IOQM(Pre RMO)_JY CLASSES_NUMBER THEORY_DAY 2

[00:00:09]Hi hello good evening Jri Krishna.

[00:00:13]So today day two. So yesterday we discussed about uh what is iOS [clears throat] and their properties. Based on the properties we did some questions. Okay.

[00:00:32]So now we'll continue the exercise today. We'll continue the exercise today.

[00:00:46]See how many of you uh received your material?

[00:00:52]Just raise your hands. How many of you received your materials?

[00:00:56]Because without material we can't do anything.

[00:02:24]Okay, let us start.

[00:02:44]>> [snorts] >> So write down the question all of you.

[00:03:02]Write down the question all of you.

[00:03:15]Let uh f be a real valued function.

[00:03:29]Let f be a real valued function.

[00:03:33]f(x) is equal to xq + 3x² + 6 x + 14 defined for all for all real x for all real x.

[00:03:54]It is given that a b are two real numbers.

[00:03:58]Let a comma b belongs to r. Let a comma b belongs to r such that f of a = 1.

[00:04:12]f of b = 19.

[00:04:16]Then a + b² is what?

[00:04:21]This is the question.

[00:04:24]So once read the question carefully.

[00:04:29]Let f be the real valued function. f of x is given f of x is given.

[00:04:38]So f of x is given xq [clears throat] + 3x² + 6 x + 14. And uh one more information is given. This is f of x is given and f of b is given f of a is given. We have to find a + b whole square. See one thing you have to observe for any question whatever the question whatever the question in mathematics the question will be divided into three parts.

[00:05:10]Question will be divided into three parts. So read the question carefully.

[00:05:15]This is called solving techniques.

[00:05:22]solving techniques. So for [clears throat] the given question divided into three parts. So what are the three parts? First one is what is given by reading the question carefully we can observe we can identify what is given right.

[00:05:45]The second part is what to be find.

[00:05:50]What to be find question.

[00:05:57]Now the third part is how the given data used to get required data by using this uh in this third point.

[00:06:22]mostly mostly either linear equations or exponents or simplifications simplifications by using number system.

[00:06:39]So what is given what do we find and how the data is bridge up how the data is used to to be find data. So while solving this either one or two steps only regarding to are related to that particular topic from next step onwards we have to use either linear equations or exponents or simplifications.

[00:07:01]So if you thorough in these three areas you can solve any question H. So, so what is our question?

[00:07:18]f(x) = xq + 3x² + 6x + 14. So even though it is number theory, even though it is number theory, so just we have to use algebraic operations.

[00:07:33]See a + 3 a² b + 3 a b² + b cube is there. It can be written as a + b whole cube. But we have to find a + b. So it should be converted in that form.

[00:07:49]Right? So xq + 3 into x² into 1 + 3 into x into 1 square + 1 cube.

[00:08:01]Right? Uh plus see once observe here x cube is there 3x² is there out of 6 x here 3x is there remaining 3x plus see 1 is there here 13 you have to write three here + 10 once observe.

[00:08:23]So what is f of x?

[00:08:26]x + 1 whole cube.

[00:08:29]See this is x + 1 whole whole cube. From here to here from the two terms we can take out three as common. This is x + 1 + 10. This is f of x.

[00:08:43]Now see f of a.

[00:08:46]So a + 1 cube + 3 into a + 1 + 10 it is given 1 f of a. Now find f of b. a + 1 sorry b + 1 cube + 3 into b + 1 + 10 is equal to 19.

[00:09:13]Got it? So by using these two results see once observe here what is f of a a + 1 whole + 3 into a + 1 is = - 9.

[00:09:29]f of b is equal to b + 1 cube + 3 into b + 1 is = 9.

[00:09:41]So while adding these two while adding these two let it be 1 this is two if you'll take 1 + 2 [clears throat] if you'll take 1 + 2 what we'll get here a + 1 whole q + b + 1 q + 3 into a + 1 + 3 into b + 1 is equal to 0 is equal to zero. So then you can solve it right down.

