IOQM(Pre RMO)_JY CLASSES_Number Theory_day 03_congruency_cyclicity

[00:01:58]Good morning children.

[00:02:00]So yesterday because of uh because of some my personal problems class was not held. So that uh now we'll continue the class.

[00:03:28]Okay, after uh two two three questions we'll continue the next concept. Chapter divisibility rules and cyclicity. We will discuss before that solve this question. Number of factors of uh number of positive divisor of number of positive divisor of 2008 whole cube.

[00:04:21]Listen carefully. 2008 whole cube + 3 into 2008 * 2009 + 1 cube.

[00:04:39]Right? That is the question. We have to find number of factors of that expansion.

[00:04:52]Okay. So, 2008 whole cube + 1 cube + 3 into 2008 into 1 into 2008 + 1. This can be written as a cube and is of the form see a + bq + 3 a b * a + b. This is nothing but uh a + b whole cube. So here a is 2008 and b is 1. So 208 + 1 whole cube.

[00:05:34]So this is 209 whole cube. 2009 whole cube. So is it divisible by uh it is not divisible by 2 3 5. Let's try with seven.

[00:05:52]7 7 2's 14 6 0 7 7 8 56 4 9 7 7 9 7 7 Whole cube. So that is again 7 into 7 into 7 4 7 1 whole cube whole cube whole cube because this is a + b whole cube again one more cube is that 9.

[00:06:29]Okay. So now it is 7² power 9 into 41 power 9. Okay. So this is 7^ 18 into 41^ 9. So what is number of factors formula?

[00:06:47]You know number of factors prime power alpha 1. So alpha 1 + 1 * prime power alpha 2 that is alpha 2 + 1. So 19 into 10. So there will be 190 factors.

[00:07:07]Okay.

[00:07:09]Do it all of you.

[00:09:35]See one more question.

[00:09:38]Let n be the largest integer.

[00:09:42]Let n be the largest integer.

[00:09:54]Let n be the largest integer that is the product of that is the product of three distant primes.

[00:10:20]three distant primes D, E, and 10 D + E. 10D + E where D comma E single digits.

[00:10:47]Then maximum largest number. Then what is the sum of digits of n?

[00:10:56]Then sum of digits of n.

[00:11:02]All of you try to solve this question.

[00:11:04]Sum of digits of n.

[00:11:12]Sum of digits of n.

[00:11:32]product of three distant primes. Product of three distance primes two uh de. So this is 235 D comma E belongs to 2357.

[00:11:49]Now using these digits 23 37 53 73 with the possible primes with the possible primes. So the given question is n is largest integer. n is largest integer. So that here we have to consider see 7 into 3 5 into 3 21 15 this is also 21 uh this is five sum. So product so 73 into 7 into 3 what is the product here?

[00:12:47]73 into 21. So 73 6 14.

[00:13:00]So 153 n max.

[00:13:05]So sum of digits what is sum of digits? 1 + 5 + 3 + 3 which is 12.

[00:13:19]Simple question only.

[00:13:21]Solve it.

[00:15:20]Okay, all of you try to solve this question.

[00:15:30]If a + b + c is equal to 66, a b + b c + ca is equal to 1 0 7 1 where a b c are r primes where a comma b comma c are primes.

[00:16:02]Find ABC into B into C.

[00:16:39]all of you try to do this question here. You have to use the concept 1 + a + b + c + a b + b c + c a + a b c.

[00:17:04]This is 1 + a.

[00:17:07]If B is common from these two, uh, B is common, 1 + A plus C is common, 1 + A plus U, BC is common 1 + A.

[00:17:36]So 1 + a as common 1 + b + c + b c.

[00:17:42]So 1 + a * 1 + b * 1 + c.

[00:17:48]Here given that a b c r prime given a b c r prime.

[00:17:58]So you can solve the question.

[00:18:01]So try to do homework.

[00:18:04]Write down. Make it fast.

[00:18:22]Smallest positive integer x such a x² - x + 5 x² - x + 5 is not a prime.

[00:18:49]So by verification we can solve this.

[00:19:02]This is also how to do homework.

[00:19:19]m belongs to n m² + n² + 2 mn - 2013 m - 2013 n - 2014 = 0 number of pairs of mn Right. So simple question only. This is m + n² - 2013 as common m + n - 2014 =0.

[00:20:08]Right? Let it be a a² - 2013 e - 2014 =0 by factorizing this is a - 2014 * a + 1 is = 0 right. So a a means m + n m + n = 2014 m + n is equal to 1 because of m and natural numbers this pairs. So from starts from 1 here 1 + 2013 and so on 2013 + 1. So total how many pairs? 2013 pairs will come that is also simple question.

[00:21:02]So remaining questions uh are there in our worksheet almost uh based on this concept near about 20 questions are there that you have to solve.

[00:21:18]Near about 20 questions are there in the worksheet. kindly collect the worksheet from our branches and solve the worksheet.

