If A = 10! + 12! + 14! + 16! + ... + 100!, then the highest power of 2

Question

If A = 10! + 12! + 14! + 16! + ... + 100!, then the highest power of 2 in A is

A. 7

B. 8

C. 9

D. 10

E. 11

Jansaida · Answer

Since we need to find the highest power of 2 and A is the summation of those factorials.

We will look at 10!.

(10)/2 --------> quotient=5
5/2-------------> quotient= 2
2/2--------------> quotient=1

So in total there are 8 2's in 10!. Since A is the summation of all the given factorials, we need to take power of 2 as common from every factorial.

Every factorial is bound to have 2^8 in their factorial. If we go beyond 8 then 10! will not contain extra 2's.

IMO B is the answer.

Lampard42 · Answer

When 10! in A is taken common then (1+12*11+14*13*12*11+.....) then number inside the bracket will be odd (Odd+99*even) 
No. of 2 in 10! is 10/2+10/4+10/8=5+2+1=8. So A will have only 8 multiples. IMO B

CareerGeek · Answer

A = 10! + 12! + 14! + 16! + ... + 100!
--> A = 10!(1 + 12*11 + 14*13*12*11 + . . . . . . . . . )
--> A = 10!(1 + Even)
--> A = 10!(Odd)

10! = 2*4*6*8*10*k , where k is any odd integer
--> 10! = 2*2^2*2*2^3*2*m, where m is any odd integer
--> 10! = 2^8*m

--> A = 2^8*m*odd

So, highest power of A possible = 8

IMO Option B

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LeoN88 · Answer

Extrapolation, I solved for 10! + 12!
Realized 2^8  (1 + 2^2) , Now there will always be +1 inside the bracket to turn the RED term odd. So max power of 2 is always 8.

Shorub · Answer

10! * (1+11*12+11*12*13+....)

10! * (1+even+even+....even)

10! * (odd)

By factorising 10! and finding the powers of 2, we observe the answer is 8.

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EgmatQuantExpert · Answer

Solution 
  Given:  
 • A = 10! + 12! + 14! + 16! + ... + 100!

To find:  
 • The highest power of 2 in A

Approach and Working Out:   
 • A = 10! + 12! + 14! + 16! + ... + 100! = 10! * (1 + 11*12 + 11*12*13*14 + ……… + 11*12*13*….*100) = 10! * (odd + even + even + even + … + even) = 10! * odd number
• Thus, the highest power of 2 in A = the power of 2 in 10! =    +   +   = 5 + 2 + 1 = 8

Hence, the correct answer is Option B.

Answer: B

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-NP-e1533292144892.png

Archit3110 · Answer

A = 10!*(1+11*12+11*12*13+ ....) we get 10! * (odd+even ) ' 10! * odd integer
so 10! /2 +10!/4+10!/8 = 5+2+1 ; 8
IMO B

ScottTargetTestPrep · Answer

We can factor out 10! from 10! + 12! + 14! + … + 100!, and we have:

10!(1 + 12 x 11 + 14 x 13 x 12 x 11 + … + 100 x 99 x … x 11)

We see that all the terms in the parentheses are even except the first term, which is odd. Thus, the sum of all the terms in the parentheses is odd.

So, we need to determine only the number of 2’s in 10!.

10 = 2^1 x 5

8 = 2^3

6 = 2^1 x 3

4 = 2^2

2 = 2^1

Thus, there are 8 two’s in 10!.

Answer: B

bumpbot · Answer

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If A = 10! + 12! + 14! + 16! + ... + 100!, then the highest power of 2

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If A = 10! + 12! + 14! + 16! + ... + 100!, then the highest power of 2

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