need help witn math prob

Ok i have an extra credit probelm I would like to solve, as I really need the extra credit to boost my grade up, here it is.

" A factory worker uses machines to sort cards into pile. On one occasion a machine operator obtained the following curious result. When a box of cards was sorted into 7 equal groups, there were 6 cards left over; when a box of cards was sorted into 5 equal groups, there were 4 cards left over; and when it was sorted into 3 equal groups, there were 2 left over. If the machine cannot sort more than 200 cards at a time, how many cards were in the box?''

 

hmmm this is a toughy.....im workin on it as we speak

edit: ah crap man, this is driving me ca-razy!!!!!!!!! lol wish i could have figured it out......when is it due? i might try it tomorrow if its not due b4 then

 

Yep due tomorrow

 

How much money you got, kid?

Here's how it goes. You write all the multiples of 7, +6, up to 200. Which would be up to 7x27, because 7x27(#of cards in group)=189+6(leftover cards)=195 total, and the machine doesn't sort more than 200 at a time. So you write down on a page across, in columns, 1,7,13 because 1(# of cards in group) x7,+6(leftover cards) =13. So you write 1,7,13 in a column. You follow by filling up the page with these multiples. 2,14,20/3,21,27/4,28,34, etc.

Then you look at the final #'s. 13,20,27,34,41,etc. Here's the trick. Any # that is divisible by 3 adds up to a # that is divisible by 3. 3,6,9, 12/1+2=3. 3 is divisible by 3. 15/1+5=6. 6 is divisible by 3. 134/1+3+4=8, which is not divisible by 3. So you take the final #'s you ended up with, which would be 13,20,27,etc. subtract 2 from them, add the #'s that remain, and figure out which would be divisible by 3. 20-2=18. 1+8=9, which is divisible by 3. 62-2=60, 6+0=6, which is divisible by 3.

So after you do this, you cross out all the #'s that aren't divisible by 3. The remaining #'s you -4 from, and look to see if they're divisible by 5, which is easy. Multiples of 5. 83-2=81. 8+1=9, which is divisible by 3, so it could be the #, but 83-4=79, which is not a multiple of 5, so it couldn't have been sorted into 5 equal groups. After you do this, there remains only one #, which is the answer.

 

Seems to me like you could use "Mod" to find your answer in this one.

 

You can simply this problem by noting that "5 equal groups, with 4 cards left over" means the last digit of the number of cards must be a 4 or a 9.

 

there were 104 cards in the box

7x + 6 = a
5y + 4 = a
3z + 2 = a

x= 14, y=20, z=34; a=104

the joys of quadratic equations

all the best

 

104 mod 7=6
104 mod 5=4
104 mod 3=2

 

So I take it I'm the only one who doesn't think he deserves to get extra credit toward his grade by CHEATING?

 

As long as my trig teacher isnt browsing this forum I consider it to be my helping hand

 

Sometimes we all need a little guidance. I just note that noone gave him both the answer and how to get there until it was after he would probably have class. Most math professors/teachers require you to show work, so I didn't feel that helping him would be harmful. Plus, they at least took an interest in extra credit and didn't ask outright for someone to do it all for them. Agreed in spirit, though.

 

dude

it is one or the other.

giving both is cheating.

 

Ok mom.....geeze its just extra credit, calm down compadre....

 

Gross analogy.........

 

If you object to this method of help then why did you provide me help? Im just an idiot in need, thats all.

 

Yeah, I gotcha big fella.
I just don't think giving someone everything after their deadline has passed constitutes as cheating. Otherwise any villainous teacher that actually went over the solutions to problems to help students learn what they missed should be fired immediately. btw, love your avatar!

Bladesofhalo- Did you figure it out in time for class?

 

Actually i did figure it out, we got similar problems in class today and I was able to do em, no sweat.

 

Cheating:The act of deceiving:FRAUD, DECEPTION I fail to see were this occurred Bladesofhalo just dropped by the Geek library to do some research, and he learned what the Teach wanted, if he found the answer to the problem in a book is that "Cheating".

 

Sorry to be rude.

I dont object. In maths it is important you understand what and why are you doing something. Just following the "mechanical" method will leads to unwanted errors, and when something slightly different is asked it will cause confussion.

Good for you for working it out yourself well done!

 

Ummm...asking people for the answer without figuring it out for yourself, then using that answer for personal gain is cheating. Which is why one often hears of kids "cheating" on tests by copying the answers from someone else.

I also noted that no one gave him good information until after he would have had to turn it in, but I don't think it's appropriate to ask us to help him cheat. He as much as admitted that's what he was going to do when he said as long as his teacher doesn't frequent this forum he'd be fine. Which quite obviously means he knows he would have gotten in trouble had his teacher found out. Which also quite obviously means he knows he was doing something wrong.

That said...I'm glad he figured it out for himself in time to get the credit. That's the way to do it, rather than asking someone to give you the answer.

 

Just wanted to add my 2c....

Cheating might get a person better grades and avoid the hard slog of the learning process, but seriously, all you cheat is yourself, as you still don't understand not only how to do it, but how to work a situation through to achieve the end result. It may just be a math problem today for some, but in a few years this kind of positive thinking and approach to problems becomes an essential life skill to coping with those lemons we keep mentioning