[00:10:27]So up to here you can easily so then now this is to be simplified. So, a + b whole cube, sorry, a + 1 whole cube plus b + 1 whole cube + 3 into a + 1 + b + 1 is equal to z.

[00:11:40]Make it fast.

[00:11:42]Right up.

[00:12:09]I'm not sure.

[00:12:51]So it is PQ + QQ + 3 into P + Q is equal to Z. What is PQ + Q + Q into P² + Q ² - PQ? This is aq + bq formula plus 3 * p + q is equal to 0. What we can take out as common? P + q as common p² + q ² - pq + 3 is equal to zero. The product of two brackets is zero. What does it mean? Either of them should be zero. Either this bracket zero or this bracket zero. So clearly this one not equal to zero.

[00:13:44]So that p + q is equal to z. So what is p? a + 1 plus what is q? b + 1 is = 0.

[00:13:56]So a + b is = -2. What we have to find?

[00:14:00]A + b square is equal to -2 square which is 4 that we have to find. Actually this question belongs to polomial concept but in number three it is given previously.

[00:15:55]Let PQ are set to be relative prime or co-prime.

[00:16:11]relative to prime or co-prime if and only if gcd means hcf of pq is 1. Here we have to observe either p or q.

[00:16:33]No need to prime. No need to prime.

[00:16:38]It may be prime that is secondary but no need to prime either P or Q no need to prime for a relative pair of relative numbers relative primes a pair of relative primes either of them no need to prime so the based on this concept write down this point all of you based on this concept we'll do one more question Write down all of you.

[00:17:20]Simple relative prime concept.

[00:17:24][clears throat] The question is let a be are two relatively prime integers let a comma b are relative prime numbers relative prime pair Let a comma b be a relative prime type where a greater than b greater than zero and [clears throat] aq - bq by a - b whole cube is equal to 70 by 3 sorry 73 by 3 then what is a minus b see how beautiful question it is question a - bq by a minus b whole cube is equal to 73 by 3 then we have to find a minus H.

[00:19:05]So at all.

[00:19:22]So a - bq by a - b is equal to 73 by 3.

[00:19:30]So you know these two formula a - * a power 2 + a + b power 2 by a - b cube a - b * a - b² a² - 2 a + b² this is equal to 73 by 3 [clears throat] or else so what we have to find a minus b² a minus b we have to find out. So keep like that a minus b only change a minus b along. Now these two gets cancel.

[00:20:08]Now numerator is a - b² minus a - b² + 3 a b by a - b² is equal to 73 by 3. Are you okay? Because here - 2 a b + 2 a b + 2 a b + a + 3 a b - 2 a b along with a square + b square becomes a minus b whole square and the here add and subtract 2 a. So 2 a + a 3 a b - 2 a along with a square + b square becomes a - b whole square. Now cross multiply it 3 * a - b² + 9 * a b is = 73 * a - b whole square.

[00:21:02]So 9 * a is equal to 73 * a - b² - 3 * a - b square.

[00:21:13]So 70 * a - b square is equal to 9 * a b. So since uh 70 comma 9 co-prime so both 70 and 9 are not divisible by same number so that these two are co-prime that implies compulsory a minus b whole square equal to 9 and a is equal to 70 what we have to find a minus b. So a minus b is equal to<unk> 9 that is 3.

[00:21:56]Okay.

[00:21:58]Right down.

[00:23:12]Okay. No change. Not chain.

[00:24:01]>> [clears throat and cough] >> One more question based on relative primes.

[00:24:07]One more question based on relative primes.

[00:24:19]Three positive integers greater than one.

[00:24:27]Three positive integers greater than one have a product of 27,000.

[00:24:40]have a product of 27,000 and are pair wise relatively prime.

[00:24:56]are pairwise relatively prime relatively prime. Then what we have to find sir? What is their mean to the nearest whole number?

[00:25:18]What is the mean to the nearest whole number?

[00:25:29]That is the question. Write down all of you.

[00:25:37]Three positive integers greater than one. Relative prime concept.

[00:25:42]See what is relative prime. Already we discussed about relative prime.

