Grandfather's Bill

This is a problem from the 1947 Stanford Competitive Examination []:

Among grandfather's papers a bill was found

72 turkeys $_67.9_

The first and the last digits of the number that obviously represented the total price of those fowls are replaced here by underscores, for they have faded and are now illegible.

What are the two faded digits and what was the price of one turkey?

Solution

Reference

  1. G. Polya, J. Kilpatrick, The Stanford Mathematics Problem Book, Dover, 2009
  2. G. Polya, Patterns of Plausible Inference, Ishi Press (July 13, 2009)

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Copyright © 1996-2017 Alexander Bogomolny

Among grandfather's papers a bill was found

72 turkeys $_67.9_

The first and the last digits of the number that obviously represented the total price of those fowls are replaced here by underscores, for they have faded and are now illegible.

What are the two faded digits and what was the price of one turkey?

Solution

The key to the solution is the the criteria of the divisibility by 9 and by 8, for, after all, 72 = 9×8.

A number is divisible by 8 iff and only if the number formed by its three,one,two,three,eight last digits is divisible by 8. From grandfather's bill, we have a three-digit number 79x, where x is the unknown last digit. However, 79x = 720 + 64 + 6 + x = 784 + (6 + x). Observe that 784 is divisible by 8 so that to have 79x divisible by 8, 6 + x must be a multiple of 8, implying x = 2. We thus found the last digit. The grandfather's bill is now "72 turkeys $_67.92".

What is remains is to find the first digit. Divisibility by 9 comes to rescue. A number is divisible by 9 iff the sum of its digits,the sum of its digits,the sum of odd digits,the last digit is divisible by 9. Let the unknown first digit be y. Then the sum

y + 6 + 7 + 9 + 2 = y + 24

must be divisible by 9, meaning that y + 6,y + 3,y + 6,y + 9 must be divisible by 9. From here y = 3 and the price the grandfather paid for the turkeys was $367.92.

The price of the single turkey was $367.92 / 72 = $5.11.

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Copyright © 1996-2017 Alexander Bogomolny

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