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HKD1710
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An even number x divided by 7 gives some quotient plus a remainder of 6. Which of the following, when added to x, gives a sum which must be divisible by 14?
(A) 1
(B) 3
(C) 7
(D) 8
(E) 13
(A) 1
(B) 3
(C) 7
(D) 8
(E) 13
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HKD1710
The easiest method is just plugging in a number.
Such an even number x could be say 20 ( = 14 + 6)
You need to add 8 to it to make it divisible by 14.
So answer (D)
You don't have to worry about getting another answer in some other case since this is a PS question. In every case, you will always get 8 as the answer.
Alternatively, think in terms of grouping.
Even number x divided by 7 (groups of 7 balls each) leaves 6 balls (remainder). How many groups are there? Since the number is even, when you remove 6 balls, you still have even number of balls left so it must be even number of groups of 7. You can pair them out to make groups of 14 balls each. You will be left with 6 balls. You need to add 8 balls to make another group of 14 balls.
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As x is even , say x= 7*2a + 6 ( say 2a is the quotient)
or , x= 14 a + 6
14a is always div by 14. So the Q is which option when added to 6 , the sum is divided by 14 ?
It would be 8 ( we donot even need to add and check .... as it should be an even one ..... and the only even option is D) 8 ).
Hence Ans D.
or , x= 14 a + 6
14a is always div by 14. So the Q is which option when added to 6 , the sum is divided by 14 ?
It would be 8 ( we donot even need to add and check .... as it should be an even one ..... and the only even option is D) 8 ).
Hence Ans D.
General Discussion
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HKD1710
The least number which when divided by 7 gives a remainder of 6 is the number 6 itself (Which is even)
Now, for the number to be completely divisible by 14 we must add 8 to it...
Thus, Answer must be (D) 8
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HKD1710
x/7 = q + 6/7
x = 7*q + 6
since x must be even so 7*q = even
If q = 2 then 14+6 = 20
x = 20 + 8 = 28
Answer choice D
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HKD1710
\(x= 7q+6\)
test values for \(Q\)
7*1+6 = 13
7*2+6 = 20 no need to go further since 20+8 is divisible by 14
D
HKD1710
Hey everyone, first thanks for the practice!
I'm wondering if we could see this problem another way ( I ended up getting the right answer with my reasoning but i'm not 100% sure of it).
Here it is:
We are told that the number is even, and that a remainder of 6 is given when divided by 7.
Now we are also told that the number must be divisible by 14: Thus, the end number has to be even (divisible by 2) and divisible by 7.
To solve this:
even number + even number = even number --> eliminate all answer choices except D.
Is it correct? I also wonder how i should takcle this in case 2 even numbers would be present in the answer.
iac00
I dont think that it would be right all the time.
In this particular question, only 1 answer option was even, hence it was easy to choose the answer.
eg, if there were 6, 8 and 10 also in the options then the option may not be true.
We know that
x=7q+6--- where q=2a for some number a
hence, x= 14a+6
to have a number completely divisible by 14
it should by
x+b= 14a+14--- b is the number to be added to x for 14 to completely be able to divide.
14a+6+b=14a+14
6+b=14
b=14-6=8
with this we know that 8 is the correct even number, and you can discard the other even options.
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HKD1710
even no X=7.a+6
it can be noted that for all even factors of 7 viz, 14, 28 so on would give a remainder 6 for even integer X.
so X can be 20,34 which when added by 8 would be divisible by 14... hence option D
HKD1710
Take even number when divided by 7 leaves remainder 6-(6, 20, so on)
now plug the answer choices in 6 or 20 -so that result must be divisble by 7
On plugging , you will find only 8 to satisfy the conditon
Hence OA:D
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HKD1710
We can let x = 20 (notice that 20/7 = 2 R 6). Let’s try each answer choice, to see which number, when added to 20, is divisible by 14.
Choice A. Since (20 + 1)/14 does not equal an integer, then A is incorrect.
Choice B. Since (20 + 3)/14 does not equal an integer, then B is incorrect.
Choice C. Since (20 + 7)/14 does not equal an integer, then C is incorrect.
Choice D. Since (20 + 8)/14 =2, then D is correct.
Alternate Solution:
We note that an integer is divisible by 14 if and only if it is divisible by both 2 and 7.
Since x divided by 7 produces some quotient and a remainder of 6, we can let x = 7q + 6 for some integer q. We notice that if we add 1 to x, we obtain x + 1 = 7q + 7; which is divisible by 7 but not by 2, since x is even and x + 1 is odd. If we add 7 to x + 1, we will obtain x + 8 = 7q +14; which is divisible not only by 7 but also by 2, since even + even = even. Thus, adding 8 to x will produce a number that is divisible by 14.
Answer: D
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Method - Plugging in values.
The number which is even and nearest multiple of 7 is 20.
20 /7, gives the remainder 6.
6, when added to 8, gives 14, which is divisible by 14.
The number which is even and nearest multiple of 7 is 20.
20 /7, gives the remainder 6.
6, when added to 8, gives 14, which is divisible by 14.
An even number x divided by 7 gives some quotient plus a remainder of 6. Which of the following, when added to x, gives a sum which must be divisible by 14?
x = 7q +6
Plug in some numbers: x = 20
20 = 7(2) + 6
Now lets check the answers:
A. 20 + 1 /14 = False.
B. 20 + 3 / 14 = False
C. 20 + 7 / 14 = False
D. 20 + 8 / 14 = True.
x = 7q +6
Plug in some numbers: x = 20
20 = 7(2) + 6
Now lets check the answers:
A. 20 + 1 /14 = False.
B. 20 + 3 / 14 = False
C. 20 + 7 / 14 = False
D. 20 + 8 / 14 = True.
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HKD1710
Just plug in a number for x. Let's make it 6. What do we need to add to 6 to make it divisible by 14?
Answer choice D.
Posts: 29
I used 13 to plug. 13/7 is 1 with r6. If you add 1 to 13, that becomes 14 and is thus divisible by 14. What am I doing wrong?
tomloveless
I suppose you missed the constraint: x is an even number.
Quote:
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Dang I did, appreciate it. I was driving myself mad trying to figure out what I missed. Long day.
Krunaal
Posts: 20
Why doesnt 1 work then? If i assume x=55 then R=6 on Denom 7.
Adding 1 gives 56 which is divisible by 14. Also divisible by both 2 and 7
Adding 1 gives 56 which is divisible by 14. Also divisible by both 2 and 7
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ofc! thank you
Bunuel
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