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Two men, David and Clifton, and their wives, Kim and Allison, go out shopping for books. Each person paid for each book a number of dollars equal to their number of books. David bought 1 more book than Kim, while Allison bought only 1 book. If each couple spent the same two-digit sum, who is Kim's husband?

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For some arrangement of M-F couples the sums of two sets

of squares lies in the interval [10, 99]. Two of the integers

being squared differ by 1, and one of the integers is 1.

Find Kim's husband.

There are two possibilities: David and Clifton.
Assume Kim's husband is David. Then,

9 < d2 + k2 = c2 + 1 < 100

There are no solutions. The closest ones are
(d k c sum) = (9 8 12 145 >99) and (2 1 2 5<10)

David is not Kim's husband; Clifton is
.

Verify OP claim that a solution exists for the Clifton-Kim marriage.

9 < (k+1)2 + 1 = c2 + k2 < 100
9 < 82 + 1 = 42 + 72 = 65 < 100

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OP says: "Each person paid for each book a number of dollars equal to their number of books."

Which is meant?

  1. Each person paid for each book bought by that person a number of dollars equal to the number of books bought by that person.
  2. Each person paid for each book bought by that person a number of dollars equal to the number of books they all bought together.

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Two men, David and Clifton, and their wives, Kim and Allison, go out shopping for books. Each person paid for each book a number of dollars equal to their number of books. David bought 1 more book than Kim, while Allison bought only 1 book. If each couple spent the same two-digit sum, who is Kim's husband?


Say Kim bought x books for x^2 dollars.
David bought x+1 books for (x+1)^2 dollars.
Allison bought only 1 book for 1 dollar.
Clifton bought y books for y^2 dollars.
If David and Kim were the couple, then they spent x^2 + (x + 1)^2 = two digit number = say z…....[1]
So x < 7 and x > 1
Then remaining couple Allison and Clifton would spend 1 + y^2 = z…………………………..…......[2]
So y can’t be greater than 9, i.e. z is not greater than 82.
Then from [1], x < 6 or z cannot be greater than 61.
Then from [2], y can’t be greater than 7, i.e. z can’t be greater than 50.
Then from [1], x < 5 or Z is not greater than 41.
And so on…
But we do not find suitable values of X and Y which satisfy both [1] & [2].

Note this can be prooved by equation y = SQRT 2x(x + 1), where we get only suitable values of x as 1 to get a whole number y. But from [1], X can't be lesser than 2, to get double digit number Z.
Therefore David and Kim are not the couple, so David and Allison are the husband & wife.
And Clifton is husband of Kim.
Then we have the two equations
(x + 1)^2 + 1 = z………………………………………………………………………………………..[3]
X^2 + y^2 = z…………………………………………………………………………………………….[4]
From [3] & [4], y^2 = 2x + 2 OR y = √(2x + 2)..…………………………………………………….[3]
[3] can be satisfied only when x= 7, or 1 to get y as a whole number.
But from [1], to get Z a double digit number X must be greater than 1.
Therefore X = 7, and y = 4.
Answer: David bought 8 books, his wife Allison bought 1 book. The spent total (64 + 1) = 65 dollars.
Clifton bought 4 books and his wife Kim bought 7 books, and they also spent 16 + 49 =65 dollars.

Edited by bhramarraj

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