InterviewSolution
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InterviewSolution
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Prove that one of every three consecutive positive integers isdivisible by 3. |
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Answer» Any three consecutive positive integers must be of the form n, (n+1) and (n+2), where n is any natural number i.e. n=1,2,3… Let, a=n,b=n+1 and c=n+2 Order triplet is (a,b,c) (n,n+1,n+2), where n=1,2,3… (i) `"At n=1 (a,b,c)=(1,1+1,1+2)=(1,2,3)"` `"At n=2, (a,b,c)=(1,2+1,2+2)=(2,3,4)"` `"At n=3, (a,b,c)=(3,3+1,3+2)=(3,4,5)"` `"At n=4, (a,b,c)=(4,4+1,4+2)=(4,5,6)"` `"At n=5, (a,b,c)=(5,5+1,5+2)=(5,6,7)"` `"At n=6, (a,b,c)=(6,6+1,6+2)=(6,7,8)"` `"At n=7, (a,b,c)=(7,7+1,7+2)=(7,8,9)"` `"At n=8, (a,b,c)=(8,8+1,8+2)=(8,9,10)"` We observe that each triplet consist of one and only one number which is multiple of 3 i.e. divisible by 3 Hence one of any three consecutive positive integers must be divisible by 3. |
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