[00:25:47]PQR are set to be relative prime. If the GCD is one and this is the important point is no need either of them is prime. No need either of them is prime.

[00:26:00][clears throat and cough] So this is the question the point three positive integers three positive integers have a product of 2,00 27,000 and pair wise relative primes. So simple questions are 27,000 prime factoriation 27 means 3 cq into,000 means 2 cq into 5 cq which is 10 cq simple. So three numbers what are the three numbers 27 8 125 pair wise relative primes. So now means average 27 + 8 + 125 by 3. So 27 35 160 by 3. So 5 is 3 is.1.

[00:27:01]So nearest integer is 53 with the required answer.

[00:27:08]So simple question only solve it all of you.

[00:27:32]Find the number of counter examples for the given statement.

[00:27:48]So find the number of counter examples for the given statement. So you can give how many examples for the given statement? What is the statement here?

[00:27:57]So this is the statement.

[00:28:00]If n is odd positive integer, if n is an odd positive integer, n is an odd positive integer.

[00:28:17]The sum of whose digits is four.

[00:28:22]The sum of whose digits is four.

[00:28:30]The sum of whose digits is four and none of the digits is zero.

[00:28:48]Then n is prime.

[00:28:53]n is prime.

[00:28:57]n is odd positive integer whose digits sum is four. The number of none of the digits is zero. See what is given.

[00:29:11]What is given? n is an odd positive integer. So n must be odd and n must be prime. n must be odd. n must be prime.

[00:29:23]And the sum of whose digits is four. Sum of whose digits is. See we will check here 4 + 0. But it fails in the group. None of the digits is zero. None of the digits is zero.

[00:29:45]What fails? Next. 1 + 3 3 + 1. So 13 31 are primes.

[00:29:57]13 or and 31 are primes.

[00:30:02]Okay. So how many we'll get here? Two.

[00:30:06]Now 1 + 1 + 2.

[00:30:10]1 + 1 + 2. But here digit repeats fails.

[00:30:15]Digit repeat fails. So what are the possibilities for four? These three only 4 + 0 1 + 3 or 3 + 1 1 + 1 + 2. So number of required numbers. How many counter examples are there? Only two.

[00:30:37]That is the concept.

[00:30:42]See one statement is given.

[00:30:45]According to this statement, how many examples counter examples you can generate? That is the question. So how many counter examples we can generate?

[00:30:54]Two, 31 and 13. Right?

[00:33:26]Okay. Now see next question.

[00:33:35]Very simple question.

[00:33:44]Using the digits.

[00:33:49]Using the digits 1 2 3 4 5 6 7 9. Using the digits 1 2 3 4 5 6 7 9 to form four twodigit primes.

[00:34:10]Four two digits primes.

[00:34:14]with each digit used exactly once.

[00:34:18]With each digit used exactly once, then the sum of then the sum of four prime numbers is is 10 * a then find a this is also previous year question for two marks for two marks this is previous year year question so two marks So two 4 6 remaining five If one set is it either 2, 4 or 6, it is divisible by two.

[00:37:05]In the same way, if one is it is five, it's divisible by Five.

[00:37:18]So on's place 2 4 6 first given digits 2 4 6 5 given digits.

[00:37:33]So two remaining a 1 3 uh sorry 7 9 [clears throat] suppose 21 divisible by 21 divisible by 7 2 3 maybe and 27 divisible by 3 2 9 maybe. So either 2 3 or 2 9. Okay. Now 4 4 3 because 4 1 or 4 3 or 4 7 H. Next these are the possibilities what we are checking. 6 6 1 6 6 7 6 3 divisible by 3.

[00:38:36]Next 5 uh 53 or 59.

[00:38:43]Okay. H. So if you'll observe here, if you'll observe here, 29 29.

[00:38:55]Okay. 9. So this is not possible.

[00:39:00]43.

[00:39:03]No 53. Not possible.

[00:39:06]Okay. Next. If you'll take uh uh four already used so 53 must be required.

[00:39:18]53 59.

[00:39:21]So each digit used exactly once.

