Latest TCS Aptitude Question SOLUTION: find unit digit of 7^(1!+2!+.......+200!)

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find unit digit of 7^(1!+2!+.......+200!)

Asked In TCS abhishek mishra (10 years ago)

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Harsha Lanka  (10 years ago)

Sum of the factorials=------------13(last two digits will defnitely be 13)
so
7^----------13 is the question
according to cyclicity(7^1=7, power2=9(lastdigit),power 3=3,power 4=1 this will be repeated to 5,6,7 powers and so on)
when we divide the power by 4 we get remainder as 1(divisiblity rule of 4= last 2 digits must divisible by 4)
but here last two digits 13%4 will give 1 as remainder
so it will be converted into
7^1
7 must be last digit...

if it is wrong..my work will be waste

Is this solution Helpfull? Yes (29) | No (7)  

Harsha Lanka  (10 years ago)

1!+2!+3!+4!+5!+.....+200!
calculate manually
(consider each and every factorials last two digit)
1
2
6
24
23
20
40
20
80
00(from 10! on wards we get last two digits as 00..think logically)
by adding these we get 13 as last two digits
Sum of the factorials=------------13(last two digits will defnitely be 13)
so
7^----------13 is the question
according to cyclicity(7^1=7, power2=9(lastdigit),power 3=3,power 4=1 this will be repeated to 5,6,7 powers and so on)
when we divide the power by 4 we get remainder as 1(divisiblity rule of 4= last 2 digits must divisible by 4)
but here last two digits 13%4 will give 1 as remainder
so it will be converted into
7^1
7 must be last digit...

if it is wrong..my work will be waste

Is this solution Helpfull? Yes (24) | No (3)  

Palash Bansal  (10 years ago)

1!+2!+3!+...+n!=(n+1)!/n
1!+2!+3!+...+n!=(201)!/200
201! has so many 0's in last so 201!/200 will also have..
and if last 2 digits are 00 then it is surely divisible by 4 (divisibilty rule)
so 7^4n is the form of ques
and unit digit of 7^4n is 1
so 1 is the answer

Is this solution Helpfull? Yes (11) | No (11)  

mithun Abhimanyu  (10 years ago)

1!+2!+3!=9/4=rem 1 other term will be zero..
7^1=7
finally ans is 7

Is this solution Helpfull? Yes (5) | No (2)  

susmita dutta chowdhury  (9 years ago)

1!+2!+......+200!=[200(200+1)]/2=20100
so,7^20100
nw 20100 is divisable by 4.
so odd^4k unit digit must be 1.so naswer will be 1.

Is this solution Helpfull? Yes (4) | No (3)  

Harsha Lanka  (10 years ago)

7^------13 will be the power
acc to cyclicity:
13 %4 will give 1 as remainder
7^1=7

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

Vibhu Gaur  (10 years ago)

I am not sure about this approach so plz correct me if its wrong

1!+2!+3!.....200! can be written in reverse as 200! + 199!+198!...1!. Now taking 200! common out of this equation
200!(1+1/200+1/199*200...)
now this is equal to 200!*1.005 =200!(approx)

200! is completely divisible by so it takes d form 4*n
7^4*n=1 at unit place so ans is 1.

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

Abhijit  (10 years ago)

@HARSHA LANKA.....Could you plz specify how did you find the last two digits of the factorial series.....

Is this solution Helpfull? Yes (2) | No   

Guru Pranav  (10 years ago)

simply the sum of power will be 7^(....0033)
so, according do cyclicity the power repeats after every 4.
here dividing power by 4 we get remainder by 1.
So, 7^1 =7 Ans.

Is this solution Helpfull? Yes (2) | No   

Ashok kumar Gajula  (10 years ago)

please explain clearly harsha lanka

Is this solution Helpfull? Yes | No (1)  

mavuri tejaswi  (10 years ago)

please explain it clearly... i didn't get you

Is this solution Helpfull? Yes | No (1)  

jiten choudhary  (9 years ago)

We know reparation of last digit in 7 is 7,9,3,1 if there is a power of multiple of 4 then last digit is 1 . All the factorial greater then 4! Are divisible of 4 so 4!+5!+6!........200! Sum of these will be multiple of 4 . Now 1!+2!+3! =9 Will decide the last digit of number 9%4=1 so remainder is Ist in repeated power of 7 that is 7 so..

Ans. is 7

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