When 10 is divided by the positive integer n, the remainder is n - 4.

SOLUTION

When 10 is divided by the positive integer n, the remainder is n - 4. Which of the following could be the value of n?

(A) 3
(B) 4
(C) 7
(D) 8
(E) 12

Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

Original question says that when 10 is divided by the positive integer n, the remainder is n-4, so \(10=nq+(n-4)\) and also \(n-4\geq{0}\) or \(n\geq{4}\) (remainder must be non-negative).

\(10=nq+n-4\) --> \(14=n(q+1)\) --> as \(14=1*14=2*7\) and \(\geq{4}\) then --> \(n\) can be 7 or 14.

Answer: C.
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Let’s test each answer choice:

A) n = 3

10/3 = remainder 1, which does not equal 3 - 4.

B) n = 4

10/4 = remainder 2, which does not equal 4 - 4.

C) n = 7

10/7 = remainder 3, which does equal 7 - 4.

Answer: C
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I simply did it this way:

\(y = 10n + (n - 4)\)

And then I just plugged in values for n. So:
\(10(0) + (0-4) = -4\)
\(10(1) + (1-4) = 7\) <--my answer

Is this correct reasoning, or simply a lucky coincidence?

hi erikvm,
here it is a lucky coincidence..
here you have found the value of y and not n...

\(10(1) + (1-4) = 7\)... here n=1 and y=7.... but the question asks us the value of n...
the equation that you have formed is wrong ..
it says when 10 is divided by the positive integer n, the remainder is n-4.. so the equation will be 10=xn+n-4..

what we have to find is that " when 10 is divided by n, the remainder is n-4"
now substitue the values
(A) 3.... 10/3 rem=1, which is not equal to n-4=3-4=-1.... not correct
(B) 4.... 10/4 rem=2, which is not equal to n-4=4-4=0.... not correct
(C) 7.... 10/7 rem=3, which is equal to n-4=7-4=3.... correct
(D) 8.... 10/8 rem=2, which is not equal to n-4=8-4=2.... not correct
(E) 12... 10/12 rem=10, which is not equal to n-4=10-4=6.... not correct

hope it is clear..
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Here's a slightly trickier question that is good for further practice on this OG QR2 question:

https://gmatclub.com/forum/when-81-is-divided-by-the-cube-of-positive-integer-z-the-remainder-is-197386.html#p1522792

Wish you all the best!
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Hi,

Here are my two cents for this question. This solution is same as all others but want to share one important aspect of thinking in case of remainders.

My solution would appear little lengthy, but if you can visualize certain things about remainders, we could solve this question this question in about 33 Secs.

Here we are told that when \(\frac{10}{n}\) we have remainder in the form (n-4)

So we can write this \(\frac{10}{n}\) as 10 = nk +(n-4) ------(1)

where 0\(\leq{reminder}\)<n ; where 0\(\leq{n-4}\)<n ( This is the first hint toward rejecting incorrect choice A, B )

which says 4\(\leq{n}\)

here we can have two cases
Case 1: if n>10
then nk =0 and we can re-write the equation (1) as .
10 = n-4
where n =14
This is our second hint for rejecting Answer Choice "E" Consider a case where n = 12 so we have \(\frac{10}{12}\)= 12*0+ 10) here remainder is 10 . But using the equation given to us in the question stem that remainder is n-4 , which in this case would be 12-4 =8 So we can reject answer choice E


Case 2:
If n< 10,
then we have k is a positive integer.
\(\frac{n}{10}\) = nk +(n-4) ------(1)
and simplifying we get
10+4= nk+n
14=n(k+1)
so we have
14= 7(1+1)
so we have n=7

Hence form given choices we have answer to the question which is choice "C"
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We can always plug in one by one and test which works.
(A) The remainder is 3 - 4 = -1 which doesn't make sense.
(B) 10/4 gives a remainder of 2, n - 4 = 4 - 4 = 0 however so not B.
(C) 10/7 gives a remainder of 3. n - 4 = 7 - 4 = 3. Bingo.

Ans: C
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APPROACH #1: I'd say that the fastest approach is to simply test answer choices

(A) 3
The question tells us that we get a remainder of n - 4
So, if n = 3, then the remainder = 3 - 4 = -1, which makes no sense (the remainder must always be greater than or equal to 0)
Eliminate A

(B) 4
Plug n = 4 into the given information to get: When 10 is divided by 4, the remainder is 4 - 4 (aka 0)
This is not true. When we divide 10 by 4, we get reminder 2
Eliminate B

(C) 7
Plug n = 7 into the given information to get: When 10 is divided by 7, the remainder is 7 - 4 (aka 3)
WORKS!

Answer: C
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APPROACH #2: Apply the rule for rebuilding the dividend

There's a nice rule that says, "If N divided by D equals Q with remainder R, then N = DQ + R"
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3

Given: When 10 is divided by the positive integer n, the remainder is n - 4
Since we're not told what the quotient is, let's just say that it's q.
In other words: When 10 is divided by the positive integer n, the quotient is q, and the remainder remainder is n - 4

When we apply the above rule we get: 10 = nq + (n-4)
Add 4 to both sides of the equation to get: 14 = nq + n
Factor to get: 14 = n(q + 1)

Importance: Since n and (q + 1) are INTEGERS, n must be a FACTOR of 14.

Check the answer choices.... only answer choice C (7) is a factor of 14

Answer: C

Cheers,
Brent

RELATED VIDEO

The QUICKEST Way that I have found when dealing with remainders that are n +- some value is by dealing with the problem as below:

\(\frac{10}{n} = Q + \frac{(n-4)}{n}\)

Is the same thing as...

\(\frac{10}{n} + \frac{4}{n} = Q + \frac{n}{n}\)

\(\frac{10}{n} + \frac{4}{n} = (Integer)\)

The only value that 14 could be divided into to create an integer is 7.

Given that when 10 is divided by the positive integer n, the remainder is n - 4 and we need to find which of the following could be the value of n

Theory: Dividend = Divisor*Quotient + Remainder

Lets solve the problem using two methods

Method 1: Substitution

Let's take each option choice and see which one satisfies the question

(A) 3

=> n = 3. But remainder cannot be 3-4 = -1 => NOT POSSIBLE

(B) 4

=> 10 when divided by 4 gives 2 remainder ≠ (4-4 =0) => NOT POSSIBLE

(C) 7

=> 10 when divided by 7 gives 3 remainder = (7-4 =3) => POSSIBLE
In Test, we don't need to solve further, but I am solving to complete the solution

(D) 8

=> 10 when divided by 8 gives 2 remainder ≠ (8-4 =4) => NOT POSSIBLE

(E) 12

=> 10 when divided by 12 gives 10 remainder ≠ (12-4 =8) => NOT POSSIBLE

So, Answer will be C

Method 2: Algebra

10 when divided by n gives n - 4 as remainder

10 -> Dividend
n -> Divisor
a -> Quotient (Assume)
n-4 -> Remainders
=> 10 = n*a + n-4 = na + n - 4

Putting a =1 we get
10 = n*1 + n - 4
=> 10 = n + n - 4 = 2n - 4
=> 2n = 10 + 4 = 14
=> n = \(\frac{14}{2}\) = 7

So, Answer will be C
Hope it helps!

Watch the following video to learn the Basics of Remainders


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