Question in Math

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engineers feel

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hello guys

This is my question


Find three consecutive odd integers such that 3 times their sum is 5 more than 8 times the middle one
 
Homework, eh?

First you write down what you know, ie the question in form of math notation:

For integers, a,b and c, where a<b<c, and (a+1) = mod 2, (b+1) = mod2, (c+1) = mod2. Solve for a,b,c, when 3(a+b+c) = 5 + 8b.

Alternatively, if you read the question, you're not meant to calculate the answer, just find a,b,c so that the statement is true. For example, is it true for a=1, b=2, c=3?
 
OutToLunch said:
you're kidding, right?

this the level of math at your college?
Click to expand...


hello guy

what's up man?

I haven't told you that the question is from my book

I've written it for enjoying

that's all
I hope to understand me

see ya
 


good for you

you're all right
but in my opinion if you take a look at the question you'll see that
three odd intergers


thanks again
 
engineers feel said:
good for you

you're all right
but in my opinion if you take a look at the question you'll see that
three odd intergers


thanks again
Click to expand...

Whops, my bad. The maths statement still holds ( x + 1 = mod 2, gives x an odd number). a, b, and c for values 1, 3, and 5, respectively.
 
I saw that the statement is true for a=3, b=5, c=7. But that was before I realised it the statement is 3(a+b+c)=8b-5, and not 3(a+b+c)=8b+5...
 
It is too easy to be a engineering collage level problem. As a rule I get such things wrong so let me give it a try. If I am wrong I am sure there will be no shortage of people with the right answer.

Find three consecutive odd integers such that 3 times their sum is 5 more than 8 times the middle one

The queston asks for three consecutive odd integers,
I would thake this to be numbers like
x, x+2, x+4 where x is odd.


(3 * (X + (x+2) + (x+4) ) ) -5 = ((X + 2) * 8)
3x + 3x + 6 + 3x + 12 = 8x + 16 + 5
x + 18 = 21
x = 3

check
3 * (3 + (3 +2) + (3+4) ) = ((3 + 2) * 8) + 5
3 * (3 + (5) + (7) ) = ((5) * 8) + 5
3 * (15 ) = (40) + 5
45 = 45

so the numbers are 3, 5, and 7
 
johankj said:
That is correct, apart from it is 3(a+b+c)+5=8b not 3(a+b+c)=8b+5.
Click to expand...
no, 3(a+b+c)=8b+5 is correct.

the problem stated that 3 times the sum is 5 more than 8b.

3(a+b+c)+5=8b
would translate into 8b is 5 more than 3 times the sum.
 
no, 3(a+b+c)=8b+5 is correct.

the problem stated that 3 times the sum is 5 more than 8b.

3(a+b+c)+5=8b
would translate into 8b is 5 more than 3 times the sum.
Click to expand...

Ah. Right you are..
 
As always I complicate things, congruent modulo was not the way to go. (However, I conjecture that it would give you all solutions to the problem if they exist...)
 
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