Math Help!!! Please

[KathiWithAnI]

Ok, I teach math and I can't work out this stupid problem that my poor 7th grade nephew was given for homwork......
I know the answer because my sister looked it up in the back... but I can't for the life of me figure out the pattern/formula for my DN.. maybe it's just too late at night
anyway,
What is the least positive number such that when the number is divided by 7, the remainder is 4, when the number is divided by 8, the remainder is 5 and when the number is divided by 9, the remainder is 6.
scroll down for the answer

please help with the formula if you can!
TIA
Kathi














501

 

[KathiWithAnI]

.

 

[mudnuri]

Totally lost me, but I'll bump it up for you....

DAN- the man with many links- where are you? You must have a link for this!!

Brandy

 

[KathiWithAnI]

Thanks Brandy-- it's driving me nuts, I KNOW I should be able to work this out, but so far tonight, I keep working in a circle!
Kathi

 

[Serena]

 

[Dan Murphy]

 

[LittleMissMickey]

Alright, I tried thinking, but am very tired. If I am not mistaken, this owrks:

7*8*9=504, subtract 3, 501

That would leave you with remainders of 4,5,6, respectively.

It works with other numbers, too.

If I'm wrong, I'm sorry... I got my best friend in on the action, too, and she says we should be correct!

Try it out!!

Hope this works!

Ashley

P.S. I am feeling very victorious, because I usually kind of stink at math.

edited to fix typo!!

 

[LittleMissMickey]

P.S. Please PM me if you think I'm right, or if you know for certain I am right! Perhaps I COULD teach 7th grade math after all, and here I've been afraid of it all this time...

Ashley

 

[KathiWithAnI]

That does give you the 501 and it's what my sister keeps saying too..... BUT if you didn't know that 501 was the answer what would make you subtract 3?? I'm trying to find the algebraic formula to give to my DN...Thanks so much for taking the time to answer, I really really appreciate it!!
Kathi

 

[#1 Disney Fan]

The question that remains, is how do you know to subtract that "3" from 504 to get 501. Here's my best take at it:

Find the lowest common multiple between 7,8,9. (504 I believe.)

Working with the 7: 72 x 7 = 504. Thus using one less multiple of 7: 71 x 7 = 497. You want 71 x 7 with a remainder of 4. 497 + 4 = 501. If you were to divide 501 by 7, you would get 71, with the added remainder of 4.

The 8: 63 x 8 = 504. Thus one less multiple of 8: 62 x 8 = 496. However, you want a remainder of 5, thus 496+5 = 501.

The 9: 56 x 9 = 504. Thus one less multiple of 6: 55 x 9 = 495. However, you want a remainder of 6, thus 495+6 = 501.

 

[Dan Murphy]

Originally posted by #1 Disney Fan
The question that remains,....

Sounds like engineer talk.

 

[danacara]

I see a general equation that's really a proof, for you the math teacher, there must be some extraordinarily elegant way to show this using some rule but I don't know it offhand. You could do this using nonlinear optimization if you have operations research software, for sure.

Here's the proof, skip it if you want, the equation that works for problems that are just like this one is at the end. If you give the seventh grader the equation and the explanation at the end he should be able to solve it.

It's tough to indicate coefficients here, sorry,

Given a positive integer sequence of divisors A=a1, a2, ..., aN, such that a(n+1) = a(n) + 1,
(in other words, in this problem the divisors were 7, 8, 9. They are consecutive integers, increasing by 1)

Given a positive integer sequence of remainders R=r1, r2, ..., rN, such that r(n+1) = r(n) + 1,
(in other words, in this problem the remainders were 4, 5, 6. Consecutive integers, increasing by 1)

And given that aN>rN over all cases (in other words, the divisor always has to exceed the remainder. It doesn't make any sense to divide a number by 7 and have a remainder of 10, for example)

It can be shown that the lowest positive integer Z that will yield those remainders when divided by those divisors is:
Z = (a1)((a2*a3*...*aN)-1)+r1
There are more elegant ways to write that, but this is the most useful way for a seventh grader. These examples will help:

So plugging in numbers to show an example - in this problem a1=7, a2=8, a3=9, and r1=4, r2=5, r3=6, so
Z = (a1)((a2*a3)-1)+r1
= (7)((8*9)-1)+4
= (7)(71)+4
= 501.

This would work if you extended the problem ... say if you had the exact same math problem but added another condition, that when the number is divided by 10, the remainder is 7. Now you'd have a4=10, r4=7, and
Z = (a1)((a2*a3*a4)-1)+r1
= (7)((8*9*10)-1)+4
= (7)(719)+4
= 5037, and that's correct

No matter where you start the divisors or remainders, as long as the divisors in the problem increase by 1 and the remainders increase by 1, and the divisors exceed their respective remainders, that formula will work. Hope it's clear enough. I'll write it in words:
Z = (the first divisor) times (all the other divisors multiplied together, then subtract one) plus (the first remainder).

