1000th post giveaway! Coin math problems!!!
haletj
Posts: 2,192 ✭
The first reply answering a question correctly (with proof or sufficient explanation) gets the prize for the question. (I'm a mathematics grad student studying number theory by the way 
).
Question 1: You have some number of coins, at least 30 but no more than 35, that have values that are consecutive powers of 2, i.e. you have one coin worth $1, one coin worth $2, one worth $4,$8,$16...etc. Oh no, someone stole some of your coins! Thankfully only 1/5 of the value of your collection was lost. Exactly how many coins did you have originally and how many were stolen?
Prize: 2004 pcgs ms69 collectors club Silver Eagle
Question 2: You have a roll of 1982 zinc pennies. Someone (like me!) cherrypicked the roll for the best coin, and put another 1982 cent back into the roll, but oops, they put in a copper cent by accident instead. Oh no, now they can't tell at all which one is the copper cent by looking at or holding any of the 50 pennies. Thankfully they have a very precise balance beam scale (recall the copper cent is heavier), but they are allowed to only use the scale 4 times (uhm, then it'll break and be useless). How can they use the scale only 4 times yet determine which of the 50 pennies is the copper one?
Prize: A Gem++ BU set of the 7 1982 penny varieties
).Question 1: You have some number of coins, at least 30 but no more than 35, that have values that are consecutive powers of 2, i.e. you have one coin worth $1, one coin worth $2, one worth $4,$8,$16...etc. Oh no, someone stole some of your coins! Thankfully only 1/5 of the value of your collection was lost. Exactly how many coins did you have originally and how many were stolen?
Prize: 2004 pcgs ms69 collectors club Silver Eagle
Question 2: You have a roll of 1982 zinc pennies. Someone (like me!) cherrypicked the roll for the best coin, and put another 1982 cent back into the roll, but oops, they put in a copper cent by accident instead. Oh no, now they can't tell at all which one is the copper cent by looking at or holding any of the 50 pennies. Thankfully they have a very precise balance beam scale (recall the copper cent is heavier), but they are allowed to only use the scale 4 times (uhm, then it'll break and be useless). How can they use the scale only 4 times yet determine which of the 50 pennies is the copper one?
Prize: A Gem++ BU set of the 7 1982 penny varieties
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Comments
Split into 3 groups...
18 coins
18 coins
14 coins
Weigh 18 vs 18 (1 weighing)
(if equal, then go to weigh the 14 (see section on this below))
If not equal then concentrate only on the heavier group
If unequal then -
Split the heavier grouping and weigh 9 vs 9 (2nd weighing)
Follow the same logic...keep the grouping with the heavier coin
Split the heavier group into 3 groups of 3 each
If you weigh 2 groups of 3 against each other and they are equal (3rd weighing) you then weigh the last group of 3 one vs one. (4th weighing) If they are equal, then the remaining cent is the copper one by deduction. If the 3 vs 3 has one that is heaving (that was the 3rd weighing) then you take that one and weigh 1 coin against another and use the same logic as above....you WILL have the heavier copper coin by the 4th weighing.
So, now we go back to the 18vs18 being equal and we need to focus on the remaining group of 14 (one of which is the heavier copper coin). So far, only 1 weighing has been used.
Split into 2 groups of 7 each and weigh (2nd weighing)
Focus on heavier group of 7
Split into 3 vs 3 (with 1 remaining).
If 3 vs 3 is equal (3rd weighing) then remaining one is copper
If not equal, grab the heavier one and weigh 1 coin against another (4th weighing). If unequal, the heavier one is copper. If equal, remaining coin is copper.
Nice......
Ron
I've been told I tolerate fools poorly...that may explain things if I have a problem with you. Current ebay items - Nothing at the moment
You had 32 coins and 16 of them were stolen.
2^32 - 1 = sum of coins which must equal a number divisible by 5 = 4294967295
divided by 5 = 858993459 (the amount of money stolen).
Turn 858993459 into binary yields = 110011001100110011001100110011
The ones are the coins that were stolen (# of 1s equals 16).
-KHayse
You are also by the way, very rich. by adding up the sums in powers of two, you are left with a total of
33 784 961 426 for 35 coins - 16772864402 for 34 coins and 8266815890 for 33 coins - the only sum from 30-35 that is divisible by five. eight of the other coins total a value of 1653363178 - which is one fifth of 8266815890
You only gave us 6 options for # of coins so I performed the 2 ^ n -1 on each of them to see which was divisible by 5 (which
must be the case if 1/5 of the value was stolen).
-KHayse
Khayse and put together actually got the posted times the same (though, Khayse' got put first).
Then, on the edit, and explanatory post, they actually had the same time stamp again.
wow!
I've been told I tolerate fools poorly...that may explain things if I have a problem with you. Current ebay items - Nothing at the moment
I didn't feel that was worthy of the "mathematics" behind it, and I was too tired to think, so I didn't try.
Good to see some others posting though
I've been told I tolerate fools poorly...that may explain things if I have a problem with you. Current ebay items - Nothing at the moment
Never thought that would be necessary, that someone would be awake and on my heels at this time.
Glad we got different answers in case there was any controversy.
So where's teacher to grade our papers?
-KHayse
I tried to tackle the first question but I soon realized I was in way over my head.
Excellent giveaway.
Ask me no questions, I'll tell you no lies.
I started with excel on the first one but realized that excel may disqualify me for what Haletj wants and, believe it or not, I want the 7coin 1982 pennies (yep...I said PENNIES
So, the 1982 lincolns would be cool for us.
I've been told I tolerate fools poorly...that may explain things if I have a problem with you. Current ebay items - Nothing at the moment
-KHayse
ps Doesn't look like haletj is around. Now I have to try to go home and sleep and wait for the ruling.
#1 ) XF-45
#2) PR67CAM
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My damned psych instructor got me good one day. He says "Tom, what's 1/2 divided by 1/2?" My response wasn't quick and I blurted out several incorrect answers. All were wrong. If I'd have just remembered Jethro Bodine's unique way of cyphering, "Goesintas," I'd have blurted out 1.
With no power vested in me I declare that both Bohicaman and khayse are correct.
Joe.
Yes, Bochiman and KHayes are correct!
For problem 1 I like KHayes trick of putting it into binary! I didn't think of that. My solution was more theoretical...
If it were N coins, the sum would be 2^N - 1 which we want = 0 mod 5, or equivalently 2^N = 1 mod 5. Since 2^4 = 16 = 1 mod 5... we have 2^32 = 1 mod 5 is the only possible N from 30 to 35. So N=32 coins...
Now our sum of 32 numbers factors into 5 factors, 1+2+4+8+...+2^31 = 2^32 - 1 = (2^16 + 1)(2^8 + 1)(2^4 + 1)(2^2+1)(2+1)
To get a fifth of this total value, we divide out by the 2^2+1 = 5 factor, leaving us 4 factors which expands into 2^4 = 16 terms of the sum. Therefore he still has 16 coins remaining and 16 coins were stolen.
For problem 2 you could have up to 81 coins and be able to find which one was heavier. Split into 3 groups of 27, use the scale to determine which group of 27 had the copper cent, repeat with groups of 9, and then 3. N uses of the scale could determine which coin out of 3^N coins!