Latest Elitmus Aptitude Question SOLUTION: Q1. Find the Highest power of 2 in 1! + 2! + 3! + .......+ 600! Q2. what max. value of n in 160^n divides 100!. A). 19 B). 24 C).26 D). 28 Q3. What is the number of consecutive z

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(#M40039392) ELITMUS Numerical Ability question Number System Keep an EYE

Q1. Find the Highest power of 2 in 1! + 2! + 3! + .......+ 600!

Q2. what max. value of n in 160^n divides 100!.
A). 19 B). 24 C).26 D). 28

Q3. What is the number of consecutive zeros in the end of 1000! ?
A). 248 B). 249 C). 250 D). 251

Asked In Elitmus Ravi Ranjan (9 years ago)

Unsolved Read Solution (13)

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Read Solution (13) :

Paramita Saha  (9 years ago)

Q3. ans> no of consecutive 0 is 249
1000/5=200
200/5=40
40/5=8
8/5=1
so, total no of consecutive 0 is (200+40+8+1)=249

Is this solution Helpfull? Yes (40) | No   

avinash jha  (9 years ago)

ans for qestion 2:
160= 2^5*5
100/2=50/2=25/2=12/2=6/2=3/2=1.....so 50+25+12+6+3+1=97/5=19
whereas 100/5=20/5=4.......so 20+4=24 which is more than 19. so correct ans is 19.

Is this solution Helpfull? Yes (23) | No (3)  

DHEERAJ ASHISH  (9 years ago)

(1)
highest power of 2 is 0..because everytime we take the sum it would be an odd number.
for eg
1!+2!+...+5!=155
1!+2!+......+6!=875

Is this solution Helpfull? Yes (14) | No (1)  

Shirsendu Bikash Das  (9 years ago)

1.We now, odd+even = odd. Since 1! is odd so the highest power of 2 is zero

2 & 3. Highest power of n in N! is [N/n]+[N/n^2]+.......

Is this solution Helpfull? Yes (6) | No (1)  

Deepak Raut  (9 years ago)

(2) Ans:- 160 prime factor is 2 and 5,but for 160 we need five 2's and one 5, check for no of 5 in 100!, using Highest power of n in N! is [N/n]+[N/n^2]+.... So we get 24 as answer. Now checking for 2 we get 97 using same formula. But we need pair of five 2's. So 97/5 gives 19 pair of five 2's which is less than 24 no. of 5. So 19 answer.

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Anjana Kumari  (9 years ago)

ANS 1)596
2)24
3)249

Is this solution Helpfull? Yes (5) | No (3)  

BRAJESH YADAV  (9 years ago)

Q.3. number of zeros in 1000!=[1000/5]+[1000/25]+[1000/125]+[1000/625]=249
Where [x] is x is nearest integer.

Is this solution Helpfull? Yes (4) | No   

puja kumari  (9 years ago)

1000/5=200
200/5=40
40/5=8
8/5=1
200+40+8+1=249

Is this solution Helpfull? Yes (3) | No (1)  

ASHISH KISHORE  (9 years ago)

1)1! will add 1 to sum nd thus altimately sum is odd n so odd no is not divisible by 2 so ans is 0

2) 160= 2^5*5
100/2+100/4+100/8+100/16+ 100/32+100/64=97
that is it contain 2^97 i.e (2^5)^19
100/5+100/25 =24
total 19 2^5*5 combination
so ans is 19

3). 1000!/5 + 1000!/25 + 1000!/125 + 1000!/625= 249

Is this solution Helpfull? Yes (2) | No   

bablu kumar  (9 years ago)

Q3
1000/5=200
1000/5*5=40
1000/5*5*5=8
1000/625=1
bcz 1000

Is this solution Helpfull? Yes (1) | No (1)  

Natasha Arora  (9 years ago)

q3. 1000!/5 + 1000!/25 + 1000!/125 + 1000!/625= 249

Is this solution Helpfull? Yes (1) | No   

Raashid khan  (9 years ago)

Q1)answer is 9..as 512! is the max. no

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Anjana Kumari  (9 years ago)

1) highest power iz 9
2) 24
3) 248

Is this solution Helpfull? Yes | No (1)  

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