IOQM_Pre RMO_Number Theory_Congrueny_Cyclicity_day 04

[00:00:08][clears throat] Good [clears throat] morning Jri Krishna.

[00:01:28]So today we will continue congrences as well as cyclicity.

[00:01:43]So Next property A congrent to B mod.

[00:01:58]Then A power K congrent to B power K modm for all positive integer K.

[00:02:16]for all positive integer K. So write down this property.

[00:03:00]Okay. Simple one. A power [clears throat] k - b power k.

[00:03:08]So a - b * this is a power k -1 + a power k - 2 b + and so on + b power k - 1.

[00:03:23]So it is a - b * t.

[00:03:29]So what does it mean? A power k minus b power k is a minus b multiple Okay. So therefore a power k congrent to b power k model.

[00:03:50]Right. Next property.

[00:03:53]A + b congrent to codm b c congrent to d modm.

[00:04:05]B congregant to B mod. Then A congrent to A + D congrent to C model.

[00:04:19]A plus D congrent to C mod. So how it is possible sir? 1 [clears throat] C A + B congrent to C mod M B congrent to D mod right I'm subtracting A + B minus B congrent to C minus D model okay now A congregant to C minus D modm so A + D congreent to C minus D + D model.

[00:05:00]So A + D congrent to C model.

[00:05:06]Simple property only.

[00:05:13]A congrent to B mod. A congrent to B mod.

[00:05:20]uh and uh d divides m and d divides m then a congrent to b mod d a congrent to b mod d.

[00:05:49]So A congrent to B mod M means A congrent to B mod M means M divides A minus B m divides A minus B and D divides M. That implies D divides A minus B. D divides A minus B. So A congregant to B mod D of you. Okay.

[00:07:03]A congrent to B modium, C congrent to D modium. m then ax + c y congrent to bx + dy.

[00:07:30]It is general property only. By using the above properties we can make it standard result only.

[00:08:14]completed. Huh?

[00:08:28]Okay. Now see next one.

[00:08:35]Let p of x is equal to p x^ n + p1 x n -1 plus uh p2 x n -2 and so on plus pn be a polomial.

[00:09:00]Okay. be a polomial with integral coefficients.

[00:09:18]Integral coefficients. Your coefficients are p1, p2 and so on are coefficients which are integers.

[00:09:26]and uh a congregant to b mod then p of a congrent to p of b mod.

[00:09:44]So it is very beautiful and interesting result.

[00:10:06]CA congrent to CB modelm CA congrent to CB modem and uh GCD of C comma M is equal to one gcd of C comma 1 HCF of these two is one because means it is co-prime then a congrent to bodium what is next property So these are some uh [clears throat] important and basic properties of congrencies. I think uh these uh properties are enough to solve our exercise.

[00:11:35]Enough to solve our exercise.

[00:11:38]So now we are going to discuss about uh cyclicity.

[00:12:23]Cylicity.

[00:12:35]So in cyclicity directly we can do the uh we can form the table form.

[00:12:44]N= to what?

[00:12:46]1 n 2 power n 3 power n 4 n 5 n 6 power n 7 n 8 power n Okay.

[00:13:05]n = 1 2 3 4 n = 1 2 3 4 period.

[00:13:17]So [clears throat] one power uh see these are all uh we are considering units place and with our units place we have to observe these two points only one power one uh 1 power 1 ends with one power two ends with one power 3 ends with one power 4 ends with One right period is one for every uh number one power. So this is two power n 2 power 1 2 2² 4 2 2 cube 8 2 4 4 16 ends with 6 2 4 16 ends with 6. Again if you'll take 2^ 5 32 ends with 2. 2^ 64 ends with four. 2^ 7 128 ends with 8. that you have to observe.

[00:14:18]2^ 7 128 ends with 8. 2^ 64 ends with 4.

[00:14:23]2 5 ends with 2. 2 power 4 16 ends with six.

[00:14:31]2 8 256 ends with six. So 2 cube ends with 8. 2 square ends with four. 2 1 ends with two. So 2 4 8 6 2 4 8 6 2 4 8 6 All on one's places will come uh will follow this pattern. 2 4 8 6 so period is four for every right now 3 power n 3 power 1 ends with 3 3 square ends with 9 3^ 3 27 ends with 7 3^ 4 4 81 ends with 1 again 3^ 5 243 ends with 3^ 6 729 ends with 9 3^ 787 ends 7 3 power 8 6 5 6 1 ends with 1.

[00:15:29]Well, all of you right? So, it's period 1. Now, four 4^ 1 4 4 square 6 ends with six. 4 cube 64 ends with four.

