A model of this drawback appeared on the 1947 Stanford Aggressive Examination.
I purchased 72 similar objects. Every merchandise had the identical price, and the fee was an entire variety of {dollars}. The whole price was $_679_ (you have no idea the primary or final digit). How a lot did every merchandise price?
As ordinary, watch the video for an answer.
A pleasant logic puzzle from the Stanford Aggressive Examination (1947)
Or maintain studying.
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“All might be properly if you happen to use your thoughts to your selections, and thoughts solely your selections.” Since 2007, I’ve devoted my life to sharing the enjoyment of sport idea and arithmetic. MindYourDecisions now has over 1,000 free articles with no advertisements because of group assist! Assist out and get early entry to posts with a pledge on Patreon.
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Reply To “Not possible” Value Puzzle
(Just about all posts are transcribed rapidly 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 contemplate divisibility guidelines. Discover 72 = 8 × 9, so the entire price needs to 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 will 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 entire price is 36792, and every merchandise prices 36792/72 = 511.
Proof of divisibility by 8
Suppose ok ≥ 3. Any complete quantity might be written with 3 or extra digits, utilizing main 0s if vital (12 = 012), as follows with the ci equal to the digits 0 to 9:
cok(10ok) + cok – 1(10ok – 1) + … + c2(102) + c1(101) + c0(100)
Since 1000 = 8 × 125, any increased energy of 10 can also be divisible by 8 as a result of 10ok(1000) = 10ok(125 × 8). Thus now we have:
cok(10ok) + cok – 1(10ok – 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 are able to write a quantity in base 10 as:
cok(10ok) + cok – 1(10ok – 1) + … + c2(102) + c1(101) + c0(100)
Since 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1, and many others. each energy of 10 is 1 greater than a a number of of 9. Thus 10ok ≡ 1 (mod 9).
cok(10ok) + cok – 1(10ok – 1) + … + c2(102) + c1(101) + c0(100)
≡ cok + cok – 1 + … + c2 + c1 + c0 (mod 9)
Thus a quantity is divisible by 9 if the sum of its digits are.
Supply
Puzzle on Reduce The Knot
https://www.cut-the-knot.org/Define/Arithmetic/DivisionBy72.shtml
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