Hope that helps

 

[totalia]

lol. I'm lazy. I'd just try different numbers till they all got the right answer and narrow it down.

 

[crs7568]

Wow! This reminds me why I hate math so much. My daughter is in 3rd grade so it looks like I only have a few years to find a great tutor.

 

[snarfer1]

Originally posted by danacara
I see a general equation that's really a proof, for you the math teacher, there must be some extraordinarily elegant way to show this using some rule but I don't know it offhand. You could do this using nonlinear optimization if you have operations research software, for sure.

Here's the proof, skip it if you want, the equation that works for problems that are just like this one is at the end. If you give the seventh grader the equation and the explanation at the end he should be able to solve it.

It's tough to indicate coefficients here, sorry,

Given a positive integer sequence of divisors A=a1, a2, ..., aN, such that a(n+1) = a(n) + 1,
(in other words, in this problem the divisors were 7, 8, 9. They are consecutive integers, increasing by 1)

Given a positive integer sequence of remainders R=r1, r2, ..., rN, such that r(n+1) = r(n) + 1,
(in other words, in this problem the remainders were 4, 5, 6. Consecutive integers, increasing by 1)

And given that aN>rN over all cases (in other words, the divisor always has to exceed the remainder. It doesn't make any sense to divide a number by 7 and have a remainder of 10, for example)

It can be shown that the lowest positive integer Z that will yield those remainders when divided by those divisors is:
Z = (a1)((a2*a3*...*aN)-1)+r1
There are more elegant ways to write that, but this is the most useful way for a seventh grader. These examples will help:

So plugging in numbers to show an example - in this problem a1=7, a2=8, a3=9, and r1=4, r2=5, r3=6, so
Z = (a1)((a2*a3)-1)+r1
= (7)((8*9)-1)+4
= (7)(71)+4
= 501.

This would work if you extended the problem ... say if you had the exact same math problem but added another condition, that when the number is divided by 10, the remainder is 7. Now you'd have a4=10, r4=7, and
Z = (a1)((a2*a3*a4)-1)+r1
= (7)((8*9*10)-1)+4
= (7)(719)+4
= 5037, and that's correct

No matter where you start the divisors or remainders, as long as the divisors in the problem increase by 1 and the remainders increase by 1, and the divisors exceed their respective remainders, that formula will work. Hope it's clear enough. I'll write it in words:
Z = (the first divisor) times (all the other divisors multiplied together, then subtract one) plus (the first remainder).

Hope that helps

W O W

That is some proof!

Thanks!

 

[KathiWithAnI]

danacara,
Thanks! I didn't even try sequences last night... DUH.. I knew there was a pattern and I kept setting it up trying to use 3 variables and solve for each and then substitute and it just doesn't work with this one!
I'll email this to my sister, of course she'll have a heartattack LOL

I told her she BETTER let us know what her son's teacher was looking for and I'll post here.
I know this isn't the type thing they are working on at school... can't wait to see thier answer. But you have put my mind to rest and I appreciate that

Kathi

 

[danacara]

glad it helped, Kathi with an I

 

[#1 Disney Fan]

I kept setting it up trying to use 3 variables and solve for each and then substitute and it just doesn't work with this one!

Well, you CAN set it up as a system of equations, it's just way above a 7th grade level, as you have to write it like such:


x = 4(mod7)
x = 5(mod8)
x = 6(mod9)

And then you solve use modulo division while solving the system of equations. For instance, you begin by writing:
x = 4(mod 7) = 7a+4.
However, x is also equal to 5(mod8).
Thus 7a+4 = 5(mod8) = 8b + 5.
7a = 8b + 1
a = 1/7(mod 8) Now you need to determine 1/7 (mod 8) to continue on... which can be done using modulo tables. And I won't continue on, because a) my lunch is almost over, and b) I'm sure this is way more complex than needed or wanted!

However... here's a page I just found specifically on theChinese Remainder Theorem Problem.

 

[KathiWithAnI]

Thanks, glad to know I wasn't nuts, just a bit off base LOL
Also, I emailed my sister the link to this thread and she was absolutely floored that so many people jumped in and tried to help, even just keeping this bumped! She said THANK YOU so much and I told her Welcome to the DIS.,.. that's just how we are
Kathi

 

[#1 Disney Fan]

Well Dana and I both have degrees that focused heavily on math, though probably different areas. I don't know about her, but I certainly don't get to use it very much in my job, so I enjoy a chance to problem solve now and then (even if it is 7th grade homework! )