What is the largest possible number of cards i could have had in the hat in the beginning?there are several cards in my hat. a positive integer is written on each of the cards so the numbers on...

What is the largest possible number of cards i could have had in the hat in the beginning?

there are several cards in my hat. a positive integer is written on each of the cards so the numbers on different cards are different. I take two cards from the hat and replace them with a new card on which i write the product of the difference of the numbers on these two cards and their sum. e.g- if the numbers on 2 cards were 6 and 2, the new number would be (6-2)(6+2) =32. After i repeat this a few times, there's only 1 card left with the number 2764 on it. What is the largest possible number of cards i could have had in the hat in the beginning?

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Working backwards:

What cards could produce 2764? We have (a+b)(a-b)=2764. Now the factors of 2764 are 1,2,4,691,1382,2764. Since 2764 is even, at least one of a+b or a-b is even; adding and subtracting b results in the same parity for a+b and a-b,(Assume a is even; if b is even then both a+b and a-b are even, if b is odd then both a+b and a-b are odd. Likewise, if a is odd then if b is odd both a+b and a-b are even and if b is even then a+b and a-b are odd), so both a+b and a-b are even. This occurs if a,b are both even or both odd. a+b is a factor of 2764 and a-b is a factor of 2764.

Try 1,2764. This results in a+b=2764,a-b=1. Solving this system yields 2a=2765 so a is not an integer.

Try 2,1382. Then a+b=1382,a-b=2==>2a=1384==>a=692,b=690. This works:(692-690)(692+690)=2764.

Try 4,691: Then a+b=691,a-b=4==>2a=695 and a is not an integer.

(1) The only possible cards to get to 2764 are 690,692.

Repeat this process for 690 and 692. In working the process you will note that the sum of the factors must be even. So 690 has no cards: the factors are 1,2,3,5,23,30,138,230,345,690. Now a+b=690,a-b=1 ==> 2a=691,a+b=345 a-b=2 ==>2a=347,a+b=230 a-b=3 ==>2a=233 etc... all resulting in noninteger a.

For 692 the factors are 1,2,4,173,346,692. The only pair of factors whose sum is even are 2 and 346. This means a+b=346 a-b=2==>2a=348==>a=174,b=172.

(2) The only possible cards for 692 are 174 and 172.

Repeating the process we find:

(3) No cards are possible for 174. The only cards for 172 are 42 and 44.

(4) There are no possible cards for 42. The cards for 44 are 10 and 12.

(5) There are no cards for 10. The cards for 12 are 2 and 4.

(6) There are no cards for 2. The cards for 4 are 2 and 0, but zero is not a positive integer.

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The cards must be 2,4,10,42,174,690 -- there are 6 cards in the hat.

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Check: Cards 2 and 4 yield: (4+2)(4-2)=12

The cards 10 and 12 yield: (12+10)(12-10)=44

The cards 42 and 44 yield: (44-42)(44+42)=172

The cards 172,174 yield: (174-172)(174+172)=692

The cards 690 and 692 yield: (692-690)(692+690)=2764

 

 

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