How many consecutive terms (minimum no. of terms without leaving any of the term starting from fir

M4maths Previous Puzzles

15 Dec 2010 perfect square

How many consecutive terms (minimum no. of terms without leaving any of the term starting from first term)
of the sequence given below are required to make the product, a perfect square?
1, 3, 5, 6, 8, 9, 10, 12, 15, 20, 24, 48, 72, 100, 124, 150, 200, 240, 300, 500,

OPtion

1) 12
2) 10
3) 11
4) 9
5) 13
6) 14
7) 15
8) 16
9) 17
10) 18

Solution

The product of first 12 terms of the sequence is a perfect square.

Correct Option: 1 Best Solution (4)

Dr.Dipin Singh   14 years AGO

product of 1, 3, 5, 6, 8, 9, 10, 12, 15, 20, 24, 48, is equal to 518400^2.

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Dimple   14 years AGO

1* 3* 5* 6* 8* 9* 10* 12* 15* 20* 24* 48= 3*5*6*4*2*3*3*5*2*6*2*3*5*4*5*24*24*2 .

Making pairs of 3*5*6*4*2*3*3*5*2*6*2*3*5*4*5*24*24*2 we get
(3*5*6*4*2*3*5*24)^2

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mohd anas   14 years AGO

8=2*4,10=5*2,12=4*3,15=5*3,20=4*5,24=4*6,48=16*3
now
multiply all these and leave the squares we get
2*5*2*3*5*3*5*6*3=(2*2)*(5*5)*(3*3)*5*6*3
again leave the squares
5*6*3
now taking the terams
1,3,5,6 and multiply in above sequance
we get 1*3*5*6*5*6*3=3*3*5*5*6*6
which a perfect square
hence we need to multiply till 48 and number of terms are=12

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Asheesh Jain   14 years AGO

1*3*5*6*8*9*10*12*15*20*24*48=268738560000=perfect square

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