Brain Teasers
How Many Dollars
Sandy and Sue each has a whole number of dollars. I ask them how many dollars they have.
Sandy says: "If Sue gives me some dollars, we'll have the same amount of money. But if I give Sue the same number of dollars, she'll have twice as much money as I have."
Sue says: "And if you remove the first digit of my wealth and place it to the end, you'll get Sandy's wealth."
If neither of them has more than 1 million dollars, how many dollars do they each have?
Sandy says: "If Sue gives me some dollars, we'll have the same amount of money. But if I give Sue the same number of dollars, she'll have twice as much money as I have."
Sue says: "And if you remove the first digit of my wealth and place it to the end, you'll get Sandy's wealth."
If neither of them has more than 1 million dollars, how many dollars do they each have?
Hint
First get the ratio, then get their amount.Answer
Let X and Y be Sandy's and Sue's amounts of money, respectively. Let Z be the number of dollars in Sandy's statement.Then X+Z=Y-Z and 2(X-Z)=Y+Z. Solving this we get X=5Z and Y=7Z.
According to Sue's statement, let A be the first digit of Sue's wealth, B be the remaining digits, and n be the number of digits, then Y=A*10^(n-1)+B and X=10B+A.
7X=5Y
7*(10B+A)=5*(A*10^(n-1)+B)
(5*10^(n-1)-7)A=65B
Since 65B is divisible by 5 and 5*10^(n-1)-7 is not, A must equal 5. We get:
5*10^(n-1)-7=13B
The least value of n such that 5*10^(n-1)-7 is a multiple of 13 is 6:
499993=38461*13
The next is 12, which makes them have more than 1 million dollars. Hence n=6 and B=38461.
Therefore, Sandy has 384615 dollars and Sue has 538461 dollars.
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Comments
I started it then got lost- too long since my algebra!! Good one though
Huh?
I liked the part about the eggs.
Easy but fun.
wow, great teaser. it left me with a headache, though.
Dec 05, 2005
Hi...terranmarine? Just wondering, how can this be easy, huh!... please explain to me in an easily understandable way...my email [email protected]
ive developed a tumor... ouch
What's Going On????? Where is X? and who is Y? and How did you arrive at any numerical figure with the data, No the information you gave? I don't see any Numbers to add to the end, or beginning. Is it Over?
OK, we count the number of instances that each has stated money. multiply the two over the one instance in which Sue would make an exchange with her friend. I see the 5 and now the 7, the 10 and 1. I am so very confused and loss. Good Try!!!!!!!
A bit harder than it first appeared, it could be right, I didn't even follow the explanation all the way to the end.
I got the answer in a different way.
First I solved for x to determine that Sue has 7x dollars and Sandy has 5x, so:
Sandy = 5/7 * Sue
If I represent Sue a decimal between 1 and 10 then if A is the first digit of Sue, then:
Sue = A + Sandy/10
Combining these two equations gives the geometric series:
Sue = A + (5/7)*A/10 + (5/7)((5/7)*A/10)) + ...
Sue = A + A*(1/14) + A*(1/14)^2 + A*(1/14)^3 + ...
Factoring out A gives:
Sue = A * (1 + 1/14 + (1/14)^2 + (1/14)^3 + ...)
Sue = A * (1 / (1 - (1/14)))
Sue = A * (1 / (13/14))
Sue = A * 14/13
Since Sandy is a multiple of 5, I know that the last digit of Sandy's money and the first digit of Sue's money is a five. Replacing 5 for A gives:
Sue = 5 * 14/13 = 70/13 = 5.38461538461538461...
Since I know Sandy must end in a 5, Sandy is 384615 and Sue is 538461.
First I solved for x to determine that Sue has 7x dollars and Sandy has 5x, so:
Sandy = 5/7 * Sue
If I represent Sue a decimal between 1 and 10 then if A is the first digit of Sue, then:
Sue = A + Sandy/10
Combining these two equations gives the geometric series:
Sue = A + (5/7)*A/10 + (5/7)((5/7)*A/10)) + ...
Sue = A + A*(1/14) + A*(1/14)^2 + A*(1/14)^3 + ...
Factoring out A gives:
Sue = A * (1 + 1/14 + (1/14)^2 + (1/14)^3 + ...)
Sue = A * (1 / (1 - (1/14)))
Sue = A * (1 / (13/14))
Sue = A * 14/13
Since Sandy is a multiple of 5, I know that the last digit of Sandy's money and the first digit of Sue's money is a five. Replacing 5 for A gives:
Sue = 5 * 14/13 = 70/13 = 5.38461538461538461...
Since I know Sandy must end in a 5, Sandy is 384615 and Sue is 538461.
Thanks for your puzzle.
I managed to solve it with glpsol (a free linear and mixed integer programming solver, available for most platforms). I uploaded the model file that I used to pastebin for anyone who might be interested to play with it.
http://pastebin.com/rspuwsN2
I managed to solve it with glpsol (a free linear and mixed integer programming solver, available for most platforms). I uploaded the model file that I used to pastebin for anyone who might be interested to play with it.
http://pastebin.com/rspuwsN2
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