Modular Arithmetic| Mensuration| Step-by-step explanation | GMats Hub

追加: 2026-05-14

[00:00:02]Hello my wonderful student. You're welcome to class.

[00:00:06]My name is Mr. Um I want to welcome those that are just um seeing me for the first time. You're especially welcome to the class. So here we do on mathematics. If you have any problem um in your mathematics, you can always bring the problem to this problem. Now um today we want to look at question modularics and meation. Okay. So I want you to sit back. Let's uh do one or two things and don't forget um get your writing materials ready and uh also don't forget to invite your friends. Let them know the class is about starting. All right.

[00:00:49]So as you are joining the class, en sure you say hi, read the whole class and en sure you also tap the like button. So those of you that have not subscribed to this channel, en sure you do that, okay?

[00:01:02]So that um you'll be notified anytime we are here. All right. So thank you very much. So um this is the first question we are going to start with for today.

[00:01:13]Okay. So um I want to believe we have a good number um of you in the class before we commence the class. So once you join the class say hi. So greet the class.

[00:01:28]All right. Show me you're welcome to class.

[00:01:31]Gra and you're welcome to class. All right.

[00:01:40]So let's inform others that the class is about starting right. Okay. You're welcome to class.

[00:01:58]All right. Now um now let's look at the first uh question we have here under modular arithmetic. Now the question express this in mode seven. Don't forget whenever I want to write any number in a particular mode it must be less than the mode the number mode okay so now that means the highest number we're supposed to have here should be what should be six all right it must not be up to the mode all right so how do we go about this because this is a very big number 10^ 3 so let's look at it So how do we go about that?

[00:02:45]Okay. So um to do this the first thing is I just pay attention to uh the base here which is three.

[00:02:56]So I find a way of writing this such that is going to have a remainder of one. Right? So now this is what I'm saying. Now just raise this one to the power of you know when you write 3^ 1 we have that to be one.

[00:03:14]Okay.

[00:03:15]And also 3^ 2 gives you 9.

[00:03:23]All right. So if you're writing that in mode seven what's it going to give you?

[00:03:29]That's going to be two because the same thing as 7 * 1 + 2 in mode seven. So that means um 3^ 2 which is 9 is the same thing as what as two.

[00:03:50]Okay. So the next one is 3 to power 3 and that gives you 27. Okay. So 27.

[00:04:00]All right. So um 27. Now this is going to be 7 times 7 * 4.

[00:04:09]Okay. Now 7 * 3 7 * 3 that's 21. Okay.

[00:04:15]So we have a remainder of six.

[00:04:18]Okay. So that means we have six here.

[00:04:23]All right. So now uh so that's six. The next one is 3^ 4.

[00:04:31]Okay. So now let's look at it. What do we have? So uh that's going to be now just look at this. This one is the same thing as 27 here. So you can simply multiply this by because now I'm increasing this power by one. is the same thing as saying multiplying 3 power 3 by 3. So have 3 4.

[00:04:58]So that will give me 6 * 3. Make the work easier for me. So now when you multiply this what do you have? Now you have 18. So and 18 in mode 7 that is 7 * what * 2 that's 14 remain what? four.

[00:05:22]Okay. So that means the answer is four.

[00:05:27]All right. So now let's move down to the next one. We have 3^ 5. Okay. So we have this which is also the same thing as 3^ 4. So now we are increasing it. We are multiplying both sides by what? By three. So when you multiply this by three, you have this. Multiply this by 3. Okay. So you have 4 * 3. So what do you have? You have 12. 12 can be written as 7 * 1 + 5.

[00:06:03]That means we have five. Get that. Okay.

[00:06:07]So now the next thing you see I can increase this.

[00:06:12]Okay. So that's going to be what? That's 3 to power of 3^ 6.

[00:06:20]3^ 6 is the same thing as 3 5 * 3. All right. Sorry, I'm multiplying both side by three, right? All right.

