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Mathematics 14 Online
OpenStudy (anonymous):

Please help!!!!!!! Sam counted the lines of a page in his book. Counting by threes gave a remainder of 2, counting by fives also gave a remainder of 2, and counting by sevens gave a remainder of 5. How many lines were on the page? Show work.

OpenStudy (anonymous):

I think this needs modular arthemetic. So @mukushla could help

OpenStudy (anonymous):

x MOD 3=2 x MOD 5=2 x MOD 7=5

OpenStudy (anonymous):

What's mod?

OpenStudy (anonymous):

@mukushla u busy?

OpenStudy (anonymous):

yeah we can use mods but lets think easier

OpenStudy (anonymous):

let the number of pages be n

OpenStudy (anonymous):

Ok

OpenStudy (anonymous):

Counting by threes gave a remainder of 2 so n=3k+2

OpenStudy (anonymous):

CRT have fun

OpenStudy (anonymous):

counting by fives also gave a remainder of 2 so n=5j+2

hartnn (hartnn):

tried that 4 unknowns.,3 equations....

OpenStudy (anonymous):

what am i doin???

OpenStudy (anonymous):

Ya. agreed with @hartnn

OpenStudy (anonymous):

@satellite73 come here plz :)

OpenStudy (anonymous):

Anyone found the answer?

OpenStudy (anonymous):

I think there are infinite number of answer

OpenStudy (anonymous):

What's one of them?

OpenStudy (anonymous):

47

OpenStudy (anonymous):

I have figured it out that the last digit must be 7

OpenStudy (anonymous):

i think they mean to find minimum number of pages and so going with trial and error

hartnn (hartnn):

agreed, minimum and trial and error.

OpenStudy (anonymous):

I think they are looking for the number of lines in one page.

OpenStudy (anonymous):

257

OpenStudy (anonymous):

Explain

OpenStudy (anonymous):

The last digit must be 7

OpenStudy (anonymous):

have u figured that out

mathslover (mathslover):

47

OpenStudy (anonymous):

Actually I think I can give the general term too.

OpenStudy (anonymous):

I am confused, what's the right answer?

OpenStudy (anonymous):

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