[00:15:40]4^ 4 4 256 ends with 6.

[00:15:45]So two five power n for any five power it ends with five only for any 6 ends with six only 7^ 1 7 7 square 49 ends with 9 7 cube 3 3 4 3 ends with 3 7 4 2 4 0 1 ends with 1.

[00:16:09]So periodicity four 8 power 1 ends with 8 uh 8 cube 4 [clears throat] 8 square 8 cube 52 52 ends with 2 52 8 4 6 8 4 2 12 2 48 Right.

[00:16:40]4 0 9 6 ends with 6. So 4 9^ 1 9 9 square 81 9 cube 729 9^ 4 4 is there here 65 61 1 4 periodicity 2.

[00:16:58]In other words also we have to write this table. In other words also we have to write this table.

[00:17:09]So how can we write in other words?

[00:17:21]So first of all see five power 5 power n ends with five for all n belongs to any natural number.

[00:17:40]Right? Next 6^ n ends with 6 or any n belongs to n right now see if you'll observe four power power one three odd to four even for four for even powers it is six for odd powers it is So four power even ends with six.

[00:18:19]Four power odd four power odd ends with four. In the same way 9 power even ends with what is 9 power even ends with one.

[00:18:37]9 power odd ends with 9. 9 power odd ends with 9. So 5 6 5 6 4 9 completed. 5 6 4 9 completed.

[00:18:52]5 6 4 9 completed. Now the remaining things. See [cough and clears throat] when a number divided by four When a number divided by four what are the possible remainders?

[00:19:08]When a number divided by four possible remainders.

[00:19:21]When a number divided by four, possible remainders are 1 2 3 0 divisible by four. Zero means divisible by four. Now it is in the form of division algorithm. 4 into n + 1 remainder. 4 into n + 2 remainder. 4 into n + three remainder and four multiple there is no remainder.

[00:19:52]Now by using these three these four results it can be applicable for 2 3 2 3 7 8.

[00:20:04]So 2^ 4n + 1 ends with ends with means units place 2.

[00:20:14]2^ 4n + 2 ends with 4. 2 4n + 3 ends with 8. 2^ 4n + 4n + 4 means 4n ends with 6.

[00:20:35]Next 3^ 4 n + 1 ends with 3. 3 power 4 n + 2 ends with 9. 3^ 4n + 3 ends with 7. 3 power 4 n ends with 1.

[00:20:58]>> [snorts] >> H. Next.

[00:21:05]7^ 4n + 1 ends with 7.

[00:21:11]7^ 4n + 2 ends with 9. 7^ 4n + 3 ends with 3. 7^ 4n + 4 ends with 1.

[00:21:30]8^ 4n + 1 ends with 8. 8^ 4n + 2 ends with 6. 8^ 4 n + 3 ends with 2. 8^ 4 n + 4 uh means 4 n ends with this is four. Sorry.

[00:21:57]This is two. This is six.

[00:21:59]So all of you write down the cyclicities.

[00:22:02]All of you write down the cyclicities.

[00:22:04]First you have to write this table form and then these values.

[00:22:09]Then we'll go to exercise.

[00:22:12]Make it fast. All of you make it fast. Make it fast.

[00:24:17]voice. Make it fast.

[00:25:35]>> [snorts] >> Somebody sir completed.

[00:27:09]So now we'll see some questions then we'll go to format theorem and its applications format theorem little firmat theorem and its applications we can go okay See 2026 power 202 uh 2000 27 power 2026 ends with 2026 power 2027 ends with that is the So how can we solve this 2026 power 20 27^ 2026?

[00:28:19]See it is nothing but ends with what is 7^ 2026 ends with is equal to 7 power 20 26.

[00:28:43]So 7 power see when it is divided by 4 4 into 4 5 is 20 4 6 24 2 remainder 4 5 is 20 4 6 24 two remainder see what is 7 power this one observe here power two 7^ 2 ends with 9.

[00:29:15]So ends with uh 9 since 7^ 4n + 2 ends with 9.

[00:29:28]So therefore units place of 227^ 226 is 9.

[00:29:43]For example, 20 27^ 226 plus 2029 power 2030 ends with another question 2027 power 2026 26 + 229 2030 ends with. So 726 + 9^ 230 ends with what? So 7^ 226 already is there 7^ 4n + 2 ends with 9 also 9 power even. What is 9 power even?

[00:30:40]One sub here 9 power even ends with one.