[00:39:34]So 59 not possible. So 53. Okay.

[00:39:40]So 43 not possible. Now the thing is about 41 and 67. So if you will take 40 1 you have to consider 67 or 47 61. So what is the sum? 29.

[00:39:57]So 41 67 53. So add this 10 + 10 20 2 4 8 14 19. So 190 is equal to 19 into 10. So 10 a that implies a is equal to that is equal to 10 a a= 19 what we have it is a logical question only This is beautiful question.

[00:41:04]Let m comm n be two positive integers.

[00:41:15]m n2 positive integers such that 75 m is equal to n^ 3. Then what is minimum value of m + n?

[00:41:29]What is minimum value of m + n?

[00:41:36]It it must be a perfect cube. It must be a perfect cube.

[00:41:49]So 75 this is 5² into 3^ 1 5 square into m is equal to because of 5 and 3 relative primes. So it must be 5 cube into 3 cq. This can be written as 5 into 3 cube. This is n cq. So n is equal to 15. What about m? See 5 square is there.

[00:42:18]So 5^ 1 is required. 3^ 1 is there here.

[00:42:23]3 power 1 is there. So what we require?

[00:42:27]3².

[00:42:28]So that implies m = 95 is 45.

[00:42:34]So what we have to find m + n. This is 45 + 15 which is least value is 60.

[00:42:44]Do it all of you.

[00:42:50]Make it fast.

[00:44:15]Okay, next question is it is also based on relative prime.

[00:44:24]Let k greater than zero.

[00:44:31]I K is equal to 1 0 0 0 and so on 0 64 this is K number of zeros K number of zeros that is question here K is K is number of zeros between 1 and six. [clears throat] Number of zeros between 1 and six.

[00:45:22]Let n of k be the number of factors of n of k be the number of factors of two in the prime factorization of number of factors of two in the prime factorization of I K the prime factorization of I K.

[00:45:58]What is the maximum value of N of K?

[00:46:02]What is maximum value of N of K?

[00:46:12]It's not a difficult question. It is moderate. Easy to moderate. But k greater than zero i k 1 followed by k zeros and 64 that is the number.

[00:46:33]So n of k is number of factors of two means how many times two will come?

[00:46:41]How many times two will come? Number of factors of two. How many times two will come?

[00:47:44]Okay.

[00:47:47]See suppose 1 0 1 0 64 is there. So 1 0 is there. This is 10^ 3 + 64.

[00:48:04]1 0 is there 10^ 3 + 64 suppose 1 0 0 64 two zeros are there this is 10^ 5 + uh 10^ 4 + 64 suppose 1 0 64 there are three zeros 10^ 5 + 64 see if suppose three zeros 3 + 2 2 0 2 + 2 1 0 1 + 2 so Now I K is equal to 1 followed by K 0 1 followed by K 0 64. This is 10^ k + 2 + 64 10^ k + 2 + 64.

[00:48:57]Okay. Huh? So this is 2^ k + 2 into 5 power k + 2 + 2^ 6 2 power 6. So this can be done in two ways.

[00:49:14]So if you'll take 2^ 6 as common 2^ 6 by 6 by 6 means 4 - k into 5 power k + 2 + 1 this is one way and the other one is 2 power k + 2 has common 5 power k + 2 + 2 power 6 - k minus this is by 6 this is k - 4 this is 4 - k okay uh see here how many six are there here so number of factors as per number of factors concept yesterday we discussed number of factors see P1 power alpha 1 that is alpha 1 + 1.

[00:50:20]So here it is number of factors 6 + 1 7 or else k 2 + k + 2 it must be less than or equal to h simply you can take this 6 + 1 7 factors so n of k maximum n of k is 7.

[00:50:58]I know Right.

[00:52:54]So it is based on number of factors.

[00:52:56]Question based on number of factors for some positive integer n.

[00:53:04]Write down all of you. some positive integer n or for some positive integer n. The number the number 110 n cq has 110 factors. This is the concept based on number of factors. 110 factors.

[00:53:35]Also this is called positive devices including one end itself.