[00:06:33]Sorry, I'm multiplying both sides by three, right? So I have three. So since we have so we have time one. Okay. Okay.

[00:06:41]So when you multiply this by three you have 3 7 multiply this one by three you have three. So and that is that is three. Okay. So now let us look at one more. So we have 3^ 3^ 8 which means I'm multiplying both sides. So I have 3 * 3. So 3 * 3 is what? 3 * 3. That gives me 9. And 9 can be written as what? 9 can be written as 7 * 1 + 2. So that means we have two.

[00:07:18]Okay. All right. So don't for my objective is I want to get an expression for 3^ 10. So why don't you why didn't I do it this way? I will write it as 3^ 8 * 3^ 2 = 2. Now we have this we have an expression here. What? 2 * 3^2. So now what did I do here? Now I both sides both sides. This one I multiply it by 3^ 2 and this I multiply this by 3^ 2.

[00:07:59]Okay. So now when you do that now this one is going to give you 3^ 8 + 2. All right. So now this one is 2 * 3^2. I think we have gotten an expression for that before and that is the same thing as two. Can you see that now? 3^ 2* 2 in mode 7. Okay. So we now have 3^ 10 = 4 in mode.

[00:08:32]So and this is going to be the answer to the right. So can you see that now? So that's how to solve this problem. All right. So is there any question based on what I've done here? So if you have any reservation based on what I've done, okay, so can you drop that in the comment section before I move down to the next thing, right?

[00:09:05]Okay.

[00:09:12]All right. Okay. So, uh I hope there's no question, no observation based on what I did here. Okay. So, that's that. Now let's move down to another question.

[00:10:26]Okay. So now uh let's look at the next question. Now find the last digit of 7 to 27. So I hope everyone of you are hearing me.

[00:10:41]All right. Okay. I hope you can hear me.

[00:10:47]Okay. So now let's move down to the question. Find the last digit of 7^ 27. So how do you solve this problem?

[00:11:00]Now just like we did in the previous uh question. So we pay attention to seven. So right to do now is just to find all the last digit of any number that ends with seven. Okay. So we want to find the last digit. So that's what we're going to do.

[00:11:24]So now let's look at it. So we start um by having all the digits. So we start by saying one power say 7. So that is seven.

[00:11:36]Okay. So last digit last digit of two that gives you 9. All right. So you have 49. So 49 last digit of 49 we have 9.

[00:11:52]Okay. So the next one is 7^ 3. Okay. Now since this is going to be a big number so I can just say okay I'm multiplying both sides by seven. So that is 7 * 9 and that gives you 63.

[00:12:11]Do you get that now? So you can see this is 7^ 2 is the same thing as 9. All right. So just multiply both sides by seven. So this by seven you have this.

[00:12:22]This by seven you have this. So that means the last digit is what? Three. You have three.

[00:12:30]Okay. So you may not need to stress yourself of getting 49 * 7 again. All right. So now let's look at the next one. So let's multiply. So that we have 7^ 4 that gives us what? That is 7 * 3. Okay? Because we multiply both sides by. So what do you have for this one? So you have what? 21. So we have what? One. Okay. So that's one.

[00:13:04]All right. So now let's look at another one. So we have 7^ 5.

[00:13:10]Okay. So now that mean I'm multiplying both side by seven. So when you multiply this one by seven you have it. Multiply this by seven you have seven. So now can you see now that we now have a circle of 7 9 31. So we have started another series seven. So the next one we going to have going to be n. All right? Because I'll be multiplying both side by seven. So I have 49. So I will have nine as the last digit. Okay? So and so on. So now I can get the circle. So the circle is 7 9 31. So we have circle of circle of seven 9. Okay. So now can you can you see it?

[00:14:12]So now the length of this circle now what's the length? The length is four.

[00:14:16]We have 1 2 3 4. So length is four. The circle is what? Is four. So that means that we take this to be in mode four.