[00:21:25]Now we'll uh discuss about divisibility rules and cyclicity.

[00:21:32]You know divisibility rules. Write down very Congratial and cyclicity.

[00:24:16]So this is m divides a m divides a this is say m by a m upon a Means m is divided by a.

[00:24:49]M divides a means.

[00:24:54]A is divisible by m.

[00:25:01]Observe the difference here. m divides a and m is uh a is divisible by m. A is divisible by M. That is in other words A is divisible by M. In other words words M is a factor of uh this is an exact divisor of A.

[00:25:28]Exact divisor of A. M divides A. So a is equal to what is a k * m where k belongs to z uh k belongs to n multiple of m that is a is multiple of mic difference m and m upon a that is mid but here m divides a means a is divisible by m means m is a factor of a so a is a multiple of m this is very uh important point right Oh, let a comma b be two integers and m is any positive integer.

[00:26:53]M is any positive integer.

[00:26:57]M is any positive integer and AB are two integers.

[00:27:04]Okay.

[00:27:11]When a divided by m a divided by m if the remainder is remainder is b remainder is b then see this is the important point a is set to be A is set to be congrent B.

[00:27:52]A is set to be congrent B modulo M.

[00:27:59]This is what we have to discuss. A is set to be congrent to B modulo M. So that uh a congrent to b mod then m divides a minus b m divides a minus b that is when a is divided by m b is the remainder. So a is equal to k m + b.

[00:28:38]This is remainder. This is quotient.

[00:28:44]This is divisor.

[00:28:47]This is dividend.

[00:28:52]Okay. So this is about congrent. A congrent to b modulo m means m divides a minus b. Here a minus b is divisible by m.

[00:29:10]In other words, already we have written here a b are two integers. So here uh uh where a divided by m the remainder is b. Okay. So write down the concept hungry.

[00:30:30]Here we'll see one more example. There is a small example here.

[00:30:51]Very small example.

[00:30:53]Find the least positive integer.

[00:31:06]Find the least positive integer mod 11 to which 282 is congrent.

[00:31:32]282 is congrent.

[00:31:36]So we didn't understand this terminology but you can solve it. In other words, if I tell in other words you can solve it easily. But this terminology is a little bit difficult to us to understand. Find the least positive integer mod 112 which 282 is congrent. So it is somewhat difficult to us to understand but see 282 congrent to x 11 very good. So chra as per this uh statement you know dividend you know uh divisor you have to find remainder. So 11 282 2's 22 62 55.

[00:32:46]So 282 congregant to 7 mod 11. So what is the least number?

[00:32:55]Remainder is 7.

[00:33:00]Write down. Then we'll see the properties of uh then we'll see the properties of congrens. Make it fast.

[00:35:44]Okay, we'll [clears throat] see the properties.

[00:35:58]Two inteious are congrent modulo m.

[00:36:04]Two inteious are congrent modulo m. Write all of you. Two integers are congrent modulo m if and only if if and only if they leave the same remainders.

[00:36:37]They leave the same remainders when divided by m when divided by m.

[00:36:59]So let a is equal to q1 m + r1.

[00:37:08]B = Q2 M + R2 Q2 M + R2 H. Two inteious are congrent to modulo M. Two inteious are congrent to modulo M.

[00:37:30]So let A congrent to B mod M.

[00:37:37]A congrent to B mod divides A minus B M divides A minus B. So in other words it is M divides CA and M divides B.

[00:37:57]Then only M divides A minus B.

[00:38:01]Now see here M [clears throat] divides A minus B. Q1 M + R1 minus Q2 M + R2.

[00:38:15]This is division algorithm.

[00:38:18]This is what we have written. Division algorithm divi dividend equal to divisor into quotient plus remainder. Diviser into quotient plus remainder. This is division algorithm. So now come to the point m divides M into Q1 - Q2 plus R1 - R2. [clears throat] See by using this property m divides a + b means m divides a m divides b right.

[00:39:07]So by using this property m divides m * q1 - q2.

[00:39:15]So this is a multiple of this is multiple of multiple of m very clear and uh m divides r1 - r2 m divides r1 - r2 so m divides r1 - r2 R2 means R1 - R2 divisible by M. R1 - R2 divisible by M.

[00:40:02]See R1 less than M but R1 less than M R2 less than M as per this definition.

[00:40:15]So remainder is always less than divisor. Reminder is always less than remainder less than divisor.

[00:40:25]Remainder less than divisor.

[00:40:28]H R1 - R2 divisible by M.

[00:40:33]What does it mean?

[00:40:36]R1 - R2 divisible by M. So R1 - R2 is multiple of M. R1 - R2 is multiple of M.

[00:40:50]So R1 - R2 greater than= M.

[00:40:58]But here R1 less than M R2 less than M.

[00:41:01]How it is possible? R1 - R2 less than M.

[00:41:05]No way. Here R1 - R2 less than M. R1 - R2 less than M is not true.

[00:41:25]equal.

