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

[Jonathan-Joe]

I suppose the point is that not every every numerical problem can easily be resolved to an equation (or even set of equations). That is often the case for "real life" problems but here we have a contrived "brain teaser", designed to be solved using insights into the nature of numbers.
It can be difficult enough for teachers to teach "number crunching", thank goodness some are "mischievious" enough to encourage "outside the box" thinking.

[factor-fiction]

Sorry vrach, but if I've understood the birds question correctly I get an answer which involves a fraction of a bird!
Using 56 gives an answer of 56 +28 +14 +1=99.
The next whole number is 57 but your formula gives an answer of 57 +29.5 + 14.25 +1=101

[dr b]

The question was a little hard to interpret, but I read it as, there are X birds and it would take X more + .5X more + .25X more +1 to get to 100, so 2.75X + 1 = 100 --> X=36.

[vrach]

Exactly. X+X+0.5X+0.25X + 1 = 100.Yes, jj.That was the kind of brain teaser that would make a child want to become a mathematician.
Thanks again,all you guys.

[geeknick]

Hi All,

I understand most algebra problems that have exponents (equivocate the bases and solve). But I ran into a problem from the Veritas Algebra book and the feedback provided from the Veritas book/forums wasn't up to par.

I'd like some real step by step guidance on this problem:

3^x - 3^x-1 = 2(3^13)

the answer is 14. I've worked through the problem as it has been laid out in the book but I cannot understand the methodology entirely in order to apply it to another problem.

The book denotes that you can factor out a 3^x OR (more efficiently) factor out a 3^x-1.

3^x-1 (3-1) = 2(3^13)
3^x-1(2) = 2(3^13)
x-1 = 13
x = 14

I don't fully understand this logic, can someone please clarify the underlying math logic/rules applied that is going on here please?

Thanks in advance,

NICK Media URL: http://solvealgebraequations.wordpress.com/2011/10/31/take-tutorvista-help-to-enlighten-algebraic-equations/
Description:

[geeknick]

Hi All,

I understand most algebra problems that have exponents (equivocate the bases and solve). But I ran into a problem from the Veritas Algebra book and the feedback provided from the Veritas book/forums wasn't up to par.

I'd like some real step by step guidance on this problem:

3^x - 3^x-1 = 2(3^13)

the answer is 14. I've worked through the problem as it has been laid out in the book but I cannot understand the methodology entirely in order to apply it to another problem.

The book denotes that you can factor out a 3^x OR (more efficiently) factor out a 3^x-1.

3^x-1 (3-1) = 2(3^13)
3^x-1(2) = 2(3^13)
x-1 = 13
x = 14

I don't fully understand this logic, can someone please clarify the underlying math logic/rules applied that is going on here please?

Thanks in advance,

NICK Media URL: http://solvealgebraequations.wordpress.com/2011/10/31/take-tutorvista-help-to-enlighten-algebraic-equations/
Description:

[bibblebub]

Using the convention where . represents multiplication:

3^x = 3.3^(x-1)

therefore 3^x - 3^(x-1) = 3.3^(x-1) - 3^(x-1) = 2.3^(x-1)

so 3^x - 3^(x-1) = 2.3^13 is the same as 2.3^(x-1) = 2.3^(13)
i.e. (x-1) = 13
thus x = 14

[factor-fiction]

Another (very similar) way of doing it is as follows. I have used the letter n instead of x because it's easier for me to type and it avoids confusion with the multiplication sign.

3ⁿ - 3ⁿ⁻¹= 2 (3¹³)

Multiply throughout by 3
3(3ⁿ) - 3ⁿ = 2(3¹⁴)

Take out 3ⁿ as a factor on left hand side:
3ⁿ (3-1)= 2 (3¹⁴)
i.e. 3ⁿ (2)= 2 (3¹⁴)

Divide both sides by 2:
3ⁿ = 3¹⁴
Therefore n = 14