[00:14:30]All right. So what are we doing with that mode? So now you come here the power we have here now express it in mode four.

[00:14:39]Okay. So now we have 25. Express 25 in mode four.

[00:14:46]Mode four. So what is going to be the answer to that? All right. So when you divide 4 by 20 25 so that's 4 * 6 okay that's 24 remainder one okay so mode four so that means you have one mod 4 okay so now one of these digits is the last digit for this and we can see one so we pick the first one. So, so therefore seven is the last is the last digits the last digit of seven.

[00:15:35]So that is the answer to the problem. So I hope you all understand that. Okay. So any question any reservation based on what we just did there.

[00:15:54]Okay.

[00:16:02]Uh the board is low. Okay.

[00:16:08]All right. So that is uh the answer to that. So can we move down to another question?

[00:16:54]All right. So, um let's take this Okay. Uh meation the bucket is 12 cm in diameter. So from here to here is 12 cm. So now in diameter. So D.

[00:19:31]Okay. I think um the network must be given. Okay. So, all right. So, that is uh the answer to the first part of the question. All right. So, now um the question go further saying that we should find the total surface area of the bucket if the two ends are close.

[00:19:59]So, I want to believe you can hear me.

[00:20:02]Um, are we back?

[00:20:05]Yes. Because the network was bad initially. All right.

[00:20:27]Okay. So uh let's continue with the second part of the question. So uh all right so the question says find the total surface area of the bucket if the two ends are closed. Now don't forget the bucket looks like this.

[00:20:58]The bucket looks like this.

[00:21:02]Okay. So, and if you are calculating the total surface area while the two ends are close, that means you going to find the area of the upper part area of the lower part and the area of the C surface. So, that's what we're going to do now. So now the first thing we're going to do here now is that let's find the area of the C surface. Okay. So now let's do that. So um what's the formula now to calculate the C surface? So we have um here area of C surface C surface surface that gives us area of C surface of the bigger cone. All right. So area of c surface that is pi r minus area of what do we have now? So that is this will be capital r is l by r l.

[00:22:23]Okay. So that is u what we going to do now. So let us find um because we can factor out pi. So when you factor out pi you have r l minus um r.

[00:22:43]Okay. So now let us keep our formula here. We have the area. So it's going to be pi r - r. So this is the formula we're using.

[00:22:59]So now we need to get the slant length.

[00:23:03]L is the slant length. This capital letter L is for the bigger SL length and the small letter is for the smaller cone. All right. So now let's do that.

[00:23:18]So don't forget we have to complete this.

[00:23:22]Get that. Let's complete this.

[00:23:25]So completely bring it down to this.

[00:23:40]Okay, want to get the slant length here.

[00:23:43]So now here we already know this place down. Let's bring out the triangle here.

[00:23:51]Now this.

[00:23:53]Okay. So now we have x and then now this is this is 24. Okay.

[00:24:06]Now the longer side is 24. Right? So um and this smaller one is what? It is what is 14.

[00:24:18]Okay. So now don't forget this is 3.5.

[00:24:24]Why this is six? So we can make use of pythagoras theorem now to get this side here. All right. So can we do that now?

[00:24:34]So now we have um let's call let's look at the bigger one. The bigger one. So now this side let's keep this. Now this side is 24. Now this is 16. All right. So we want to get this side. So we are going to be making use of pythagoras theorem we have x² = now you had the squares of the two sides that's going to be 6² + 24² okay so and then we have x² = 36 um plus now what the square of 24 So we have 24 squar is 5 76.

[00:25:30]All right. So now we have x² = Now let's add that and find the square root. So we have 36 + 5 76.

[00:25:43]So what do you have? We have 612 and you find the square roots.

[00:25:50]So when you find the square root so we have 24.

[00:25:55]So the slant length x is for the bigger one. So we have um 24 7.

[00:26:07]Okay. So now let us look for the second one.

[00:26:14]All right. For the second one the triangle looks like this.

