Pile of Cards Puzzle
There are n cards numbered with consecutive whole numbers starting with 1 and up to n . It is divided into two piles. What is the smallest value of n that guarantees that one stack will contain two cards whose numbers sum to a perfect square?
The answer is 15.
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2 solutions
Im confused. Why would this be anything more than 4? 4 is a perfect square. So why wouldn't you just put 1 and 3 in a pile and 2 and 4 in another. The first pile's sum equals a perfect square(4).
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But we need to account for any distribution of the n cards into two piles. With 4 cards, if we distributed them so that one pile had 2 , 3 and the other 1 , 4 then in neither pile are there two cards which sum to a perfect square.
Divide the Whole set into two sub sets with the conditions that have to be satisfied: 4 = 1 + 3 , 9 = 1 + 8 = 2 + 7 = 3 + 6 = 4 + 5 , . . . ( 1 ) Choose member for each set that satifies (1) until 14 we get: S 1 = [ 1 , 2 , 6 , 1 3 , 9 , 4 , 1 1 ] S 2 = [ 1 0 , 3 , 7 , 8 , 1 2 , 5 , 1 4 ] So add 15 to each set we always get the result. R = 1 5
the statement simply says cards with consecutive whole numbers starting at 1 ending at n separated into two piles with one having the sum of a perfect square with the least cards used. whats n? that would be 3. 1, 2, 3 separated 1+3, 2 sum for both stacks 4, 1. 4 is a perfect square. all that other bs for the answer is completely wrong.
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I had the same answer. 4. I dont understand why there is a need to go beyond that
Add numbers one by one, trying to avoid making perfect squares. Just before you get stuck, you have a division of the largest possible deck into two stacks without perfect square.
Put 1 in stack 1.
Put 2 in stack A. (Later we decide whether A = 1 and B = 2, or vice versa.)
Put 3 in stack 2, to avoid 1 + 3 = 4.
Put 4 in stack X. (Later we decide whether X = 1 and Y = 2, or vice versa.)
Put 5 in stack Y, to avoid 4 + 5 = 9.
Put 6 in stack 1, to avoid 3 + 6 = 9.
Put 7 in stack B, to avoid 2 + 7 = 9.
Put 8 in stack 2, to avoid 1 + 8 = 9.
Put 9 in stack A, to avoid 7 + 9 = 16.
Put 10 in stack 2, to avoid 6 + 10 = 16.
Put 11 in stack X, to avoid 5 + 11 = 16.
Put 12 in stack Y, to avoid 4 + 12 = 16.
Put 13 in stack 1, to avoid 3 + 13 = 16. Now we decide that stack Y = stack 2, otherwise we would have 12 + 13 = 25.
Put 14 in stack Y = 2, to avoid 11 + 14 = 25. Now we decide that stack A = stack 1, otherwise we would have 2 + 14 = 16.
Now there is no place to put 15. In stack 1, we'd get 15 + 1 = 16; in stack 2, we'd get 15 + 10 = 25.
Therefore it is not possible to do this for n ≥ 1 5 . But with n = 1 4 there is a (unique!) possible division into two stacks without any two cards summing to a perfect square, namely [ 1 , 2 , 4 , 6 , 9 , 1 1 , 1 3 ] , [ 3 , 5 , 7 , 1 0 , 1 2 , 1 4 ] .