[00:41:30]So that implies R1 equal to R2.

[00:41:36]What we have to prove? They leave the same remainders when divided by M.

[00:41:43]A R divided by M A divided by M remainder R1 B / M remainder R2. These two remainders are same. This is very useful statement.

[00:41:57]In other words, it is also a definition of congregant. It is also a definition of congregant.

[00:43:49]Rad. [snorts] The second one is when a number and it's cube divided by Six leaves remainder same in each case leaves remainder same in each case. That is the next point when a number divide and it's cube.

[00:45:57]So let n be the number n cq minus n congregant to leaves remainder six in each case that we have to uh leaves remainders uh when a number and it's cube divided by 3 divided by 6 leaves remainder Same in H case.

[00:46:39]What about n - n? So n has common n² - 1. So this is n + 1 into n into n -1.

[00:46:49]This is product of r consecutive integers.

[00:47:03]Product of our consecutive integers is divisible by what?

[00:47:11]This is important point is product of all consecutive integers is divisible by what?

[00:47:24]Answer choice.

[00:47:27]Product of all consecutive integers is divisible by what?

[00:47:40]No, no, no. R consecutive R consecutive integers divisible by R factorial R factorial R factorial so here it is three consecutive integers so n cq - n divisible by nq - n divisible by 3 factorial that is 6 3 factorial means 6. So nq - n divisible by 6 nq - n congrent to 0 mod 6.

[00:48:24]So n cq congrent to n 6 n cq congrent to nongren to ba n having same remainder when divided by six. 3 factorial means six.

[00:49:02]Third property.

[00:49:05]If A congrent to B mod A congrent to B modium, then B congrent to A modium.

[00:49:18]A congrent to B modium, then B congrent to A modium.

[00:49:33]So a congrent to b m means m divides a minus b then m divides minus of b minus a then m minus of b minus a b minus a so that implies b congrent to a model AGant to A M that is M divides A minus A 0 divisible by anything 0 divisible by M 0 by M.

[00:50:28]A congrent to B mod M congrent to C mod then A congrent to C modium. This property is very important property.

[00:50:47]A congrent to B modm, B congrent to C modm. Then Cong A congrent to C modm.

[00:50:53]See here it is M divides A minus B. M divides A minus B. Here M divides B minus C. M divides B minus C. Then M divides. So Mus B is multiple of M.

[00:51:14]B minus C is multiple of M. So m divides a minus b + b minus c. See we have written here this property m divides a + b means m divides a m divides b right. So by using this property m divides so now these two gets cancel m divides a minus c that implies a congrent to cm hold it all of you write down these five properties make it fast make it fast all of And one more thing sir if uh uh today evening also if possible I'll take the take the class otherwise Tomorrow because of Sunday um morning 6:00 to 8 I will take the class 2 hours class 1 hour 1 hour 6 to 7 and then again uh one more link will be sent for 7 to 8 so it's possible today I'll take the class otherwise tomorrow morning your class timings will be 6 to 8 after 6 to 8 online class you can go to your Sunday class Sunday class uh as usual there is no change in the Sunday class schedule that is from 9 to 1 I think uh 9 to 1 so but 6:00 to 8 you have online class after 6:00 to 8 you can go otherwise 6:00 to 7 morning and evening 6 to 7 I'll take the class tomorrow Group schedule.

[00:53:42]Okay. Write down. Make it fast.

[00:53:45]We're going to change A congrent to B modium.

[00:54:43]A congrent to B modium.

[00:54:53]Then a + congrent to b + c modm as well as a congrent to bc model for c is any integer whether it is positive or negative.

[00:55:19]Simple agree into B mod M. Okay.

[00:55:29]M divides A minus B. M divides A minus B. So it can be replaced as M divides A + C minus of B + C. Here C gets cancel. C gets cancel.

[00:55:45]So, a + congregant to b + c mod m.

[00:56:05]question.

[00:56:19]Next, a congrent to b mides a minus b. So, a minus b is multiple of m.

[00:56:33]So therefore m divides c * a minus b because of a minus b is a multiple whatever the c nothing to worry because a minus b * c is divisible by m. So m divides a c minus b c. So a congrent to b c mod that is sixth property. Now seventh property a congrent to b m congrent to d mod.

[00:57:14]Okay this means m divides a minus b m divides c minus d. By taking these two see the previous property this property uh their sum also m divides a + c minus b + d. So therefore the property is a + congrent to b + d modm.

[00:57:49]So a congrent to bm congrent to dm m then a + c congrent to b + d modm right down. So okay tomorrow we'll continue the class if possible I'll take the class in the evening otherwise tomorrow we'll continue now the time is almost 7 uh 6:30 because of uh you have to school so that h right thank you write down uh if possible evening I'll continue otherwise tomorrow morning 6 to 7 first class Tomorrow uh tomorrow two sessions will be there. First session we'll take at 6:00 a.m. 6 to 7 a.m. Get ready for the class. Right. Thank you.