Uncategorised question #3 The total cost for 88 adult tickets was \(\$*293*\). Because the printing machine was not functioning well, the first and the last digits of the 5-digit number were missing. If the cost for each ticket is $P, where P is an integer, find the value of P

so the first and the last number were same??

This is a divisibility problem.

@experimentX It didn't mention..

147

Very easy actually, do you want me to post the solution?

Not to be a pain but @FoolForMath, we're not supposed to just give the answer, are we?

Not in general but @callisto doesn't always want full solution.

Oh, special case, eh?

Your solution wouldn't happen to give \(*293*\) numbers such that it's divisible by 11 would it?

And force \(93*\) to be divisible by 8.

For *293* to be divisible by 88 it must be divisible by 8 and 11 Thus 93* is divisible by 8 the last digit must be 6 and *2396 and the similarly the first digit must be 1 hence 12396 divide by 88 you will get 147.

89232

Ah Ha! I knew it.

mistake 12936

@pfenn1: That rule is generally intended for answer snipers and users who are just looking for answers and for cases like someone is trying to explain a concept and someone else come by and drop the answer due to this the OP left ruining the teaching experience of that user concerned. Once you spend time here you will know who wants answer and who wants to understand the problem For example you may or may not post full solution in Fool's problem of the day :)

Thanks for explaining that @FoolForMath.

@FoolForMath Thanks!!! I understand that now :) Sorry everyone. I don't mean to just get the answers, but I really want to know how to solve these questions. Thus, numerical answers without explanations won't help much, maybe just a 'hint' for me (Actually, I don't have to do these questions. It's just for my interest only) Thank you for all your help!!!

haha ... easiest method :D for i = 1:9 for j=0:9 k = i*10000 + 2930 + j; if mod(k, 88) == 0 disp(k) end end end

programming?

haha cheating!!

I don't expect to use a computer :P

well ... bad for me :(

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