If k and m are integers and k = m*(m+4)*(m+5), k must be divisible by

Question

If k and m are integers and k = m*(m+4)*(m+5), k must be divisible by which of following(s)?

I. 3 
II. 6 
III. 8

A. I
B. I and II
C. II and III
D. II
E. III

generis · Accepted Answer

Plug in m = 1, m = 2, m = -7

(1)(5)(6) = 30 
divisible by 3 and 6

(2)(6)(7) = 84
divisible by 3 and 6

(-7)(-3)(-2) = -42
divisible by 3 and 6

Answer B

niks18 · Answer

Mathematical approach

k=m*(m+4)*(m+5) = (m+3-3)*(m+4)*(m+5)

or  k = (m+3)*(m+4)*(m+5)-3(m+4)*(m+5)

Now,  (m+3) ,  (m+4)  &  (m+5)  are three consecutive integers, hence   MUST   be divisible by  3  &  6

-3(m+4)*(m+5) , has  3  as a factor and  (m+4)  &  (m+5)  are consecutive integers, hence either of the two will be a multiple of  2 . Hence the product   MUST   be multiple of  3  &  6

Therefore  k    MUST   be divisible by  3  &  6

Option  B

thangvietnam · Answer

if m is divided by 3, so the expression is divided by 3
if m is not dividec by 3, m+1 or m+2 is divided by 3, and m+4 or m+5 is divided by 3.
in conclusion, the expression is divided by 3

m+4 and m+5 are consecutive numbers , so, one of them must be divided by 2
so, the expression is divided by 3

for 8
if m is divided by 2, but not divided by 4, m+4 is divided by 2. and m+7 is odd. so the expression is not divided by 8

answer is b.

thangvietnam · Answer

if m is divided by 3, so the expression is divided by 3
if m is not dividec by 3, m+1 or m+2 is divided by 3, and m+4 or m+5 is divided by 3.
in conclusion, the expression is divided by 3

m+4 and m+5 are consecutive numbers , so, one of them must be divided by 2
so, the expression is divided by 3

for 8
if m is divided by 2, but not divided by 4, m+4 is divided by 2. and m+7 is odd. so the expression is not divided by 8

answer is b.

ScottTargetTestPrep · Answer

Roman Numeral I:

If m is 1, then k = 1 x 5 x 6.

If m is 2, k = 2 x 6 x 7.

If m is 3, k = 3 x 7 x 9.

In all three cases, we see that m*(m+4)*(m+5) must be divisible by 3. A more dynamic proof to show that k is divisible by 3 is to let m = 3q + r where q is an integer and r = 0, 1 or 2.

If m = 3q, obviously k is divisible by 3 since m is a multiple of 3.

If m = 3q + 1, then k is divisible by 3 since m + 5 = 3q + 6 is a multiple of 3.

If m = 3q + 2, then k is divisible by 3 since m + 4 = 3q + 6 is a multiple of 3.

Hence, k is always divisible by 3, regardless of the value of integer m.

Roman Numeral II:

We observe that m + 4 and m + 5 are two consecutive integers, hence, one of them is necessarily even. We know that k is even and from Roman Numeral I, we also know that k is divisible by 3. Thus, k must be divisible by 6.

Roman Numeral III:

We notice that m and m + 4 have the same parity (they are both odd or both even) and the parity of m + 5 is opposite to that of m (if m is even, m + 5 is odd and if m is odd, m + 5 is even). When we have a product of three integers, the product can be divisible by 8 if a) one of the integers is divisible by 8, b) one of the integers is divisible by 4 and one of the integers is divisible by 2 or c) each of the integers is divisible by 2.Since the parity of m + 5 is opposite to the remaining integers, it is impossible that each of m, m + 4 and m + 5 are divisible by 2. So if we pick a value for m such that neither of m, m + 4 and m + 5 are divisible by 4, the product cannot be divisible by 8. Such a value is m = 1, in which case k = 1*5*6 = 30. As we can see, k is not necessarily divisible by 8.

Answer: B

Mansoor50 · Answer

i understand that k is even and that k is divisible by 3 but dont follow why is this div by 6.

Also, any particular reason why you chose m as 1 2 3?

thanks

ScottTargetTestPrep · Answer

i understand that k is even and that k is divisible by 3 but dont follow why is this div by 6.

Also, any particular reason why you chose m as 1 2 3?

thanks

For the first one, we are using the rule that if x is divisible by n and m, then x is divisible by LCM(n, m). If k is even (i.e. divisible by 2) and k is divisible by 3, then k has to be divisible by LCM(2, 3) = 6.

For the second question, the values m = 1, 2 and 3 correspond to the simplest special cases of the more general argument to follow. It's sometimes helpful to look at a few special cases before attacking the general case.

CrushTHYGMAT · Answer

CrackverbalGMAT ,  avigutman  could you please share your approach for this question? How can we look at it more logically instead of plugging random numbers?

avigutman · Answer

Hi CrushTHYGMAT , I think you'll find exactly what you're looking for here: https://gmatclub.com/forum/if-n-is-an-i ... l#p2662420 This will enable you to solve such problems in under 30 seconds, with no pen or paper.

karakman · Answer

Is there any reason why we can't select m as zero? So k=0(4)(5) which would not be divisible by 6.

Krunaal · Answer

Every integer is a divisor of 0, except, by convention, 0 itself. In this case  \frac{0}{6} = 0

Here's a useful post on  Number Theory

Usernamevisible · Answer

Think in **patterns (mod / parity)** — quick GMAT way:

---

### Step 1: Test divisibility by **3**

Numbers:
**m, m+4, m+5**

👉 Mod 3 cycle is every 3 numbers. These three **cover all residues**:

* If m ≡ 0 → already divisible
* If m ≡ 1 → m+5 ≡ 0
* If m ≡ 2 → m+4 ≡ 0

✅ **Always one multiple of 3 → Product divisible by 3**

---

### Step 2: Test divisibility by **6** (needs 2 AND 3)

We already have **3**, so just check **even (2)**:

* If m is even → m and m+4 are even
* If m is odd → m+5 is even

✅ **Always at least one even → Product divisible by 2**

👉 So product has **2 × 3 = 6**

---

### Step 3: Test divisibility by **8**

Need **3 factors of 2**

Try a quick counterexample:

* m = 1 → k = 1×5×6 = 30 (not divisible by 8)

❌ Not guaranteed

---

### Final Answer:

**B. I and II**

SHREYA292K · Answer

If I take m=4 - I'll get 4*8*9 which is divisible by 8

If k and m are integers and k = m(m+4)(m+5), k must be divisible by

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