Latest TCS Aptitude Question SOLUTION: 7^1+7^2+7^3+.......+7^205. Find out how many numbers present which unit place contain 3?

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7^1+7^2+7^3+.......+7^205. Find out how many numbers present which unit place contain 3?

Asked In TCS sumit paliwal (8 years ago)

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Guduru Sreekar  (8 years ago)

Ans:51
SIMPLE,
7^1=7
7^2=49
7^3=343
7^4=2401
7^5=16807
7^6=117649,
obsereve unit place 7,9,3,1 are repeating so,
total=205%4=51

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Anindita Dutta  (8 years ago)

51.
Follow power cycle method.

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Jit Paul  (8 years ago)

answer: 51.
7^3, 7^7, 7^11... has 3 in its unit place.
thus for every power increment by 4 places, we get 3 in the unit place.
Hence, 205%4=51.

Is this solution Helpfull? Yes (4) | No (2)  

sagar kumar  (8 years ago)

7^1=7(in the base),7^2=9,7^3=3,7^4=1.............and this cycle repeats like this onward 7,9,3,1
(7^1=7 + 7^2=9 + 7^3=3 + 7^4=1)....................+ 7 ^205=(7+9+3+1+7+9+3+1................. and this cycle continue until 204th term)+7^1(205th term)=7+9+3+1=20 base is 0 so
(7+9+3+1)*54 + 7^1= 0*54 + 7 = 7 (ans)
7(1)
9(2)
3(3)
1(4)
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.(series reapts until 54 term=7^204) =20*54(20+20+20+20+20.......54 times will will give unit base as 0).......so 0+7=7(ans)
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Arvind kumar yadav  (8 years ago)

we are just considering the unit place....so
7^1=7
7^2=9
7^3=3
7^4=1
-----------
7^5=7
7^6=9
7^7=3
7^8=1
--------
so we see that in a cycle of 4 there is one 3 on the unit place
so divide 205/4=51 + reminder=1 it doesn't matter so answer will be 51........

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sunil kumar prajapati  (8 years ago)

follow power of 7 cycle is 4.
ans.51

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Dharmendra Awasthi  (8 years ago)

7*9 gives 3 at its unit place
so series will be
7*9,7*19,7*29........,7*199,
no of terms in this series are can be calculated by tn=a+(n-1)d
7*199=7*9+(n-1)*70
=>199=9+(n-1)10
=>n=20 the 20 is the correct answer

Is this solution Helpfull? Yes | No (4)  

prasad gaikwad  (8 years ago)

Please post the question clearly, this ain't a series its a sum...The question isn't clear

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Richa  (8 years ago)

Most simple way is we have units place 7^1=7,7^2=9,7^3=3,7^4=1
Thus cycles goes on repeating 3,7,11,15------ we can see common difference is 4... Thus to find upto how much time 3 is contained as unit dgt. One way is keep costing manually with difference of 4 like 3,7,11,15,19------ or one way is find a factor of 4 as last dgt is 7^205, where only 204 is lst dgt dvsble by 4 nd contains 3 in unit plce.. So ap forms 3,7,11,15,--------204 so n=51

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vikash rawat  (8 years ago)

+ sign should not be there

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UDAY SARKAR  (8 years ago)

Frst power like 7^3 = its units digit 3. SO, similarly we will get 7^3+7^7+7^11+.........+7^204. (beacuse 7 has a 4 cycle).totaly we will get 3 to 204 means 68 numbers.
ANS:68

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Binod Kumar  (8 years ago)

unit place will be 3 only on no whose power will give remainder as 3 when divided by 4.
so a series is formed in such case....
series goes like this................. 3,7,11,15......
last term will be 203 which lie in this series
so,
203= 3 + (n-1)*4
=> n=51.
hope it will help.

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Renusree  (8 years ago)

No option please follow power cycle method:)

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CHHAVI SARDA  (5 years ago)

Unit digit of first 6 terms are 7, 9, 3, 1, 7, 9 hence it forms a pattern. For every 4 terms we get one term with 3 in its unit digit. So there are total of 205/4 = 51 sets and each set contains one term with 3 as it's unit digit.
So the ans is 51.

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