@P@ is defined as the product of all even integers such r such that 0

Question

@P@ is defined as the product of all even integers such r such that \(0<r\leq{P}\). For example \(@P@= 2 * 4 * 6 * 8 * 10 * 12 * 14\). If @K@ is divisible by \(4^{11}\), what is the smallest possible value for k?

(A) 22
(B) 24
(C) 28
(D) 32
(E) 44

(A) 22
(B) 24
(C) 28
(D) 32
(E) 44

ScottTargetTestPrep · Accepted Answer

Since 4^11 = 2^22, we need to find the smallest value of k such that @k@ has 22 factors of 2.

Let’s now test the answer choices; but we will start with choice C. If it has exactly 22 factors of 2, we have found our answer. If it has fewer than 22 factors of 2, we can proceed to choices D and E. If it has more than 22 factors of 2, we can backtrack to choices B and A.

If k = 28, we see that each of the factors 2, 6, 10, 14, 18, 22, and 26 have 1 factor of 2; each of the factors 4, 12, 20, and 28 have 2 factors of 2; each of the factors 8 and 24 have 3 factors of 2; finally, 16 has 4 factors of 2. So the total number of factors @28@ has is 7 x 1 + 4 x 2 + 2 x 3 + 1 x 4 = 7 + 8 + 6 + 4 = 25 factors of 2. This is more than 22 factors of 2. So let’s backtrack to 24. If k = 24, we will remove the factors 26 and 28, which means we are removing 3 factors of 2 since 26 and 28 have 1 and 2 factors of 2, respectively. Therefore, we see that @24@ will have exactly 22 factors of 2.

Answer: B

Vidhi96 · Answer

Here's the trick, the number is not factorial thus we cannot use 44!/2 = 22 2's

here it will go by quick cal
2 * 1 = 2.... till.........12*2 = 24

then the sum of 2s will be 24

Shubhambhangre · Answer

@k@ is divisible by 4^11

4^11 = 2^22

minimum valve of k can be the number which has the minimum 2^22 factor.

Now by answer analysis.

maximum power of 2 which divide the number is:

24!/2=12/2=6/2=3/2=1/2

maximum power of 2 which divide the 24! is 12+6+3+1=22
which satsifies the condition 
hence B is correct

Posted from my mobile device

WheatyPie · Answer

44! contains many more than 22 twos... in fact it contains:

22+11+5+2+1=41 twos.

WheatyPie · Answer

@K@ will contain the same number of twos as K! because the odd numbers in the factorial don't affect the number of twos.

Thus we can use the usual technique where we divide repeatedly by 2.

If we start with the middle number, 28: @28@ contains 14+7+3+1=25 twos. We only need 22 twos, by removing the 28 and the 26 we remove 3 twos, so 24 is the correct answer.

Adarsh_24 · Answer

I think there is some typos in the question. But, I believe one way to get the solution is that

the function will give,
2^(k/2) * (k/2)! will be divisible by 4^11 or 2^22.
We can take options and try out.

1. 22
2^11* 11!

Checking how many 2s in 11!
11/2=5
11/4=2
11/8=1

So 2^11 * 2^8 => not 22 twos.

2. 24

2^12* 12!

No of 2s in 12!
12/2=6
12/4=3
12/8=1

So 2^12 * 2^10. So 22 twos.

ADisHere · Answer

Hi, the question need an edit, it should be @14@= 2 * 4 * 6 * 8 * 10 * 12 * 14

Solution
In such type of questions, the key should be recognising the pattern

the  SIGN  is much like a factorial, but excluding the odds
(anyways we are not concerned with the odds ones as we want what is the smallest possible value for 4^{11} = 2^{22}

Testing option(factorial divisor concept)
A) 22/2 = 11
22/4=5
22/8=2 
22/16=1
So we get 19, incorrect

B) 24/2=12
24/4=6
24/8=3
24/16=1
We get 22, CORRECT!!

KDM91 · Answer

The answers are not clear. Can anyone explain in detail?

KDM91 · Answer

@bunuel....can u pls explain this?

Krunaal · Answer

\frac{@K@}{4^{11}}  = integer

Above can be written as,  \frac{@K@}{2^{22}}  = integer

Now, @K@ is a product of all even integers until K,

So, if K is 3 then @K@ = 2; If K is 7 then @K@ = 6*4*2

We need to find for what minimum value of K, will  \frac{@K@}{2^{22}}  be an integer, in other words @K@ should contain minimum 22 2's

Now when we look at answer choices,

A. If K is 22, then @K@ = 22*20*18*16*14*12*10*8*6*4*2 =  2^{19} * (something) .  This option has 19 2's, it won't be divisible by  2^{22} , hence incorrect.

B. If K is 24, then @K@ = 24*22*20*18*16*14*12*10*8*6*4*2 =  2^{22} * (something) .   This option has 22 2's, it will be divisible by  2^{22} , hence correct.

Let me know in case you did not follow any specific step.

bumpbot · Answer

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