IIT JEE/MAINS MATH (CHANDRA KANT SIR)

Hello guys.. i was going through some questions for some paper work. Found these questions worth looking at ..

Q) m=(abc) & n= (def) are 3 digit numbers such that 7mn = the 6 digit number (abcdef) & neither 2 nor 3 are factors of n. Find m & n. (Russia). (abcdef)=(10^3)m + n, giving 7mn= 1000m + n...(*) (*) shows that m | n & n | m, leading to the surprising m=n. Go back to (*) to get m=n =143. Fact check: 7×143×143= 143143. Mathematics, in general & number theory, in particular, is full of idiosyncrasies.

Hello dear students and parents, With heavy heart, we would like to inform you all that we are resigning from the position of subject faculty from bakliwal tutorials effectively from today to open our own institute-Shikhshaguru Academy for jee ,neet , foundation for class 8,9,10,11,12 We wanted to take a moment to let you all know how much we’ve enjoyed our time here—and you’ve played a part in that. I know this is all of sudden since we weren’t left with any other option but to take this extreme step. It has been a very difficult decision for us, however we feel that in order to serve the students and parents better, it was the best option available. We genuinely think that through Shikshagurus we can at least double the efforts with more freedom and help students to reach greater heights. We know that all of you might be having many questions and doubts regarding our decision , hence we are arranging a counseling session followed by Q &A at both of our centres(Aundh and Fc).Our entire team will be present to answer all your queries and we hope that you all will give us the opportunity to hear our reasons. Time and duration are given below Fc centre: 1pm-2:30 pm 1st floor, status chamber wrangler Paranjape road, Off FC road, pune. https://goo.gl/maps/GkWz7GsasAMLRNUn9 Aundh centre:10-11:30 am Office number 14, 1st floor vedas centre DP road, near DAV public school, Aundh https://goo.gl/maps/vuQmGJD9ZozqS1sWA We will also inform the details of the courses we will offer at Shikshagurus and the admission process. Considering that for comp 23 its very critical time, we ensure that their preparation will not be disturbed due to this event. We are starting a Crash course at both the centre which will be in continuation of the topics which we were teaching. It will be totally free of cost (only one time registration fee of 1000rs) till mains shift 1 (24th January) For comp 24 also new batch will be announced at both centres and student will be allowed to attend one week class with only registration fee of 1000 rs. Registration can be done via 1. online mode at this link: google form https://forms.gle/Gq2CL4gUnsfzKCQq5 2.Offline mode at both centre Students and parents of pune have entrusted us since past 6 years and we hope that you all will keep having faith in us and give us the chance to be part of your dreams of becoming engineers, scientists and doctors. Thanks and regards , Team shikshagurus 1. Chandrakant Choubey 2. Bolla V Narayana reddy 3. Naveen Pandey 4. Arun Kumar Sahu Phone: 1. 8530200461 2. 8530200462 https://t.me/shikshagurus

Q)French school tournament. Solve √(x-1)+√(3x-5)+√(4x-7) = 4x-5....(1) Introduce auxiliary variables: a= √(x-1), b=√(3x-5), c=√(4x-7) Look at a^2+ b^2+c^2 =2(4x-5) -3 = 2(a+b+c)-3, via (1), which is rearranged as (a^2-2a+1)+ (b^2-2b+1)+( c^2-2c+1)= 0, ie , (a-1)^2+ (b-1)^2+(c-1)^2= 0, giving a=b=c=1, leading to x= 2. Motivate the contestants to use elegant methods rather than brute force.

The Baltic states,Lithuania, Latvia & Moravia, have lived in isolation. The Moravian, Kurt Gödel, was a friend of Albert Einstein at Princeton. Q) Baltic Olympiad. Is 712! + 1 a prime? Discussion. Wilson: If p were prime, then p | (p-1)! +1. In our case, 713 is NOT a prime, since 23 | 713. Scan numbers beyond 713 to search for a prime. 714, 715, 716, 717, 718 are composite. 719 is prime. Apply Wilson to get 719 | 718! + 1...(*). Now 718! = (712!)(713)(714)(715)(716)(717)(718) = (712!)(-6)(-5)(-4)(-3)(-2)(-1), mod 719 = (712!)(720), mod 719 = 712!, mod 719. From (*), 719 | 712! +1, showing that 712! + 1 is not prime. Exception to what we usually understand

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10 marker questions of prmo 2022

Mathematics has no geographical, religious & cultural borders. A question from Iran, the land of Q) Find all natural numbers, n, such that [n^2/3] is prime, where [ ] is the floor function. Ans: 3 or 4. Discussion: It is evident that we take three cases according to the remainder when 3 divides n. Case1: n= 3m, mεN. [n^2/3]= 3m^2, which is prime iff m=1. Thus n=3. Case 2: n= 3m+1,mεN [n^2/3]= 3m^2+2m ? = m(3m+2), which is prime iff m=1. Thus n=4. Case3: n=3m+2, mεN. This case fails?

a, b are positive integers such that 11|(a+13b) & 13 | (a+11b). Show that 143 |(6a+b). Find the minimum value of (a+ b). (Bulgaria). 11| ( a+13b) = (a+2b)+11b. Hence 11(a+2b). This implies 11| 6(a+2b)= (6a+ b) +11b. Thus 11|(6a+b). Likewise , 13 |(6a+b). Thus 143 |(6a+ b). Finding the minimum value of (a+ b) is a challenge. 6a + b = 143n ... (*), for some nεN. Add n to get 6a + ( b + n) = 144n. This shows that 6 |(b+n). Take b + n = 6m...(**), for some m ε Ν. (*) & (**) give a + m = 24n. Hence ( b + n) + (a+ m) = 6m + 24n. Isolate ( a+b), so that (a+b)= 5m + 23n. List the values of 5m + 23n in ascending order. Thus, 5m +23n = 28, 33, 38, 43,... 28 works for our hypotheses

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Q) Tokyo Institute of Technology, entrance exam, 1956. 0 0. Can we factor the expression (*)? As it stands, No. We may think of splitting 2. Thus 1+1-a-b-c+abc. Even now the factors elude us. What if we introduce a "fudge" factor ? (an additional term). Thus 1+1-a-b-c+abc-ab+ab. Rearrange terms. (1-a-b+ab)+(1-c+abc-ab). Factorisation emerges. (1-a)(1-b)+(1-c)(1-ab). All expressions in the parentheses are > 0. Hence the result. Note: Is the upper bound sharp? Observe that,as a,b tend to 1 & c tends to 0, the expression a+b+c-abc tends to 2. Indeed, 2 is the least upper bound

Q) n= abc is a 3 decimal digit number such that bac+bca+acb+cab+cba =3194...(*). Find n. (Canadian Olympiad). Discussion: Firstly, restore symmetry in the lhs of (*) by simply adding abc. Thus, abc+bac+...+cba= 3194+n...(**) Observe that (**) becomes 222(a+b+c) = 3194+n...(***), which implies 222|(3194+n). Shortlist the candidate values of n. nε{136,358,580,802}. Go back to (***): a+b+c= (3194+n)/222 > 3194/222 > 14, a big relief. Only one candidate has the digital sum > 14, viz, 358. Fact check that 385+538+583+835+853 = 3194.