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Four
Posted: March 11th, 2022, 11:35 am
by cinelli
John had created a puzzle for his local magazine. But he was annoyed to see that what he had written as 5^4 2^3 – i.e. 625 x 8 = 5000 – had been printed as the 4-digit number 5423. But then he wondered whether there are any numbers where such an error wouldn’t matter. That is your challenge – to find instances of digits a, b, c, d such that a^b x c^d = the four digit number abcd.
Cinelli
Re: Four
Posted: March 11th, 2022, 12:26 pm
by UncleEbenezer
cinelli wrote:John had created a puzzle for his local magazine. But he was annoyed to see that what he had written as 5^4 2^3 – i.e. 625 x 8 = 5000 – had been printed as the 4-digit number 5423. But then he wondered whether there are any numbers where such an error wouldn’t matter. That is your challenge – to find instances of digits a, b, c, d such that a^b x c^d = the four digit number abcd.
Cinelli
What's 0^0? With a little cheat taking the rule 0^anything = 0, we have a trivial example.
Otherwise, since the problem space is constrained to the digits 0-9, it is trivial to verify that there is no solution.
[edit] Aaargh. Misread what you wrote. a^b*c^d == 1000*a + 100*b + 10*c +d has a solution 2592.
Re: Four
Posted: March 11th, 2022, 1:42 pm
by jfgw
There appears to be only that one solution.
Julian F. G. W.
Re: Four
Posted: March 13th, 2022, 4:11 pm
by cinelli
jfgw wrote:There appears to be only that one solution.
Julian F. G. W.
Quite right. Surprising, I think.
Cinelli