50 coins! Can you make 10 different piles?

You have 50 coins, and the challenge is to arrange them into 10 piles, each with a different number of coins. Sounds simple, right? But there’s a catch.

Watch the video as we use basic math and logic to uncover the answer to this intriguing puzzle!
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[anilchandrapoudel]

n= number of piles
S= Number of coins
S(min)=n(n+1)/2
For n=10 piles, S(min)=10×11/2=55
55 or more coins needed
For n=15 piles, S(min)=15×16/2=120
So 120 or more coins are needed

[Waraqah-Ibn-Nawfal]

Simply 1+10×5=55
Similarly 1+15×7.5=120
(First one + last one / 2 x number of piles)

[nikhiljayansingla]

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 = 120
So, 120 the answer

[lilporgi]

It’s the pigeon hole principe, if you take that first one of n different number height piles, and just take some number of one pile so that it matches at least one other. Then you’ll never be able to not make 2 the same.

Because on one of those piles removing one will make it another pile, and adding be the same amount as another one to. And that’s just with ones that have the same number so once you do one of those 2 to it, it creates other ones that can’t also be solved using the same thing because it will create more and yeah.

[chemical]

The rules are so irregular that you can have pile like: 0, 1, 2, 3, 4, 5, 6, 7, 8, 14 (because there is no rule that 0 < pile < 10)

[isheamongus]

So:
1
2
3
4
5
6
7
8
9
55
Yes. All the piles starting at 1 and each one previous+1 except last one

[SushilKumar-fmbi]

it should be 5 piles, 1 coin in first row
2 coins in second row
3 coins in third row
4 coins in fourth row
5 coins in fifth row
so if we do total of ot it be 15 coins
so it should be 5 piles

[ViliamF.]

_cuts a coin in half_

(which would actually be not very easy with these chips because they have a metal core coated in plastic)

[Neonique]

120 coins we need to make 15 piles with different number of coins