A model of this drawback appeared on the 1947 Stanford Aggressive Examination.
I purchased 72 an identical objects. Every merchandise had the identical price, and the associated fee was a complete variety of {dollars}. The full price was $_679_ (you have no idea the primary or final digit). How a lot did every merchandise price?
As traditional, watch the video for an answer.
A pleasant logic puzzle from the Stanford Aggressive Examination (1947)
Or maintain studying.
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“All will probably be nicely when you use your thoughts in your choices, and thoughts solely your choices.” Since 2007, I’ve devoted my life to sharing the enjoyment of recreation principle and arithmetic. MindYourDecisions now has over 1,000 free articles with no adverts because of group help! Assist out and get early entry to posts with a pledge on Patreon.
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Reply To “Unattainable” Price Puzzle
(Just about all posts are transcribed shortly after I make the movies for them–please let me know if there are any typos/errors and I’ll right them, thanks).
The trick is to think about divisibility guidelines. Discover 72 = 8 × 9, so the full price must be divisible by each 8 and 9.
A quantity is divisible by 8 if its final 3 digits are divisible by 8. Since 8 × 100 = 800, we are able to see 8 × 99 = 792 and eight × 98 = 784. Thus we will need to have the final digit is the same as 2.
A quantity is divisible by 9 if the sum of its digits is divisible by 9. Suppose the primary 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 will need to have a = 3.
Thus the full price is 36792, and every merchandise prices 36792/72 = 511.
Proof of divisibility by 8
Suppose okay ≥ 3. Any complete quantity could be written with 3 or extra digits, utilizing main 0s if crucial (12 = 012), as follows with the ci equal to the digits 0 to 9:
cokay(10okay) + cokay – 1(10okay – 1) + … + c2(102) + c1(101) + c0(100)
Since 1000 = 8 × 125, any greater energy of 10 can also be divisible by 8 as a result of 10okay(1000) = 10okay(125 × 8). Thus now we have:
cokay(10okay) + cokay – 1(10okay – 1) + … + c2(102) + c1(101) + c0(100)
≡ c2(102) + c1(101) + c0(100) (mod 8)
Thus a quantity is divisible by 8 if its final 3 digits are.
Proof of divisibility by 9
We will write a quantity in base 10 as:
cokay(10okay) + cokay – 1(10okay – 1) + … + c2(102) + c1(101) + c0(100)
Since 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1, and so forth. each energy of 10 is 1 greater than a a number of of 9. Thus 10okay ≡ 1 (mod 9).
cokay(10okay) + cokay – 1(10okay – 1) + … + c2(102) + c1(101) + c0(100)
≡ cokay + cokay – 1 + … + c2 + c1 + c0 (mod 9)
Thus a quantity is divisible by 9 if the sum of its digits are.
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