[00:53:48]Including one end itself.

[00:54:01]The number 81 n^ 4 has d number of factors.

[00:54:18]What is d by 5?

[00:54:23]What is d by 5?

[00:54:27]It's a good question.

[00:55:07]Okay.

[00:55:19]Let n = p1 power alpha 1 * p2 power alpha 2 * p3 power alpha 3 so on number of positive divisor of n is equal to alpha 1 + 1 * alpha 2 + 1 * alpha 3 + 1 * so on that's the required concept right now what is given 110 110 n cube 110 n cq is equal to 11 into 2 into 5 into n cq into n cq has 110 factors.

[00:56:09]Listen all of you carefully. So here 110 factors. So what is 110?

[00:56:17]That is equal to 2 into 5 into 11. 2 into 5 into 11. So 2 into 5 into 11.

[00:56:38]So but as per formula, see this is the formula. As per formula, how can we split this? 1 + 1 into 4 + 1 into 1 + 1 into 4 + 1 into 10 + 1.

[00:57:07]Right?

[00:57:08]So see suppose 11 into 2 into 5 into P1 into P2 power 3 for example right so if you'll observe either of these two here P1 comma P2 belongs to either 11, five or two set.

[00:57:45]P1 P2 either 11 or five or two irrespective but any of two out of three any two must be P1 P2 P1 comma P2 from this suppose see uh 11 into suppose P1 P1 cube let it be P1 for example so P1^ 4 P1 power 4 So this is 11^ 1 1 + 1. Okay. P1^ 4 4 + 1. Okay.

[00:58:22]So P1 power 4 means here it is power one. So it is three.

[00:58:28]It is three. Okay. Now five ah so if if you'll take uh sorry it is three only uh a1 if you'll take power 3 then what we'll get here p2 power 9 and this is 1 p2 power 10 this is 10 + 1 right so therefore n is equal to p1 into p2 power 3 R P1^ 3 into P2 power 1 any of them you can take this is N right now uh 81 N^ 4 81 is 3^ 4 into N means P1 into P2 power 3 whole^ 4 so this is 3^ 4 into P1^ 4 into P2 power 34 is 12 so number of factors.

[00:59:32]So how many factors we'll get here?

[00:59:33]Number of factors. See 3 + 1 * sorry 4 + 1 * again 4 + 1 * again 12 + 1. So 5 into 5 into 13. So 25 into 13. This is number of factors D. So what we have to find dx 5 that is equal to 25 into 13 by 5 that is 65 by the required answer.

[01:00:11]The thing is this adjustment 5 into 2 into 1 by the number of factors. So 2 1 + 1 5 4 + 1 11 10 + 1 by taking this formula.

[01:00:26]So 11^ 1 1 + 1. Okay. So P1 power 3 3 this is 1. So 3 + 1 4 P1^ 4 here power 3 into 3 9 9 + 1 10 + 1 11. So that n value is either p1 into p 1 p2 power 3 or p1^ 3 into p2 power 1. So substitute here n value for 81 n^ 4 then we'll get the number of factors. So this is also PQ and beautiful application. This is also P YQ number theory.

[01:01:05]I QM.

[01:01:11]Okay. Solve it. All of you make it fast.

[01:04:27]Okay, see next question.

[01:04:40]Number of positive twodigit integers.

[01:04:48]Number of positive twodigit integers are factors of 2^ 24 - 1.

[01:05:02]Number of this is the last question for today's class.

[01:05:06]Number of twodigit positive integers which are factors of 2^ 24 minus 1. Here we have to use the formula a power n - power n = a - b * a n -1 + a n - 2 b + and so on + b n -1 this is one formula otherwise a - b that is a - b into a square + a + b² as well as a + bq a + b into a² - a b + b² and also if necessary we have to use a square minus b square that is a + b into a minus b. So the three as per requirement we have to use these three results. So first of all 2^ 12 square - 1 square that is 2 12 - 1 into 2 12 + 1.

