When 140 is divided by positive integer k, the remainder is k - 12

--ASIDE---------------
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
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In this question, we're told that 140 divided by k leaves a remainder of k - 12.
Since we aren't told the quotient (Q), let's just say that the quotient is q
In other words, 140 divided by k equals q with remainder k - 12.

We can now apply the above rule to get: 140 = kq + (k - 12)
Add 12 to both sides of the equation to get: 152 = kq + k
Factor the right-hand side to get: 152 = k(q + 1)

IMPORTANT: 152 equals then product of k and (q + 1). Since k and (q + 1) are both integers, we now know that k is a divisor of 152

152 = (2)(2)(2)(19)
So, 38 aka (2)(19) is a possible value of k

Answer: C

Cheers,
Brent

RELATED VIDEO

Approach #2
One of the great things about Integer Properties questions is that they can often be solved by testing the answer choices
So, you should give yourself about 30 seconds to find a faster (algebraic) approach to this question, and if you don't come up with anything, start testing answer choices.

(A) 16
We get: when 140 is divided by 16, the remainder is 16-12 (aka 4)
Not true.
140 divided by 16 equals 8 with remainder 12.
Eliminate A

(B) 28
We get: when 140 is divided by 28, the remainder is 28-12 (aka 16)
Not true.
140 divided by 28 equals 5 with remainder 0.
Eliminate B

(C) 38
We get: when 140 is divided by 38, the remainder is 38-12 (aka 26)
True
140 divided by 38 equals 3 with remainder 26.

Answer: C

Cheers,
Brent

When 140 is divided by k the remainder is k-12

152= 4*38

i.e. k must be 38

answer: Option C

140modk = k-12

(A) 16
140mod16 = -4mod16 = 16-4 = 12 <>16-12=4
(B) 28
140mod28 = 0 <>28-12 = 16
(C) 38
140mod38= 26 = 38-12; COULD BE POSSIBLE VALUE OF k
(D) 48
140mod48 = 44 <>48-12 = 36
(E) 51
140mod51 = 38 <> 51 -12 = 39

IMO C

Let's solve the problem using two methods

Method 1: Substitution (Substitute values of k from answer choices and check which one satisfies)

(A) 16
140 when divided by 16 gives 12 remainder ≠ (16-12 = 4) => FALSE

(B) 28
140 when divided by 28 gives 0 remainder ≠ (28-12 = 16) => FALSE

(C) 38
140 when divided by 38 gives 26 remainder = (38-12 = 26) => TRUE
In Test, we don't need to solve further. But I am solving to complete the solution.

(D) 48
140 when divided by 48 gives 44 remainder ≠ (48-12 = 36) => FALSE

(E) 51
140 when divided by 51 gives 38 remainder ≠ (51-12 = 39) => FALSE

So, Answer will be C

Method 2: Algebra (Substitute values of "a" and try to find the value of k)

When 140 is divided by positive integer k, the remainder is k - 12

Theory: Dividend = Divisor*Quotient + Remainder

140 -> Dividend
k -> Divisor
a -> Quotient (Assume)
k - 12 -> Remainders
=> 140 = k*a + k - 12
=> 152 = ak + 1

Let a = 1
=> 152 = k*1 + k = 2k
=> k = \(\frac{152}{2}\) = 76
But that is Not an answer choice given to us

Let a = 2
=> 152 = k*2 + k = 3k
=> k = \(\frac{152}{3}\) = 50.6 => Not an integer => NOT POSSIBLE
But that is Not an answer choice given to us

Let a = 3
=> 152 = k*3 + k = 4k
=> k = \(\frac{152}{4}\) =38

So, Answer will be C
Hope it helps!

Watch the following video to learn the Basics of Remainders