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# Just Enjoy!

question

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

1. Can he do so?

2. Generalize the problem,considering p pockets and n coins.

The problem is most interesting when

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

Why?

Note by U Z
3 years, 4 months ago

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1. He can not do so. Because He can't keep $$2$$ or more pockets empty (as it is given that each pocket has different number of coins) . And at the least, if we keep one pocket empty, then also minimum numbers of coins required to put in $$9$$ pockets is $$\displaystyle \sum_{i=1}^{9}=45$$.

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

- 3 years, 4 months ago

yes Sir its right , can you answer last part

- 3 years, 4 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. @megh choksi

- 3 years, 4 months ago

Yes right , thank you for replying

- 3 years, 4 months ago

1+2+3+4+5+6+7+8+9=45. He will have one empty pocket!

- 3 years, 4 months ago

Sorry @Venture HI , I was waiting for your response to the second part . nicely done ,voted up thank you

- 3 years, 4 months ago