it turns out that from 200700, I mean up to 999, the total is less than 4,177, so our conclusion is what number. The number of rooms in this hotel must exceed the thousand mark, so the number is definitely 4 digits. Well, here First, let's total up the digits that we have used, the total is 28 89. So here we try to subtract it first so that we know how much the remaining shortage is, 4177, we subtract 2889, then the remaining digits are 1288. Well, remember, 1288 is the number of digits used in the number 1000. up to the nnth number, that is the number of rooms. Okay, so we just divide it by 4, then we will get the number 322. Well, after we get the number 322, then it's clear, if we have 30 322, the number after the number 999, that means 999 is left plus 322 and the result is 13 21 so the total number of rooms in the hotel is 1321 rooms. The answer is B. Let's move on to question number 19. OK, let's go to question number 19. So here the question is a bit geometric, but maybe it's simpler here, so there is a fish pond in the shape of a triangle You can see this triangle, yes, the shape is an isosceles triangle with an angle of 904545. It turns out that to the left and below it turns out there is an L-shaped road which is gray. So here it is known that the area of the triangle or pool is apparently the same as the area of The question is what is the value of know the total length here is 3 then the length of the pool is 3 -x the base and the height of this pool is 3 -x so if we explain the area of the triangle here is 1/2 along 3 -x for the base and 3 -x for the height or 3 – we know here the width here is and the area is x^ okay, so we equate the two expressions for the area of the triangle and the area of the l shape to find the value of x that satisfies OK, let's disassemble the top one, it's 1/2, A - b^ so x² - 6x + 9 while the other one that is 6x - 2x² + x^ we multiply both sides by 2 or we simplify it first here 6x - x^ then we multiply by 2 so x^ - 6x + 9 the value will be the same as 12x - x^ we move it all to the left side so you will get 3x^ then at - 18x + 9 = 0 okay here that means x² at- with 6x + 3 = 0 to solve this problem there is a way, namely the abc formula, which friends can practice the abc formula, namely x1,2 = - B +- √ b^ - 4ac over 2A Okay so in this problem -b the b is -6 so --6 plus minus the root of b^ which is 36 - 4 times the a is 1 in front here 1 * C c is 3 over 2a 2a that's 2 * 1 OK so we get 6 +- √36 with 12 which is 24/2 so the value is 24, friends, you can take out the 4, so it's 6 +- 2√6 over 2 if divided 2 all make 3 +- √6 so the choice could be 3 + √6 or 3 -√6 but we need to pay attention again that what is being asked is What is the width of Road X which is clearly less than 3 because 3 is the total length of the road it's impossible for it to be longer in width than this length so we take the minus one so the length that satisfies x is 3 - √6 OK so the answer that satisfies is B OK, let's move on to question number 20. OK , let's go to question number 20, so here we go. here there is another number ab CD, which is a different positive integer, so friends, you have to master the key words, different and a + ba + ca + d is an odd number which is a square number. So, the value of a + b + c + d is the smallest. Well, let's look at a + b here, it's definitely different from a + c and also different from a + d because bc and d are all different. Well, if they are odd square numbers, that means the possible ones are 1, 9, then 25, then 49, which is 1 ku 3² 5² and also 7 ku so a + ba + c and a + d are possible these numbers but friends need to be careful here because a and b are both positive integers A and C are also positive numbers a + d is a positive number so the sum of them is at least two or maybe three because 1 and 2 must be different numbers. The smallest positive number is 1 and 2 so the value of 1 is the sum of a + b or a + c or a + d is not possible because if we take a as 1 then b must be 0 b is not a positive integer so the only possible ones are 9 25 and 49 which are odd numbers in the least squares so we produce an estimate of this number also the smallest Well, try it guys if you see a + ba + c and a + d if added together it is 3A Dit b + c + d or if we compare this expression with what we are looking for in the problem it is a + b + c + d but we add it again with 2A Well here the trick is to get the smallest value of a + b + c + d then the sum of these must be as small as possible but the a must be as large as possible so that what is left in a + b + c + d is very minimal. OK, we have already looked for what 3A + b + c + d is the minimum when 9 25 D with 49 and the sum is 83 this total is 83 okay now how do we make A as big as possible so we can subtract 83 from 2A so a + b + c + d is the same with 83 - 2A we take a which is large enough so that it leaves A + B + C + D which is very small OK but here you have to be careful because A cannot be more than the number 8 Why because a + b itself if we consider it the smallest is equal to 