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A version of this problem appeared on the 1947 Stanford Competitive Examination.
I bought 72 identical items. Each item had the same cost, and the cost was a whole number of dollars. The total cost was $_679_ (you do not know the first or last digit). How much did each item cost?
As usual, watch the video for a solution.
A delightful logic puzzle from the Stanford Competitive Exam (1947)
Or keep reading.
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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.
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Answer To “Impossible” Cost Puzzle
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
The trick is to consider divisibility rules. Notice 72 = 8 × 9, so the total cost has to be divisible by both 8 and 9.
A number is divisible by 8 if its last 3 digits are divisible by 8. Since 8 × 100 = 800, we can see 8 × 99 = 792 and 8 × 98 = 784. Thus we must have the last digit is equal to 2.
A number is divisible by 9 if the sum of its digits is divisible by 9. Suppose the first digit is a. Then the sum of the digits is:
a + 6 + 7 + 9 + 2
= a + 6 + 18
≡ a + 6 (mod 9)
Since a is a digit from 1 to 9, we must have a = 3.
Thus the total cost is 36792, and each item costs 36792/72 = 511.
Proof of divisibility by 8
Suppose k ≥ 3. Any whole number can be written with 3 or more digits, using leading 0s if necessary (12 = 012), as follows with the ci equal to the digits 0 to 9:
ck(10k) + ck – 1(10k – 1) + … + c2(102) + c1(101) + c0(100)
Since 1000 = 8 × 125, any higher power of 10 is also divisible by 8 because 10k(1000) = 10k(125 × 8). Thus we have:
ck(10k) + ck – 1(10k – 1) + … + c2(102) + c1(101) + c0(100)
≡ c2(102) + c1(101) + c0(100) (mod 8)
Thus a number is divisible by 8 if its last 3 digits are.
Proof of divisibility by 9
We can write a number in base 10 as:
ck(10k) + ck – 1(10k – 1) + … + c2(102) + c1(101) + c0(100)
Since 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1, etc. every power of 10 is 1 more than a multiple of 9. Thus 10k ≡ 1 (mod 9).
ck(10k) + ck – 1(10k – 1) + … + c2(102) + c1(101) + c0(100)
≡ ck + ck – 1 + … + c2 + c1 + c0 (mod 9)
Thus a number is divisible by 9 if the sum of its digits are.
Source
Puzzle on Cut The Knot
https://www.cut-the-knot.org/Outline/Arithmetic/DivisionBy72.shtml
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