- दर्शाइये किन n , n+4,n+8, n+12 तथा n+16 में से एक और केवल एक ही
5 से विभाज्य है जहाँ n कोई धनात्मक पूर्णँक
है ?
Answers
Given : n , n+4,n+8, n+12 तथा n+16 में से एक और केवल एक ही 5 से विभाज्य है जहाँ n कोई धनात्मक पूर्णँक है
To Find : Prove दर्शाइये
Solution:
let say n = 5k , 5k+ 1 , 5k+ 2 , 5k + 3 , 5k+ 4
n = 5k
n = 5k , 5k + 4 , 5k + 8 = 5(k + 1) + 3 , 5k + 12 = 5(k + 2) + 2 , 5k + 16 = 5(k+3) + 1
only n = 5k is divisible by 5
n = 5k +1
n = 5k+1 , 5(k +1) , 5k + 9 = 5(k + 1) + 9 , 5k + 13 = 5(k + 2) + 3 , 5k + 17 = 5(k+3) + 2
only n = 5(k+1) is divisible by 5
n = 5k +2
n = 5k+2 , 5(k +1) + 1 , 5k + 10 = 5(k + 2) , 5k + 14 = 5(k + 2) + 4 , 5k + 18 = 5(k+3) + 3
only n = 5(k+2) is divisible by 5
n = 5k +3
n = 5k+3 , 5(k +1) + 2 , 5k + 11 = 5(k + 2) + 1 , 5k + 15 = 5(k + 3) , 5k + 19 = 5(k+3) + 4
only n = 5(k+3) is divisible by 5
n = 5k +4
n = 5k+4 , 5(k +1) + 3 , 5k + 12 = 5(k + 2) + 2 , 5k + 16 = 5(k + 3) + 1 , 5k + 20 = 5(k+4)
only n = 5(k+4) is divisible by 5
Hence proved n , n+4,n+8, n+12 तथा n+16 में से एक और केवल एक ही
5 से विभाज्य है जहाँ n कोई धनात्मक पूर्णँक है
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