9. If the a is more than 8, for example, if we take 9, the b will be 0, so A is the biggest, it is 8, then we will leave the b with 1, the c will be 17, the d will be becomes 41, this is one example. OK, so the minimum value of a + b + c + d is achieved when 83 - 2 * 8 or 83 - 16 which will leave 67 so the answer is B. OK, let's move on to question number 21. OK, now we're in. to question number 21, this is quite a challenging question in my opinion because not all junior high school students have ever studied this symbol or combination symbol. Well, if there are n K combinations like the bracketed image, that means n factorial divided by k factorial multiplied by n - k factorial. Well, maybe there is one here. The specialty of the combination that perhaps not many friends are aware of and perhaps not many teachers teach at your school is n factorial divided by K factorial multiplied by n - k factorial is the sum below so k with n - K is always the same as the sum on top of this n as an example, for example, let me write here, yes, 20 combinations of 8, I take the example, it is 20 factorial divided by 8 factorial, 12 factorial, the special thing is that 8 + 12 is definitely the same, 20 is the property of combinations. Well, the idea of this problem is as follows For example, I'll write here 20 combinations 20 combinations 8, make sure it's in the middle here. That's multiplied by 1/ like this, so later on this problem I'll change it to 20 factorials divided by 9 * 8 factorials in* 12 factorials, so 9 is fine. I melt it into 1 with 8 so that later it becomes 20 factorials divided by 9 factorials in* 12 factorials. Well, this form is very similar to the combination form. But unfortunately the number at the bottom is 21, not 20. So here you need your ingenuity to manipulate the value of the factorial above it. so that you can get the same number as 9 plus 12, namely 21, okay, maybe for neater writing, for example, 20 combinations with K like that in this problem must be multiplied by 1/k + 1 20 20 combinations of k are 20 factorials divided by k factorial We multiply the 20 - k factorial by K + 1, now k + 1 with K we can combine to make 1 to produce 20 factorial divided by k + 1 factor 20- K factorial like that, keep in mind that the number below it is 21, the sum of k + 1 and 20 - k That number is 21. While the top number is 20 , so the trick is, friends, try multiplying it by 21/21, so that it doesn't change the value, friends, so you will get 1/21 * 21 factorial Dib k + 1 factorial 20 - k factorial OK so we can manipulate this problem into its form, so in the above problem we can manipulate its form into 1/21 open brackets. So here we take if there are 20 combinations of 0 then it will change to this one. The form of this expression is 21 combinations of k + 1, so initially we had 20 combinations of k, we will transfer them to 21 combinations of k + 1, so for the term 0, we will get 21 combinations of 1 for the next term, 21 combinations of 2, 21 combinations of 3 and so on until 21 combinations of 21, now this is also an identity that is often taught in high school material , not in middle school. So if there are n combinations, 0 is added to n combinations 1 plus n combinations of 2 and so on until n combinations of n turns out the total will be equal to 2^ n so how to explain this identity, we can use something called Newton's binomial or Pascal's segment by describing 1 + 1 dip n or friends You usually learn X + YP n like that, like x^ n + n combination 1x^ n -1 * yd and so on, that's Newton's binomial. So here we enter the X and y all sat together so we get n combination 0 and combination 1 dn combination of 2 and so on up to n combinations of n to produce the same value 2^ n. This is an identity that comes up very often in the Olympics and you have to memorize it. Okay, so in this problem the value is 1/21 multiplied, well, this looks similar to 2^ 21, similar but here he is missing one term which is 21 combination 0 he is not there 21 combination 0 that is the S value so we can change the value to 21 2^ 21 -1 this is the final answer indeed this question is quite challenging yes this is more suitable This is a high school question, but middle school friends must be required to know more about the material at the high school level. Okay, so the answer is a. Okay, let's move on to question number 22. Okay, let's go into the discussion of question number 22, so here we are given a sequence of integers X1 until X 2023 we can see in the third point the value of X1 I moved this right to the left so I get X1 + x2 + x3 and so on X4 X5 to like yes, it could be -2 -1 0 or 1. Well, the trick for doing this problem, friends, is to imagine groups of numbers, so I suppose, from between X1 to B is a number which turns out to indicate the number 0, C is a number which indicates the number -1 or shows the number -1 and lastly there is D, which is a number which is a lot, the number is -2. If we put it into the first characteristic, namely that there are 2023 numbers, then with easily we can