@@neha_845 ye lo didi dekh lo aap 😊( 2222⁵⁵⁵⁵+5555²²²²/7) 2222/7= Remainder 3 5555/7= Remainder 4 3⁵⁵⁵⁵/7+4²²²²/7 Using fermat's theorem 7-1=6 5555/6 = remainder 5 2222/6 = remainder 2 3⁵+4²/7 243+16/7= 259/7 hence it is completely divided by 7 and the remainder is 0. 🙏🙏🙏🙏
( 2222⁵⁵⁵⁵+5555²²²²/7) 2222/7= Remainder 3 5555/7= Remainder 4 3⁵⁵⁵⁵/7+4²²²²/7 Using fermat's theorem 7-1=6 5555/6 = remainder 5 2222/6 = remainder 2 3⁵+4²/7 243+16/7= 259/7 hence it is completely divided by 7 and the remainder is 0. 🙏🙏🙏🙏 Thanks a lot for a wonderful session.❤️
@@I.T.O.Deeparawat-sy5iy 55⁵⁶ iski power 57 aur ise 17 se divide Krna ho to By euler thoerum 17 ek prime number hai it means (17-1) = 16 se ab power ko divide Krna hoga . To ab 55 ko na dekh kr uske power 56 ko 16 se divide kre 56⁵⁷/16 = (8²)²⁸ × 8¹/16 (64)²⁸×8/16 0×8/16 = 0 answer
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Homework :- Ans - 31 : (a) For the first time, I read competitive maths with such depth and conceptual way from someone. And You cleared all the queries Sir.🎉❤
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@@neha_845 ye lo didi dekh lo aap 😊( 2222⁵⁵⁵⁵+5555²²²²/7)
2222/7= Remainder 3
5555/7= Remainder 4
3⁵⁵⁵⁵/7+4²²²²/7
Using fermat's theorem
7-1=6
5555/6 = remainder 5
2222/6 = remainder 2
3⁵+4²/7
243+16/7= 259/7 hence it is completely divided by 7 and the remainder is 0.
🙏🙏🙏🙏
@@gautamkumar-vs4vt thank you bhaiya
( 2222⁵⁵⁵⁵+5555²²²²/7)
2222/7= Remainder 3
5555/7= Remainder 4
3⁵⁵⁵⁵/7+4²²²²/7
Using fermat's theorem
7-1=6
5555/6 = remainder 5
2222/6 = remainder 2
3⁵+4²/7
243+16/7= 259/7 hence it is completely divided by 7 and the remainder is 0.
🙏🙏🙏🙏
Thanks a lot for a wonderful session.❤️
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Jese 55 ki power 56 iski power 57 or isko 17 se divide karna hai
@@I.T.O.Deeparawat-sy5iy 55⁵⁶ iski power 57 aur ise 17 se divide Krna ho to
By euler thoerum 17 ek prime number hai it means (17-1) = 16 se ab power ko divide Krna hoga .
To ab 55 ko na dekh kr uske power 56 ko 16 se divide kre
56⁵⁷/16
= (8²)²⁸ × 8¹/16
(64)²⁸×8/16
0×8/16 = 0 answer
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Hw- o remainder
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H. W.= after solve:3*5+4*2/7 : -2+2=0R🙏
@@preetisclasses5631 8! /11 please solve this q .
@tradewith-GP 10!/11=1/10,10×9×8!/11:90×8!/11:+2×8!/11 = 5 R (bcoz jab 5 se × karenge tabhi to 10 banega kyuki 10!/11= -1ya 10 hota h, so 8!/11 ka remainder 5 hoga)
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(23)¹⁷⁹/ 13 Iska bta do . By fermat's theorem.
@@tradewith-GP (-3)*11/13 (bcoz 179/12-Fno. = R11)
-3*3×-3*3×-3*3×-3*2/13 = -1×-1×-1×_9/13
9/13= 9 R
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Ans- 0
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2222⁵⁵⁵⁵+5555²²²² /7.2222/7=remainder 3.5555/7=remainder 4. 3⁵⁵⁵⁵+4²²²².using fermat theorem 5555/6=remainder 5.2222/6=remainder 2.3⁵+4²/7=243+16/7=259/7=remainder 0😊
259 is completed divided by 7,so the remainder will be0
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HW-31[0]Ans
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Home work answer 1
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Homework :-
Ans - 31 : (a)
For the first time, I read competitive maths with such depth and conceptual way from someone. And You cleared all the queries Sir.🎉❤
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Iska ans.kese nikla
@palaknama2005 by using Fermat's remainder theorem........
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Home work 31:- 4 Ans
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