Number System
Aptitude

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Q.

A set of S consists of,

i). All odd numbers from 1 to 55.
ii). All even numbers from 56 to 150.

What is the index of the highest power of 3 in the product of all the elements of the set $S$?

 A.

35

 B.

48

 C.

6

 D.

36

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Solution:
Option(A) is correct

The product of all element of $S$ is:
$\{3, 5, 7, 56, 58, 60$
$9, 11, 13, 62, 64, 66$
………………………
………………………
$51, 53, 55, 146, 148, 150 \}$

⇒ $\{ 3, 3×3, 3×5, .................3×17$
$3×20, 3×22, 3×24, ……………..(3×50)M\}$

⇒ $3^9×3×9×3×3^{16}×24×30×36×42×148N$

⇒ $3^{35}S$


(2) Comment(s)


Aalok
 ()

i think 37 would be answer . because b/w 1 to 55 there are 13 odd number which are multiple of 3.

$3,9,15,21,27,33,39,45,51$. where as $9 = 3^2$, $27= 3^3.$

And there are 24 number of 3 exist b/w 56 to 150

$60, 66,72,78,84,90,96,102,108,114,120,126,132,138,144,150$. where as $72 = 3^2*8$, $90=3^3*10$, $108 = 3^3*4$, $126 = 3^2*14$, $144 = 3^2 * 16$.

so $13+24 = 37$.

please reply if i am wrong.


Manasa
 ()

3,9,15,21,27,33,39,45,51 contains 9 3's and $9=3^2$, $27=3^3$ contains extra $1+2=3$ 3's

$9+3=12$ odd number multiples of 3 are there not 13

similarly 23 not 24

so ans $=12+23=35$