write that a + b + c + d is the same as 2023 because the total number is 2023 but here what is very flexible, in my opinion, friends, here the b can change because it is 0 in these equations it turns out it doesn't have any meaning, okay, so D is a kind of complement, B is a kind of complement, just the number of numbers is 0. Then for this first equation, because we just add everything up, then 1 S 1 S sat, we can add up as many as a, then bb 00, there are as many as b, then B di* 0 or we just write 0 straight away okay the number -1 appears C times so I can write C multiplied by -1 and the last one is D * -2 which is 25 for the last one which is 1² also 1 so A * 1 0^ too 0 remains 0 then -1^ is C -1² That is 1 yes okay the last -2^ is 4 so 4 * D is equal to 125 so we combine the 1st and second equations we will get that a - c - 2D = 25 while a + c + 4D will have the same value as 125. OK. Well, what's interesting here is that I can subtract and add these two equations to see approximately what value of the equation satisfies OK, but friends, you need to see also here yes, our goal is to find the sum of the powers of 3. If we are looking for the smallest value then we try to make the value -2 appear as often as possible because the value -2 -2 -2 If raised to the power of 3 is -8 so our logic is to multiply the value of d which is how many times -2 appears as many times as possible then our second priority is multiplying c because it is -1 -1 -1 like that and we minimize A so the other number is as small as possible but keep in mind the number of numbers A bc and d is all impossible So the smallest negative number is 0, it can't be negative because the number of numbers. So, let's first try eliminating the two equations. If we add them up, we will get 2D + 2A, which is equal to 150, so D + with A is equal to 75. So actually we take the d, which is quite large, 74 or 75, which is still possible, but if we subtract the bottom equation from the top equation, we try a + c + 4D = 125 while a - c - 2D = 25. If we subtract it, 6d + 2C is 100 so we get 3D + c = 50. So here it turns out the value yes, it will really limit the value of D, friends, so our assumption from the initial equation is that d can be very large, 75 74, but it has to be destroyed because 3D + C has a maximum value of only 50, so the largest D turns out to be only 16 because 16 * 3 is 48 if it's 17 then it's 51, right? the c will be negative, that's not okay. OK, if the d is 16, then the c is equal to 2 here, if the d is up to 16, then the a will be 75 - 16, which is 59. OK, so here we are know the number of 1's is 59, the number of -1's is 2, the number of -2's is 16, while the number of 0's is not important because here our task is to find the sum of the powers of 3, even if the number of 0's is how many, it is not that important. here, okay, let's immediately count -8 in* 16, because the number of -2^ 3 -8 there are 16 is -18 then there are 2 -1s so -1 * 2 is -2 then don't forget there are 59 other numbers 50 9 OK, we total OK, so -18 k minus di- 2 and plus 59, the value is minus 71, so the correct answer is option B. OK, let's move on to question number 23. OK, let's move on to question number 23, which is a geometry question, so I think you need the ability which is quite unique here , namely Power of Point. Well, here there are two semicircles which are inserted into a square here. Know that the length of EK here is 3 then the length of LH is 6 and finally the length of EG is 9 okay then The question here is what is the area of square ABCD or maybe we just need to find out what the sides of the square are. Well, first of all, what you need to know here is that there is a line that passes through the circle and that is very indicative of the Power of Point direction. So maybe A little reminder for friends of Power of Point is that if for example there is a circle and then there are two lines that intersect each other like this then you will get the property a * b = c * d so that a point divides a line into right and left or top and below, where the multiplication is constant, the value is always the same, so a * b has the same value as c * d. However, it turns out that the value is the same as something like that, maybe friends only know that it is the product of two products of the two sides, but don't know what the value is equal to. Well, in further mathematics the term is called the power of a point, that's why the theorem is called Power of point, so the power of a point, if for example this point is point X, then the power of X has a special definition , namely r² minus the distance of the point to the center, namely dx^ I will give an example, friends, so here, for example, I have a circle with radius 3, then I determine a point, for example point p here, where the distance is 2 from the center, then the power of point p is 3^ - 2^ or the value is the same as 9 - 4, namely 5, so if I divide it into two lines, a line divides it into two sides, then Mak multiplies the left sides of a and b and the right side of B is certain. 