I am really stuck on this problem. Please help?

Question

An even number x divided by 7 gives some quotient plus a reminder of 6. Which of the following, when added to x, gies a sum which must be divisible by 14?

Answer 1

a possible solution:

x/7=y+6

First solve for x
x=20

Then put x back into the equation to find y
20/7=2 (with the remainder of 6)
y=8

x+y=28 ->an even number divisible by 14

Answer 2

1. First list all Multiples of 7 until you see a pattern
0, 7, 14, 21, 28, 35, 42, 49
(adding 7 each time)

2. Second, list all Numbers that divide by 7 with remainder 6
until you see a pattern
6, 13, 20, 27, 34, 41, 48, 55
(adding 6 to each multiple of 7 above)

3. Pick out just the even numbers from this Second list =
6, 20, 34, 48
(Pattern appears to be "add 14" each time)

4. List all multiples of 14
0, 14, 28, 42, 56, 70 (add 14 each time)
These all divide by 14 evenly

What can you add to each number in Third list
to divide evenly by 14?
If x = 6, add 8, 22, 36 to get numbers divisible by 14
If x = 20, add 8, 22, 36 to get numbers divisible by 14
If x = 34, add 8, 22, 36 etc.

So the pattern looks like you have to add
8, 22, 36, 50, 64 (add 14 each time)
All these are "multiples of 7 plus 1" or 7n + 1

Does any of the choices look like this:
7n + 1

Any number or expression that is
"1 plus a multiple of 7" should work.

Answer 3

Since x divided by seven gives a remainder of 6, x + 1 will be divisible by 7 without a remainder. Add x + 1 again to it and the result will be divisible by 14. So the number to be added to x is
x + 2.to make it divisible by 14.

Answer 4

Our even number : 2x
By the divition formula:
Dividend = (divisor * quotient) + reminder
2x = 7q + 6

Substract 6 on both sides:

2x -6 = 7q

We need to look at the quotient q now. To solve our problem q must must be a multiple of 14, which means that it needs to be a multiple of 7 and 2 ( 7 *2 =14).
7q is already a multiple of 7, so now, it will depend if q is even or odd.
If q is odd, then we add +20 to our Dividend.
Our dividend is currently (2x-6)
by adding 20 ( 2x -6 +20 = 2x +14), and that will make the quotient even.
By the same idea if the quotient is even now, we need to add 34 (or add -6 if you like).

Would you prefer the answers in term of the number? You did not include any alternatives.

Answer 5

hi,
well..the even number X when divided by 7 yields a remainder 6. So X should be an even multiple of 7 + 6.

X = (7*Y)+6 where Y= 0, 2, 4, 6, 8........
X= 6 or 20 or 34 or...................

the remainder is 6 in each case and for any of these numbers to be divisible by 14, 8 should be added.

So your answer is 8

Answer 6

Even number divided by 7 gives a remainder of 6.

okay as an equation this is 2n - 6 = 7k. 2n is our even number. If we take 6 (the remainder away) this must be a multiple of 7.

For example, n = 10 gives 20 - 6 = 14. One even number which works is 20.

The next part of your question doesn't make sense. Modify and then I'll update my answer.

Answer 7

Which of the following WHAT? Looks like you're missing part of the problem.