Instead of tackling this problem with 100 coins, try a simpler version of it:
Let's say I give you a pile of 3 coins, 2 are heads up and the 3rd is tails up. You're blind-folded, but since there are only three coins, its easy to picture how they might be stacked when I give them to you.
From top to bottom:
THH
HTH
HHT
If it were the first case, THH, what should we do? Let's first move the top coin, T, into a second pile, so you have HH in one and T in the other. Now how do I make sure they both have the same number of heads up/tails up coins? I can flip that one coin I moved over, so now I have HH and H. DONE!
If it were the second case, let's try the same thing. Move the top coin, H, into a second pile, so now we have TH in one pile and H in the other. What should we do? Flip the H we just moved over!
Lastly, if we have HHT, again, move the top H over so we have HT and H. I'm going to guess you get the trick now.
Now, this problem was set up with 3 coins where only 1 coin was flipped. Can you imagine what might happen when we go back to your problem of 100 coins with 10 flipped?
This is a classic question.
Split the cards into two piles. The first pile has 18 cards, the second pile has the remaining 35 cards. Say that there are $n$ upside down cards in the first pile, and therefore $18-n$ upside-down cards in the second pile.
Now turn over all the cards in the first pile.
Best Answer
You first split the coins up into two equal piles. Then the number of heads in one pile is equal to the number of tails in the other and vice versa.
Then the next step is to turn every coin in one of the piles around. Then you'll have the same amount of heads/tails in each pile.