algebraic equation

[vrach]

A certain 3 digit number when divided by :a) 7 gives a remainder of 2. b) by 8 ...remainder of 3 and c) by 11.....remainder of 6.
I need the algebraic equation(s) to solve to get the 3 digit number.
Thanks

[factor-fiction]

The answer is 611 but finding the equations is the more difficult bit.

[Old_Geezer]

Hmm that may give me something to think about, but at first glance you seem to have more variables than equations implying there may be a range of answers. Maybe the fact that X is between 000 and 999 helps.

X = 7A + 2
X = 8B + 3
X = 11C + 6

[sunny-dave]

I think OG is correct - it's just got to be an iterative process of trial and error (excel knocks it out quickly enough to 611 as F30 says)

[vrach]

I've gotten as far as all of the above.I even tried expressing a in terms of b and substituting back into OG's equations.Boy! I got some painful truths... like 0 =0 and X= X !!!
Pass it on to your mathematically inclined friends.Maybe they'll come up with something.Thanks anyway,guys.

[factor-fiction]

Are you sure you are suppose dto use equations to SOLVE it or have you just been asked to use them to REPRESENT the problem

[dr b]

OG has the equations; add the two constraints that A, B and C are integers, and that 100<=X<=999, and you have specified the problem. There does not appear to be any way to solve the system analytically, but trial and error is a perfectly valid way to do it.

[factor-fiction]

You can also represent it using X= 100u +10v +w where u, v and w are integers between 0 and 9

[Jonathan-Joe]

For each set of numbers (7 2), (8 3) and (11 6) the difference is the same (= 5). This suggests a way of finding the required number.

Let X be the required number
X divided by 7 leaves remainder 2, so X - 5 is a multiple of 7
X divided by 8 leaves remainder 3, so X - 5 is a multiple of 8
X divided by 11 leaves remainder 6, so X - 5 is a multiple of 11
The lowest multiple of 7, 8 and 11 is 7 x 8 x 11 = 616

So X = 616 - 5 = 611.

[bigbadmarty]

nice one jj

[sunny-dave]

Now that is good - <claps hands> - nicely thought through J-J

[Jonathan-Joe]

Thanks, I suppose it's just a kind of lateral thinking.

[factor-fiction]

There's something in that, JJ but it doesn't quite work.

"X divided by 7 leaves remainder 2, so X - 5 is a multiple of 7 ". I've got one eye on Corrie, but I don't see the logic of that step. And it's not quite right- 616 is divisible by 7 , but 616-5 isn't. Do you mean x+5 is divisible by 7.
And yes, the differences of 5 works for (7 2), (8 3) and (11 6) but does it work for other pairs?
I'll continue in a bit- I've got to pop out.

[Jonathan-Joe]

Works well enough for me.

[dr b]

Yes, with X+5 instead of X-5, JJs method works fine (X+5 = 616, so X = 611). It depends on the fact that the divisor minus the remainder is the same for all 3 cases. As long as that's the case, it's general, e.g. try it with:

x/9 -> remainder 3,
x/8 -> remainder 2,
x/7 -> remainder 1,

9*8*7 - 6 = 504 - 6 = 498, which works.

[Jonathan-Joe]

Sorry for miskey: "X - 5" should be "X + 5"
So X + 5 = 616, so X = 611

[factor-fiction]

Yes, well done JJ for using the lowest common multiple idea.
The reason this works can be shown algebraically as follows.

Let's start with OG's equations:
X = 7A + 2
X = 8B + 3
X = 11C + 6

[factor-fiction]

Eh, I was just typing and the answer submitted itself. I'll try again in a second

[factor-fiction]

Yes, well done JJ for using the lowest common multiple idea.
The reason this works can be shown algebraically as follows.

Let's start with OG's equations in which A, B and C as integers:
(1) X = 7A + 2
(2) X = 8B + 3
(3) X = 11C + 6

Then add 5 to each side of each equation:
(1) X +5 = 7A + 7
(2) X +5= 8B + 8
(3) X +5 = 11C +11

Then factorise the right hand sides:
(1) X +5 = 7(A + 1)
(2) X +5= 8 (B + 1)
(3) X +5 = 11(C +1)

We can see that X+5 is a multiple of 7, 8 and 11. As JJ says, the lowest common multiple of 7,8 and 11 is 616, calculated as 7 x 8 x 11 in this case. (The next multiple is 1232 but that is well over 1000 so we can discount that and higher multiples.)

So X+5= 616
therefore X =611

[Old_Geezer]

Wish I was clever.

[vrach]

Many hands make light work!I honestly thought we were stuck.Then along came jj and developed OG's f-las.Deep bow to you all.I was trying to help an 11-yr kid all the way in Columbia with his home work.Definitely a mischievous maths teacher.(this one is much more complex than the 2 boys on a hill looking at a flock of birds....younger says:"must be 100 birds up there!". Elder one says:"Nope! It would take just as many +half as many + quarter as many + one more to be 100" How many birds were up there? Candy for babies.Enjoy!)