Once I performed a series of divisions where I got that the dividend is equal to remainder. Are my divisions flawed?
Note: I perform divisions for non-negative integers.
This problem is original.
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:3 :3 :3 :3 :3 :3 :3 :3 :3 :3 :3 :3 :3 :3 :3 :3 :3 :3
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:3^\infty :3
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( : 3 ) ∞
sir, in which class you are?
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@Dev Sharma – I am in 10th class.
FYI: Don't call me Sir. You "must" call me Nihar.
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@Nihar Mahajan – are you preparing for olympiad
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@Dev Sharma – Oh sorry , I didn't notice your comment before , since there were too many notifications. Anyways , I am preparing for olympiad (especially Maths and Astronomy).
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@Nihar Mahajan – I want to contact you on your fb account, can i send you a friend request as i am thinking to prepare for olympiad/?
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@Dev Sharma – Yeah , sure. :)
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@Nihar Mahajan – your profile pic is IMO logo??
@Nihar Mahajan – Just curious. You like maths more or astronomy ?
Same way !!
gr8 solution :XD .........
In case 2, if q = 0, D can be different from 0, but you can conclude that D < d and therefore D = r. for example dividing 4 by 6 we get 4 = 6*0 + 4.
division could still be flawed anyway... Even if the dividend were different from remainder. It should be "not necessarily flawed" instead of "may be flawed"
"0 being the dividend" case cannot be forgot.
So it may not always be flawed :)
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If we let D as the dividend , d as the divisor , q as the quotient and r as the remainder , By division algorithm we have:
D = d q + r
But since D = r , we cancel them from both the sides to get d q = 0 .
Case 1: When d = 0 , my division is definitely flawed because I cannot divide a number by 0 .
Case 2: When q = 0 , implies that the dividend itself is 0 that is D = 0 , which is possible.
Hence , some of my divisions might be flawed,.