A car has a 4-digit integer price, which is written digitally. (so in digital numbers, like on your watch probably) While the salesmen isn't watching, the buyer turns the price upside down and gets the car for 1626 less. How much did the car initially cost?
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darij grinberg
22.04.2004 17:10
Reply by darij grinberg Hmm... what do you mean by "turns the price upside down"? Is it reversing the order of the numbers or is it something of the kind 6 --> 9 ? Since, if it would be reversing the order of the numbers, then the difference would be divisible by 9, what is not the case for 1626... but maybe I misunderstand something. If it is something else, then does it enforce 1 <--> 7 or not? And digits like 4 cannot occur at all, can they? Darij Reply by Arne 1 -> 1 2 -> 2 3 -> ? (does not exist) 4 -> ? (idem) 5 -> 5 6 -> 9 7 -> ? (idem) 8 -> 8 9 -> 6
Peter
22.04.2004 19:26
you do both so the changing sheme arne replied is done, plus the order is reversed but oh well seeing that was already half the problem
tim15
28.03.2008 20:44
If ABCD is the initially cost, and dcba is the cost upside-down, then: A B C D - d c b a 1 6 2 6 So the possibilities for D,a,A and d are (X if inverted doesn't exists): D a(D-a=6) A d(A-d=1 or 2) 0 4 >X 1 5 >5 1->5-1=4 2 6 >9 2->9-2=7 3 7 >X 4 8 >X 5 9 >6 5 >6-5=1, so this is the only possibility 6 0 >0 9->0-9=-9 7 1 >X 8 2 >2 8->2-8=-6 9 3 >X A=6, a=9, D=5, d=5 6 B C 5 - 5 c b 9 1 6 2 6 The possibilities for C,b,B and c are: C b (C-1-b=2 <-> C-b=3) B c (B-c=6 or 7) 0 7 >X 1 8 >8 1 >8-1=7, so this is the only possibility 2 9 >6 2->6-2=4 3 0 >X 4 1 >X 5 2 >2 5->2-5=-3 6 3 >X 7 4 >X 8 5 >5 8->5-8=-3 9 6 >9 6->9-6=3 B=8, b=8, C=1, c=1 6815 - 5189 1626 The initially cost was 6815.
Baffledbanna74
02.08.2023 14:46
my solution and mathmax12's solution: Let the number be $abcd-dcba=999a+90b-90c-999d=1626$, doing bashy casework (which took me 8 min to do) we get $\boxed{6815}.$