Seven-Eleven

[radosr]

Bob enters a Seven-Eleven store and buys four items. The clerk enters the prices into his pocket calculator and multiplies them to get 7 dollars and 11 cents. Bob complains telling the clerk that he should add prices not multiply! The clerk apologizes, adds the prices and again gets 7 dollars and 11 cents! How much is the most expensive item Bob bought?

 

[pratikpoddar]

Information:
a*b*c*d = 7.11
a+b+c+d = 7.11
a>b>c>d>0
100*a, 100*b, 100*c, 100*d are integers

To find a.

711=3*3*79

(100a)*(100b)*(100c)*(100d) = 711000000 = 3*3*79*100*100*100
(100a)+(100b)+(100c)+(100d) = 711

Possible solution for (100a, 100b, 100c, 100d):
(316, 150, 125, 120)

So, costliest item costs 3 dollars 16 cents

-- Pratik
http://www.pratikpoddarcse.blogspot.com

 

[AlexandreH]

Information:
a*b*c*d = 7.11
a+b+c+d = 7.11
a>b>c>d>0
100*a, 100*b, 100*c, 100*d are integers

To find a.

711=3*3*79

(100a)*(100b)*(100c)*(100d) = 711000000 = 3*3*79*100*100*100
(100a)+(100b)+(100c)+(100d) = 711

Possible solution for (100a, 100b, 100c, 100d):
(316, 150, 125, 120)

So, costliest item costs 3 dollars 16 cents

-- Pratik
http://www.pratikpoddarcse.blogspot.com

Nice solution. But how did you get this: Possible solution for (100a, 100b, 100c, 100d):
(316, 150, 125, 120)?

 

[pratikpoddar]

Well 79 is prime. So, one of the 4 numbers had to be 79, 158, 237, 316, 395, 474...
So 711-(that number) would be: 632, 553, 474, 395, 316, 237...
and 711000000/(that number) would be 90000, 45000, 30000, 22500. 18000, 15000...

Since all the multiples have a lot of zeroes in them, I am more comfortable trying for 711-316=395 first. I get a solution in my first attempt. Bingo!

Hope that helps!

http://www.pratikpoddarcse.blogspot.com

 

[AlexandreH]

Thank you.