Agnishom has 10 pockets and 44 coins . He wants to put his coins into his pockets so distributed that each pocket contains a different number of coins

Can he do so?

Generalize the problem,considering p pockets and n coins.

The problem is most interesting when

\( n = \frac{(p + 1)(p - 2)}{2}\)

Why?

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TopNewestMinimum number of coins required to put into p pockets (according to the restrictions applied) \(=\dfrac{p(p-1)}{2}\)

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– U Z · 2 years, 10 months ago

yes Sir its right , can you answer last partLog in to reply

@megh choksi – Sandeep Bhardwaj · 2 years, 10 months ago

Same is the case with the last part. Minimum no. of coins required \(=\dfrac{p(p-1)}{2}\). And as you mentioned there , if no. of coins is \(\dfrac{(p+1)(p-2)}{2}\) which is one less than the no. of coins required. I want to say that \(\dfrac{p(p-1)}{2}-\dfrac{(p+1)(p-2)}{2}=1\). So, in this case too, it is not possible to put the coins into the pockets with the restrictions applied.Log in to reply

– U Z · 2 years, 10 months ago

Yes right , thank you for replyingLog in to reply

1+2+3+4+5+6+7+8+9=45. He will have one empty pocket! – Venture Hi · 2 years, 10 months ago

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@Venture HI , I was waiting for your response to the second part . nicely done ,voted up thank you – U Z · 2 years, 10 months ago

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