[00:30:44]So 9 power even ends with one. So therefore 7^ 20 26 + 9 power 20 30 ends with this one ends with 9 plus this one ends with one ends with zero.

[00:31:12]Units place of first part is 9. Units place of second part is one. The sum is 10. So ends with zero.

[00:31:21]Got it? 10.

[00:32:20]Now here uh okay in logarithms we'll do that concept number of digits in a uh expansion power number of digits we'll do You know divisibility rules. No need to discuss about divisibility rules directly. We can solve the questions.

[00:32:56]1 x 2 y 7 divisible by 9 x y number of possible pairs.

[00:33:15]This is our fifth class question in our in our syllabus in our curriculum it is fifth class question. So 1 + x + 2 + y + 7 is = see 1 + 2 3 3 + 7 10 10 after 10 what is the 9 multiple?

[00:33:37]It may be 18, it may be 27, it may be 36, it may be 45 and so on. Here we have to observe x comma y digits from 0 to 9. Digits from 0 to 9. So digits from 0 to 9. So here x + y = 18 - 10 or 27 - 10 or 36 - 10 and so on. Here we have to observe x + y max. How much sir? x + y max. You can take maximum x 9 and maximum y 9. So 9 + 9 18 the maximum more than that does not exist. More than that does not exist. So x + y = it may be 8 or it may be 17 or it may be 26 from here onwards not possible.

[00:34:45]So verify with digits because it is x y in between the numbers. So you can take zero also. So if possible 0 + 8 1 + 7 2 + 6 and so on 8 + 0. So total how many?

[00:35:03]0 to 8 9 pairs.

[00:35:07]Now x + y = 17. If it is eight, it is 9. If it is 9, it is eight. Here two pairs.

[00:35:17]So total number of required pairs.

[00:35:21]Total number of required x y is equal to 9 + 2 11 pairs.

[00:35:29]So it is basic question only. Try to solve all of you. Make it fast.

[00:36:45]I'm ready.

[00:37:24]So write down first question.

[00:37:28]Let n be a positive integer.

[00:37:33]Let n be a positive integer.

[00:37:48]Let n be a positive integer.

[00:37:52]Then number of number of possible remainders.

[00:38:03]Number of possible remainders.

[00:38:11]Number of possible remainders when 2013 power n 2013 power n - 1 1800 183 power n -781 power N + 17 74 power n / 23 divided by 23 divided by 23 Three, right? So that is a beautiful question this one.

[00:39:13]For this one, we have to use the formula. A n - b is equal to a - b * a n - 1 + b a n - 1 + a n - 2 b + a n - 3 b power 2 + so on b power n - 1 that is the expansion of a n + bn for any n integer n belongs to any natural number Sorry.

[00:39:46]Ah. N belongs to any natural number either two or three or four whatever it is. Okay. Now see 2013 power n minus 183 power n take out minus as common 1781 power n plus -744 power n.

[00:40:13]Okay. So this is a power n minus one 2013 - 18 3 into z minus j means this one this is all no need to write here 2013 nus1 plus 2013 n - 2 * 18 3 and so on like this. Now come to here 1781us 174 into y right so what is the difference here 2013 183 0 1 2 21 0 j minus it is I think 7 y is equal = to 7 * x that is 7 * whatever 7 multiple this is all uh let uh p is 7 multiple p is 7 multiple now p is equal to 2017 power uh sorry 2013 power n 2013 13 power n uh - 1781 power n minus as common 183 power n -74 power n see what happened here this is 2013 13 - 1781 * uh let it be j okay uh u is 183 - 17.

[00:42:25]So what is the difference?

[00:42:28]uh 213 781.

[00:42:35]So 2 3 this is uh 9 - 2 232 Okay.

[00:43:07]So, 232 U minus what is the difference here?

[00:43:13]183 74 9 uh 13 - 9 774.

[00:43:30]Sorry.

[00:43:34]774 183 774 9 2. So 29 V is it divisible by 29 or not?

[00:43:50]1. So see here 29 8 72 7 uh 16 + 7 23. So it is 29 multiple.

[00:44:04]It is 29 multiple. P is equal to 7 multiple as well as P is 29 multiple.

[00:44:33]Okay.

[00:44:40]As well as 29 multiple. So P is 7 multiple as well as 29 multiple because of 7A 29 co-prime P is multiple of 29 into 7 7 9 63 6 72 is 14 6 23 Okay the given number is 23 3 H. So therefore P is multiple of 23 for any N belongs to natural number.

[00:45:26]So therefore what is the possible remainder?

[00:45:31]Divisible means remainder must be zero.