5 Well, not many friends know this, but this can be used as a weapon for you to work on the problem. OK, so in this problem we can imagine that we complete this circle. Well, if we continue LH to the right side, the length will be symmetrical, that is, it will also be the same as 6 well according to the por from point H we will get that LH times the one on the right is 6 * 6 36 that will be equal to r^ minus the distance to the center of this circle OK so we can For example between the length of the center of the circle to the length of H is x so r^ - If we pull the center, let's say it's y then r^ - y^ will be equal to 3 * 3, which is 9. Now for the length EG = 9 here EG turns out to be the length x + y total between X and y so we have the info It's interesting that x + y is equal to 9. Now, using our algebraic skills, we subtract the first equation from the second equation. Here we get 36 - 9, which is 27 = r^ - r^ and then -x di- -y ^ or we get y^ - x^ Don't forget y^ - x^ we can factor it into y -x * y + x and we can get in the previous problem that y + x is 9 then we can conclude y -x = 3 okay x + y = 9, we eliminate the ation by adding, then 2y = 12, y = 6, the x will be the same as 3, so we know the y is 6, the x is 3, what is being asked in the question is how many sides the square has. where the side of the square is the diameter of the circle so we need to find out what the radius of the circle is okay here we use the first equation 36 = r^ - x² then r^ - 9 in conclusion r^ = 45 which is 36 + 9 r then will be equal to √45 Well here the side of the square is 2 times the radius so it will be 2√45 but what is being asked is the area of the square so s^ = 4 * 45 = 100 80 now this is quite an interesting and quite question challenging for junior high school questions, so the answer is c, let's move on to question number 24. OK, now we're going to number 24. So this question is quite interesting. At first glance, it's given in the Mathematics Olympiad, but in reality, it's a physics question. This is in the kinematics chapter which discusses rectilinear motion. uniform and straight motion changes uniformly Okay so here are two objects A and B where object a is moving at a speed of 5 km/h towards B so here the speed of a is 5 km/h so the distance traveled by A is 5t from perspective a well as we can see from hour 0 to 1 only object a is moving so the distance should decrease by 5 km so all of these answers show the same answer namely the graph decreases towards 50 in the first hour Well after the first hour object B is also moving at a changing speed, so when he has been walking for 1 hour, his speed will be 1 km/h, so this movement will change and we call it uniform straight-line motion or gbb. Now to find the distance that has been traveled by B, we will it's very difficult because this is actually high school material which requires calculus skills but for middle school level you can learn by drawing graphs so first we draw a graph of speed against time V against t here V This is the speed of car B here v is the speed of the object B Meanwhile, for the bottom part, this is time, so during the first hour the speed is zero because car B has not moved, but over time it will move as constant as the amount of time that has passed so we can see that an isosceles right-angled triangle will be formed. where in time T then this height is T -1 because this is a long distance here it is also t -1 to find the distance that an object has traveled We have to find What is the area under the curve or How much is the area under the graph here the area is 1/2 in* T -1 * T -1 or t -1^ So all the time up to t object B has moved along Half * t -1^ So to know how the graph looks like after the first hour then the total initial distance is 55 minus the distance from object A, namely 5t, minus the distance traveled by object B, namely 1/2 t -1², we try to dismantle the equation 55 - 5t 1/2 t - 1^ that is t^ - 2T + 1 55 -5t - 1/2t² + t - 1/2 we can simplify it to - 1/2t^ - 4T + 54 12 OK So the picture here of the ab CD option should show a This parabolic equation is a quadratic equation where the initial term is - 1/2 as friends know, quadratic equations can be facing up or facing down, well this depends on the coefficient of the highest power or the squared power here because the value is negative then we take the one with the shape facing downwards if we look at options c and d here There is an error because the movement of B itself is a straight movement changing uniformly where the graph should be curved here both are straight so no Maybe the answer is c or D As for the options a, this shows a picture where the parabola looks like it is facing upwards, this shows a graph with a which is positive. Even though in the owner of our equation the graph is negative, it looks like this, so according to the view of this graph, we can choose b as the answer to question number 24, perhaps for the points- The point in the image provided by Puspresnas is not specifically intended to show how much