Latest TCS Aptitude Question SOLUTION: Find the number of perfect squares in the given series 2013, 2020, 2027,................, 2300 (Hint 44^2=1936)

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Find the number of perfect squares in the given series 2013, 2020, 2027,................, 2300
(Hint 44^2=1936)

Asked In TCS Ishita Chakraborty (8 years ago)

Unsolved Read Solution (13)

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mohit  (8 years ago)

only 1 perfect square

Is this solution Helpfull? Yes (9) | No (2)  

ashima  (8 years ago)

there are 3
2025=45*45
2116=46*46
2209=47*47

Is this solution Helpfull? Yes (6) | No (21)  

KOKILA  (8 years ago)

ans is 1, because the series has difference 7..so 47^2 only satisfy these series.

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

poonam  (8 years ago)

Only one perfect square and which is 47×47

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

VINAYAK RAJARAM SHINDE  (6 years ago)

The given series is an AP with common difference of 7. So the terms in the above series are in the form of 2013 + 7k. We have to find the perfect squares in this format in the given series.
Given that 44^2 = 1936.
Shortcut: To find the next perfect square, add 45th odd number to 44^2.
So 45^2 = 1936 + (2 x 45 -1) = 2025
46^2 = 2025 + (2 x 46 - 1) = 2116
47^2 = 2116 + (2 x 47 - 1) = 2209
Now subtract 2013 from the above numbers and divide by 7. Only 2209 is in the format of 2013 + 7k. One number satisfies.

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

jannath mehanas  (8 years ago)

ya,i got it.it ll be on 17th march on tcs office Chennai.

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

varsha jain  (8 years ago)

guys anyone got their schedule for 2nd march 2016 off campus written exam...?

Is this solution Helpfull? Yes | No (8)  

ankita patil  (8 years ago)

2293
Bacause each square no. Is added by 7

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garima agrawal  (7 years ago)

2
45*45=2025
46*46=2116

Is this solution Helpfull? Yes | No (1)  

SrinivasGupta  (6 years ago)

ans is 3

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Tejaswini  (6 years ago)

Can anyone explain wat to prepare for coding round

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Shiv R Yadav  (5 years ago)

only 1 i.e. 47^2 = 2207

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Yogaprabha  (3 years ago)

Answer: a
Explanation:
The given series is an AP with common difference of 7. So the terms in the above series are in the form of 2013 + 7k. We have to find the perfect squares in this format in the given series.
Given that 44^2 = 1936.
Shortcut: To find the next perfect square, add 45th odd number to 44^2.
So 45^2 = 1936 + (2 x 45 -1) = 2025
46^2 = 2025 + (2 x 46 - 1) = 2116
47^2 = 2116 + (2 x 47 - 1) = 2209
Now subtract 2013 from the above numbers and divide by 7. Only 2209 is in the format of 2013 + 7k. One number satisfies.

Is this solution Helpfull? Yes | No   

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