What is the positive integer n?

Product of consecutive integers have some properties. e.g. product of any two consecutive positive integers is even, product of any 3 consecutive integers is even and is divisible by 3 so basically the product is divisible by 6.

The first statement just tells you that every product of (n+1) consecutive integers is always divisible by 16.

Say if n = 3
Is 1*2*3*4 divisible by 16? No! So n cannot be 3.

Is 1*2*3....*14*15*16 divisible by 16? Yes. Will product of any 16 consecutive integers be divisible by 16? Yes. Product of any 17 consecutive integers will also be divisible by 16. So n can take many values.

What is the smallest value that n can take?
Every set of 4 consecutive integers will have a number which has 4 as a factor and it will have another even number i.e. the product of 4 consecutive numbers must be divisible by 8.
When you have product of 5 consecutive factors, again the product must be divisible by at least 8 e.g. 1*2*3*4*5
When you have 6 consecutive factors, there must be a number with 4 as a factor and 2 other even numbers i.e. the product must have 16 as a factor e.g. 1*2*3*4*5*6
Therefore, you must have at least 6 consecutive integers i.e. n must be at least 5.

fameatop: n cannot be 4. If you have 5 consecutive factors, the product will not be divisible by 16 in every case e.g. it is not divisible in case of 1*2*3*4*5

Statement 1 tells you that n is at least 5
Statement 2 tells you that n is 4 or 5.

So n must be 5 using both
Answer (C)

Hi

From statement 1, if we consider m=2,
2*3*4*5*6 will be divisible by 16.
and here m+n=6 => n=4

So IMO the ans should be E

please correct me if I am wrong.

Statement (1) says: For every positive integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16.
If m = 1 and n = 4, 1*2*3*4*5 is not divisible by 16. So n cannot be 4.

Hi

Again my question, why are we only considering the case where m=1?
why cant we consider m = 2 or the series starting from 2
In this case, 2*3*4*5*6 is divisible by 16

We are not considering only the case m=1. For a given n, the product m(m+1)(m+2)...(m+n) must be divisible by 16 for m=1, also for m=2, for m=3, ..., for every integer m. But if for a specific value of m the product is not divisible by 16, that particular n is not an acceptable value. Since for m=1 the product is not divisible by 16 when n=4, we can conclude that n cannot be 4. The property doesn't for for m=9 either: 9*10*11*12*13. If you continue, you can find even more values for m for which the property will not hold when n=4.

It is a technique we use. We want to find the value of n which holds for every possible value of m. When you want to prove that a particular value of n will hold for every value of m, it is very hard to do. You possibly cannot calculate for EVERY value of m. So we use the reverse logic. We assume that n takes a particular value and then try to find one value of m for which it doesn't hold. Then we can say that n cannot take that particular value.

Here, we are saying that if n = 4, it doesn't hold for m = 1. It doesn't matter whether it holds for other values of m or not. As long as there is even a single value of m for which it doesn't hold, we know n cannot be 4.

When n = 5, we see that it holds for quite a few values of m. Now the question is, will it hold for every value of m? For that, we use our theoretical knowledge to say that yes, it will hold for every value of m.

Found it really understandable.. Thank You So much Sir .

if (1) is not friendly go to (2) and take out possible values..it will save lot of time and thinking..

Regards
Ishdeep Singh

Statement I is insufficient:
If n = 8 then m(m+1)(m+2)...(m+8) will have at least 4 even integers which means the product will have to be divisible by 16
same goes for n = 9, n = 10 etc....

Statement II is insufficient:
n^2 - 5n - 4n + 20 = 0
n(n-5) - 4( n - 5) = 0
(n-5) (n-4) = 0

We get two value of n, hence not sufficient.

Combining is sufficient:

If we say that n = 4 then m (m+1) (m+2) (m+3) (m+4) will not be divisible by 16 if m = 1 (1 (2)(3)(4)(5))
If we say that n = 5 then m (m+1) (m+2)(m+3) (m+4)(m+5) will always be divisible by 16. Lets try m = 1(2)(3)(4)(5)(6) as there will be a minimum of 3 even numbers with a multiple of 4 between them. Hence a factor of 16 is a must.

Hence the answer is C.

What is the positive integer n?

(1) For every positive integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16:

In other words n+1 is the number of elements in a set of any consecutive positive integers, where n+1 guarantees divisibility of the product of these elements by 16 (the product of these consecutive integers should have four factors of 2). At the beginning of the number line there is a maximal scarcity of factors of 2 and this is where multiplying the largest number of consecutive positive integers will be required to attain divisibility by 16. 1*(1+1)*(1+2)*(1+3)*(1+4)*(1+5). Therefore when n >=5, the product of all elements in the set will definitely be divisible by 16 regardless of the position of the consecutive positive integers on the number line.

n >= 5. INSUFF


(2) n^2 - 9n + 20 = 0 n = 4 or 5. INSUFF

Combined: SUFF. Answer is C.

St.1
- There are a lot of values of n.
- Specifically for every n>=5 (sufficient condition, but not a must), we will have 3 consecutive even integers. and if we have 3 consecutive even integers, m(m + 1)(m + 2) ... (m + n) is divisible by 16.
- for (n=4, and m is even) we will have m(m + 1)(m + 2) ... (m + n) is divisible by 16.
-> not suff.

St2. n=4 or n=5.
-> not suff.

St1+St2
-> for n=5, m could be even or odd - >suff.
notice that if n=4, and m is odd -> m(m + 1)(m + 2) ... (m + n) is NOT divisible by 16. so n=4 is disqualified.

First of all note that the phrase "For every positive integer m, the product m(m +1)(m + 2)...(m + n)" simply implies product of any (n + 1) consecutive positive integers. Now, the product of any (n + 1) consecutive positive integers is always divisible by (n + 1)!. So if we have to make the product of any (n + 1) consecutive positive integers divisible by 16, we have to choose n in such a way that 16 is a factor of (n + 1)!. Which is only possible if (n + 1) ≥ 6, i.e. n ≥ 5

Statement 1: Implies the product of any (n + 1) consecutive positive integers is divisible by 16. Now the product of any 6 or more consecutive integers is always divisible by 16. Hence, (n + 1) ≥ 6 => n ≥ 5.

For any value of m other than 1, the product m(m +1)(m + 2)...(m + n) is always divisible by 16 for both n = 4 and n = 5.

But if m = 1, then,
For n = 4 --> m(m +1)(m + 2)...(m + n) = 1*2*3*4*5 = 120 --> NOT divisible by 16
For n = 5 --> m(m +1)(m + 2)...(m + n) = 1*2*3*4*5*6 = 720 --> DIVISIBLE by 16
For n > 5 --> m(m +1)(m + 2)...(m + n) = 1*2*3*4*5*6*n = 720*n --> DIVISIBLE by 16

Hence for every positive integer m, the product m(m +1)(m + 2)...(m + n) is NOT divisible by 16 if n = 4. Hence, 4 cannot be a value of n.

But for every positive integer m, the product m(m +1)(m + 2)...(m + n) is DIVISIBLE by 16 if n ≥ 5. Hence, any integer greater than 5 can be a value of n.
Not sufficient

Statement 2: n² - 9n + 20 = 0
=> (n - 4)(n - 5) = 0
Hence, n = 4 or n = 5

Not sufficient

1 & 2 Together: As n ≥ 5, n must be equal to 5.

Sufficient

The correct answer is C.

For the same reason we are not starting with m=16,32 or something else because the universality of solution will not be applicable if we dont start from the least value of positive integer

For the same reason we are not starting with m=16,32 or something else because the universality of solution will not be applicable if we dont start from the least value of positive integer