distance has been traveled in time T, for example, perhaps in the fifth hour, the remaining distance may not necessarily be 23, like this image, but this image only wants to illustrate that the graph has the shape of an inverted parabola. OK, let's move on to the next question. OK, let's go to question number 25. Yes, this is quite an interesting question. We need the ability to analyze prime numbers, so the right prime number is the number that if the front digits are removed. it can produce Prime digits but what is being asked in this question is how many prime numbers are right Prime so friends have a list of all prime numbers then friends look for what turns out if you delete one or two digits in front of it because in this case we see only between the numbers 10 to 12, so if we delete the first digit or the first two digits, it still produces a prime number, which is quite interesting, friends, whatever even number it is, it can't be a prime number, so if you hope the last number is deleted, leaving the number 2, that's not it. maybe because the numbers 12 22 and so on are not prime numbers so what we need to pay attention to is the numbers that come after them are the numbers 1 3 5 7 7 and 9 but the number 5 also can't be because of what because 5 itself is prime but the numbers after that nothing is prime, so if you expect the number 125 to be like that, yes, 125 itself is not prime. It's true that if we remove the number 12, it's prime, but 125 itself is not a prime number. So it's quite unique, yes, it's a prime number, which turns out to be several the digit in front of it is deleted, it is also a prime number, so we can cross out this 5, so we just need to calculate the last digit, namely 1 3 7 or 9. So, I separate it first, so I pay attention to the numbers 3 and 7 here, now the first one is for the number 3 , if it's the last digit. 3, then you don't have to worry about the front number, for example, you hope that the number 13 will appear as prime after you delete the first number. You don't need to pay attention because you want to delete the first number, the last one or two digits will still produce the number 3, so friends, just register a list of numbers where the last digit is 3 but is prime. Well here, let's try to list the two digit numbers first, friends, we can get 13 23 33, which is divisible by 3 43 53 63, which is divisible by 3 73 83. Well, this is the number. the 2 digit numbers while the 3 digit numbers I try to write them all first 103 113 123 14 133 3 143 153 and so on we cross out the numbers which are not Prime so the ones which are clearly divisible by 3 first like 123 we cross out 153 this is also divisible by 3 183 also runs out, then we delete the numbers that are divisible by 7, namely 133, we delete the numbers that are divisible by 11, namely 143. So here we have the remaining numbers, let me list them, all of which turn out to be prime numbers. OK, now let's list the ones with 7 digits at the end. 17 then 37 7 47 67 then 77 divisible by 11 87 divisible by 3 and another one is 97 Meanwhile for three digits, we try to list them all, 107 to 197, we delete the numbers that are clearly divisible by 3, we cross out 1277, then 147 and 177 are all divisible by 3, then we also delete the numbers that are divisible by 3, then we delete the numbers that are divisible by 7 and 11 here the number 187 is divisible by 11 so we delete the remainder first then the numbers here which are Prime are like in this picture so friends between the 22 numbers here there are 22 22 numbers here are numbers prime, but even if you delete one of the first digits or the first two digits to produce the last number, the last number must also be prime because it is 3 or 7. So now the next problem is the numbers 1 and 9. The numbers 1 and 9 themselves are not prime numbers, so friends. friends have to rely on the second digit to determine whether he is prime or not. So we delete the number for the third digit which is in the hundreds and leaving two additional digits so we try to list the numbers which are Prime 11 then 31 41 also 61 71 81 not 91 too No, there are 5 choices. Well, obviously here we are only looking at the numbers 10 to 200 , so we only evaluate 111, 131, 141 and 161 and 171 from among the numbers, the numbers 111 are divisible by 3, 141 is divisible by 3, 171 is divisible by 3 and 161 is divisible by 7 so there is only one option left, namely 131 which is a prime number. Meanwhile for the ending 9 we can write 19 29 39 not 49, it is also divisible by 7 59 79 and 89 in front there must be a 1 like this, we cross out what is divisible 3 here is 129 then 189 and 159 are all divisible by 3 then here 11 turns out it is also divisible by 7 We cross out so there are 179 and 131 which are prime numbers and also right prime so there are 22 and + 2 So the total is 24 numbers OK, that's all for now. friends, discussion of OSK part 2 for junior high school level in 2023. I hope the video is useful see you on the next [Music] video [Music] OSN Mathematics Questions for SMP 2024