[00:26:20]This is 3.5 and the height is 14. So that means we have y which is the hypotenus. We have y² = 14² + 3.5.

[00:26:39]Okay. So now so now let us add the two together. 14² Okay. + um 3.5 squared.

[00:26:52]So when you do that you have 208.25.

[00:27:00]So you take square root of both side. So when you do that what do you have?

[00:27:07]So now this one gives me y = 14 4. Okay. So that's what we have. So we have the slant length. So this is our capital letter and this one is the small letter. So we can now incorporate that into this formula and get the area.

[00:27:34]Okay. So let's do that now.

[00:27:49]So uh we have big L to be um 4.3 okay 7 and small L to be what 144 4. Okay. So now let's incorporate into this formula. We have a = Now r is what? The bigger r that is 6.

[00:28:20]We have 6 * the l goten that that is what? 247.

[00:28:30]Okay. So now minus now the smaller radius that is what that is 3.5 3.5 times now we use that to multiply L here that's 14.5 14.4 so now that gives you I let's multiply this um we have 6 * 24 7. So when you multiply that you have 148 148.2 minus 3.5 * 14.4.

[00:29:27]So what do you have? 50.2.4.

[00:29:33]All right. So we have this as 527.

[00:29:38]Now let's subract that. So what do we have when we subract?

[00:29:44]Okay. So you have 97.8.

[00:29:50]Okay. So we have k = So and let's multiply the two together. So you multiply that with pi. So when you do that, what do you have? You have 30 25 in cm².

[00:30:14]All right. So now the cough surface area. So the C surface area is so C surface area is 37.25 cm².

[00:30:32]All right. So now what the question is asking for the total surface area. So that means we are going to find the area of the upper um circular sectional area and the the bottom. So now let's look at it. Now the first one we have um the first the upper one which is the bigger circle. So that is area area of circle one that's circle one the top to the up one.

[00:31:06]So that is p<unk> r² by r².

[00:31:10]Okay.

[00:31:12]All right. So area of area of s 2. Okay. Cross-sectional area the bottom that is what? Pi r². So this one is for the smaller one for the bigger one. So the total surface area.

[00:31:30]Now let's add this and this together.

[00:31:32]Then we add it to this. Okay. So you know what we're going to do now? All right. So we have um so area of the circle. Now the two circles we have um area of the two circles of the two circles.

[00:31:56]Now that gives us what?

[00:31:59]R² +<unk> R². Okay, so I can factor out this. So I can have pi.

[00:32:09]All right, let me factor that out. R is what? That's the bigger one. That is what? That's 6 square minus the smaller one. That's 3.5 3.5 squ so that gives you what?

[00:32:24]22 / 7 * now let us plus sorry so now let us add together that's um 36 + uh 3.5 squared.

[00:32:39]So when you do that what do you have?

[00:32:42]You have 48.25.

[00:32:48]Okay. And we multiply that with the pi.

[00:32:51]So times pi. So time pi.

[00:32:56]So when you multiply that with pi you have 151 58 58 cm².

[00:33:09]So this is the addition of the two circles. All right. So now the question is asking for the total now. So to get the total, so we add this and this together. So when you had that, so what do we have as our final answer? So the final answer now is going to be the addition of this and this. So let's add 3 75 plus uh 151.58.

[00:33:45]So our final answer now the total surface area total surface area now is going to give us 4 58.83 cm squared. So this is the answer to the problem. All right. So u I hope you understand that. So any question based on that any observation based on what we just did now.

[00:34:41]Okay.

[00:34:46]Okay. But um is the board clearer now?

[00:34:55]Okay.

[00:35:06]All right. Um very sorry for that. I guess it's the network. So the board was not clear initially. So um I hope we all get the explanation right. So what we did now is we find the total surface area of the bucket when the bucket is closed. So now the first thing we did was to calculate the c surface area and then we have the answer to be 37.25 cm squared. Okay. So, and the next thing