[00:45:35]Right? Or possible remainder zero. Number of possible remainders only one. Number of possible remainders only one.

[00:45:48]Right. Good question. Write down all of you.

[00:45:53][clears throat] completed. Huh?

[00:48:22]So if uh second question if 13x + if 13x + 8 y is divisible by 7 Then then prove that 9x + 5 y is al uh 9x + 5 y is also divisible by 7.

[00:49:09]is also divisible by seven.

[00:49:34]H. What is the rule for divisible by seven?

[00:49:46]Rule for divisible by seven. You not divisible by two. You not divisible by three. You not divisible by four. You not divisible by five. You not divisible by six. You know divisible by 8. You not divisible by 9. You not divisible by 11.

[00:50:00]You not divisible by 12. But what is divisible by seven role?

[00:50:04]Simple.

[00:50:10]See double the last digit of the given number.

[00:50:23]Double the last digit of the given number.

[00:50:27]then subtract it from the rest of the number.

[00:50:42]from the rest of the number.

[00:50:45]If difference is seven multiple.

[00:50:55]If the difference is seven multiple given number divisible by 7 important note point if necessary can continue this process.

[00:51:28]for the obtaining results for the obtaining results. So process until you get the result. Okay. Right? Not on the rule and try to solve this question by using that rule.

[00:52:27]So 9 x + 5 y check.

[00:52:33]So 9 x + 5 y + 13 x + 8 y because it is 7 multiple.

[00:52:40]It is 7 multiple. 13x + 8 y is 7 multiple.

[00:52:48]So add to this one replace what we'll get here 22x + 13 y there is no seven in common there is no seven in common again 22x + 13 y 22x + 13 y is to be added to 13x + 8 y what we'll get here 22 + 13 35 x + 13 + 8 21 y. This is 7 * 5 x + 3 y. So 7 multiple.

[00:53:29]So therefore where we started 9 x + 5 y.

[00:53:31]Therefore 9 x + 5 y is divisible by 7.

[00:53:42]Okay. Do it all of you.

[00:54:29]for 713 uh 7 113.

[00:54:32]If a positive difference of the last three digits of the given number.

[00:54:40]If the positive difference of the last three digit and rest of the digits.

[00:54:48]If the positive difference of positive difference of last three digit to the rest of the digits to the rest of the digits of the given number of the given number. Rest of the digits of the given number is divisible by is divisible by either 7 or 11 or 13.

[00:55:48]Then the number is divisible by Then the number is divisible by 7 11 or 13.

[00:56:05]Then the number is divisible by 7 11 or 13.

[00:56:24]check these numbers.

[00:56:28]I'll give examples here. All of you uh verify after completion of the class as homework.

[00:56:36]630616 one number 58567 another number 28105 is another number 40LE 1. So these numbers have to check afterwards not now write down the questions after completion of the class.

[00:57:00]I think uh you have today 10th class classes from 9 to I think 1 or 12 12 branches. So mostly no time be 7 a.m.

[00:57:16]within 5 minutes you'll conclude the class. Write down these questions.

[00:57:22]Write down these questions.

[00:59:23]Two power 2026.

[00:59:29]Last two digits is what? 2^ 26.

[00:59:38]You know 2^ 4 into 5 6 + 2 ends with this is in the form of 2^ 4n + 2 ends with four. So this is ends with four.

[00:59:57]But here have to find last two places.

[01:00:09]Last two places.

[01:00:12]Very good. So two congregant to 2 mod 10. Last two places means last two places means when divided by when divided by 100 remainder is last two places when divided by 100 remainder is last two places. So 2 congrent to 2 mod 100 2 square congrent to 4 mod 100 once observe these are all the required only uh 2 power 3 congrent to 8 mod 100 2 power 4 congrent to 08 04 02 16 uh mod 100 on observe here.

[01:01:21]So how can how can we conclude these results?

[01:01:31]How can we conclude these results?

[01:01:34]So 2^ 6 congregant to 2^ 5 congrent to 32 mod 100.

[01:01:50]By using these results we can raise the powers. We can raise the powers. Suppose 2 10 congregant 2. What we'll get here?

[01:02:01]2 10 congregant 2.

[01:02:06]32 into 32 1 0 2 4 32² so ends with 24 mod 100 so 226 nearby right so no change we'll continue in the morning class so today evening There is no class. Tomorrow morning by 5:30 a.m.

[01:02:37]we'll take the class.

[01:02:40]Tomorrow morning 5:30.

[01:02:43]Okay. From here onwards we'll continue tomorrow. Right. Thank you.

[01:02:52]Have a nice day.