Latest TCS Aptitude Question SOLUTION: 7+77+777+............+7777777 23 times then what is the sum?

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(#M40032441) TCS Logical Reasoning question Keep an EYE

7+77+777+............+7777777 23 times then what is the sum?

Asked In TCS KEERTHIPALLE SUDHEERKUMAR (10 years ago)

Solved mohan lal garg Read Solution (16)

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Read Solution (16) :

Mantripragada Nikhil  (10 years ago)

rewrite above series as 7(1+11+111+1111......1111111)...observe that (1+11=12,1+11+111=123,1+11+111+1111=1234...this way 1+11+111+...1111111 leads to 1234567)...value of series=7(1234567)=8641969

Is this solution Helpfull? Yes (79) | No (27)  

mohan lal garg  (10 years ago)

s=7+77+777+............+7777777 23 times
s=7(1+11+111+.............+11111 23 times
s=7/9(9+99+999+................)
s=7/9(10+100+1000.......-23(1 fromk each term))
s=7/9(10(10^23-1)/(10-1)-23)
s=70/9((10^23-1)/9-23)

Is this solution Helpfull? Yes (37) | No (14)  

AkhilK  (10 years ago)

Why this question is in blood relations category?

Is this solution Helpfull? Yes (18) | No   

susi  (10 years ago)

i am sure this will be the right ans
sn= 7/9[9+99+999+..........23 terms]
=7/9[(10-1) (10^2-1) (10^3-1)...............]
=7/9[(10+10^2+10^3+..)(1+1++.......23 terms)
using gp series
= 7/9[(10(10^23-1))/(10-9))-23]
= 70/9[(10^23-1)-161/9]

Is this solution Helpfull? Yes (17) | No (1)  

SUNNY KUMAR SHAW  (10 years ago)

taking out 7 common we have
7(1+11+111+...............+1111111 upto 23 times)
now dividing it by 9 we have:-
7/9(9+99+999+......+99999 23 times)
which can be written as
7/9((10-1)+(100-1)+(1000-1)...............(10^23-1))
7/9[(10+100+1000+.............10^23)-23]
7/9[(10(10^23-1)/9)-23]

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roopa sree  (10 years ago)

can any1 give crct ans plz

Is this solution Helpfull? Yes (5) | No (1)  

aditi bose  (10 years ago)

7+77+777+....+7777777 n times (n=23)
=7(1+11+.......)
=7/9(9+99+999+......)
=7/9[ (10-1) + (100-1) + (1000-1) +.......]
=7/9[ (10+100+1000+......) - (1+1+1+....)]
=7/9[ (10*(10^n - 1/10-1 ) - n ]
replace n by 23 as given in question
=7/81 [ 10^24 - 10 -207 ]= 8.641975309 * 10^22

Is this solution Helpfull? Yes (5) | No   

PARVY GOVIL  (10 years ago)

this question took me 20 long minutes. i wasn't able to make out a minor error , due to which i kept on trying madly...i succeeded.
a simpler solution is : 7(1+11+111+1111....+111...23 times)
add the bracket .i....e 12345679012345679012343. multiply by 7 to get 86419753086419756086401.
addition of bracket is a v.simple task:-
last column is 23 in sum..secnd last is 22 in sum..third last is 21...fourth last is 20 and so on .
thus satrting frm the extreme right. we get 23. so 2 carry forward and so on.
this is an easier method than working it out with gp .

Is this solution Helpfull? Yes (5) | No (1)  

Swathi  (10 years ago)

take 7 common
7(1+11+111+.......+1111..23times) {add the brackets} [1+11+111+1111=1234]
= 7(12345678901234567890123) {multiply with 7}
=86419752308641975230861

Is this solution Helpfull? Yes (4) | No   

preeti jijnodiya  (9 years ago)

7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7
* 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
= 86419753086419783086401 is the ans. time require 20 sec to solve

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S.PRAVEEN KUMAR  (10 years ago)

7(1+11+111+.....)
a=1, r=11 by gp a((r^n)-1)/(r-1)
7(1*(11^23)-1)/(11-1)
7((11^23)-1)/10

Is this solution Helpfull? Yes (2) | No (3)  

hima  (9 years ago)

(7+77+777+...+7..)23 times
7(1+11+111...+1..)23 times= 7(12345678909876543210321)= 86419752369135802472247

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Harish Guntukula  (10 years ago)

7(1+11+111+.......+11111 23 times)
write like this
1
1 1
1 1 1 like this 23 times
sum last one's which gives 23 and keep 3 there and 2 is added to next position
like wise do summation for 3 digits and multiply that number with 7 u will get 343*7=2401. so ans is 401.

Is this solution Helpfull? Yes | No (11)  

hemant panchal  (10 years ago)

s=7+77+777+............+7777777 23 times
s=7(1+11+111+.............+11111 23 times
s=7/9(9+99+999+................)
s=7/9(10+100+1000.......-23(1 fromk each term))
s=7/9(1111....(23 times)0 -23) //after writting 1 (23 times) put 0 at the last/unit place
s=7/9(1111....(21 times)087) //after subtracting 23 the last 3 digits of the series will always be 087, whether 1 is written 23 times/less/more, except 2 times
s=(7777.....(21 times)609)/9 // on multiplying 7 inside, the last 3 digits will always be 609

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bipin  (10 years ago)

7*23=161,so 16 is the carry and is the last term.
7*22=154,154+16=170,so 17 carry and 0 is the last term and so on

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Srinath  (10 years ago)

7(1+11+11+.....+1111111)
Multiply and divide by 9
7/9(9+99+99+......+9999999)
7/9((10-1)+(100-1)+..........+(10000000-1)) which can be written as
7/9((10+100+.....+1000000)-n)
now the part((10+100+.....+1000000)-n)comes under geometric series,therefore applying sum of series formula of GP ie a(r^n-1)/r-1.(a->first term,r->common ratio)
(7/9(10*(10^23-1))/1(10-1)-23)(ie n->23 since the problem is 23 times)
and finally the above equation is simplified to get the final answer.

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