[01:06:22]This is 2^ 6 square - 1 square into this is 2^ 4 whole cube + 1 cube a cube + bq this is 2^ 6 - 1 into 2 6 + 1 into 2 4 + 1 into 2 4² - 2 4 into 1 + 1² Now this is 2^ 3 square - 1 square into 2 6 2 cube + 1 cube into this is 17 into this is J here J is greater than 100 2^ 4 4 2 6 64 maybe 2^ 4 8 256 256 - 16 + 1 g rather than 100 no need to consider because what we have to find two digit integers which are factors. So now this is 2^ 3 - 1 into 2^ 3 + 1 into uh this one 2^ 6 aq + bq better 2 + 1 into 2² - 2 into 1 + 1 square aq + bq into 17 into j which is more than 100 Right. Now this is 2 cube - 1 7 2 + 1 9 2 + 1 9 2 + 1 5 this is uh 2^ 4 16 + 1 17 - 4 13 into 17.

[01:08:24]So 7 into 3 into 3 into 5 into 13 into 17. So twodigit numbers we have to write 17 17 into 3 17 into 5. So with 17 combination there are three 13 into 3 13 13 into 5 13 into 7 these are four got it all of you now right next uh five 5 into 7 5 into 9 5 into 3 again 3 again 3 H next.

[01:09:26]So any more?

[01:09:33]So 7 7 into 3 7 into 9 this is 2. So total 12 factors because twodigit factors this is all n.

[01:09:55]So this is n. So factors of n.

[01:10:03]These are factors of n which are two digits which are twodigit numbers.

[01:10:21]That is beautiful question.

[01:11:03]Okay, we'll do one more question.

[01:11:12]Make it fast. Boys, we'll do one more question.

[01:12:31]Right. Completed.

[01:12:46]Okay.

[01:12:47]is the next question.

[01:12:53]Let n be a positive integer for which let n be a positive integer for which let n be a positive integer for which the sum of its two smallest factors the sum of [clears throat] its two smallest factors is four. Sum of its two smallest factors is four. And the sum of its largest two largest factors.

[01:13:37]And the sum of its two largest factors is 24.

[01:13:52]2 4.

[01:13:55]Sum of its two largest factors is 2 4.

[01:14:00]Find n by 3.

[01:14:03]We don't know how many factors but okay there is a beautiful concept is there here see the solution Two smallest factors whose sum is four. What are the factors?

[01:14:48]Let a comma b such a + b = 4. So a = 1, b is = 3.

[01:14:57]Right?

[01:14:59]Let's factors be C comma D such C + D is equal to 2 4 C C + D is equal to 2 4. Clearly see observe here uh see suppose uh 24 is there for example 1 2 3 4 6 8 12 24 so smallest are 1 2 smallest 1 2 so largest is let n be the number n be the number 24 be the number largest is once go through here 24 by least one and the second largest is 24 by first least then second least in ascending. Okay. Now, so what about C? Here it is N.

[01:16:07]Here it is C. Here it is D. Here it is A. Here it is B. So this is N by A and this is N by B.

[01:16:19]So what are A? 1 3. What are A? 1 3.

[01:16:26]So C + D is equal to 2 4. C is N by 1. D is N / 3 is equal to 2 4 N + NX3 4 NX3 is equal to 24 NX3 what we have to find 51 see how beautiful question it is simply given that two smallest two two smallest obviously 1 and three only there is no way 2 + 2 but two different factors two is only one factor repeated 2 + 2. So 1 + 3 is the only case. So largest last two largest two largest factors. So first largest factor is n by least n by least.

[01:17:20]First largest second largest n by second least c right.

[01:17:32]So okay make it fast. So tomorrow we'll continue uh prime numbers. So we can conclude tomorrow prime numbers and we will start second concept in number theory that is divisibility rules and cyclicity.

[01:17:46]Tomorrow's topic is divisibility rules and cyclicity.

[01:17:53]You know divisibility rules and also you know cyclicity h. So tomorrow's topic these two we will discuss.

[01:18:10]So today's session completed.

[01:18:13]Hope you understood well.

[01:18:23]>> Okay. Right. Thank you.