Math, asked by plabivinodkumar, 9 months ago

Find the maximum value of n such that :
| 77 x 42 x 37 x 57 x 30 x 90 x 70 x 2400 x 243x343 is perfectly divisible by 21power n

Answers

Answered by amitnrw
23

Given : 77 x 42 x 37 x 57 x 30 x 90 x 70 x 2400 x 243x343 is perfectly divisible by 21ⁿ

To find : Find the maximum value of n

Solution:

77 x 42 x 37 x 57 x 30 x 90 x 70 x 2400 x 243x343

Number Divisible by 21 means divisible by 3  & 7

So lets see how many common number of factor of 3 & 7 are there

77 = 7 *11

42 = 7 * 3 * 2

37 - no factor of 7 or 3

57 = 3 * 19

30 = 3  * 10

90 = 3 * 3 * 10

70 =  7 * 10

2400 = 3  * 800

243 =  3 * 3 * 3 * 3 * 3

343 = 7 * 7 * 7

Total Factors of 3  = 11

Total Factor of 7 =  6

Hence total Factors of 21  =  6

77 x 42 x 37 x 57 x 30 x 90 x 70 x 2400 x 243x343

is perfectly divisible by 21⁶

n = 6  

maximum value of n = 6

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Answered by winterrnightz
0

Answer:

Step-by-step explanation:

77 x 42 x 37 x 57 x 30 x 90 x 70 x 2400 x 243x343

Number Divisible by 21 means divisible by 3  & 7

So lets see how many common number of factor of 3 & 7 are there

77 = 7 *11

42 = 7 * 3 * 2

37 - no factor of 7 or 3

57 = 3 * 19

30 = 3  * 10

90 = 3 * 3 * 10

70 =  7 * 10

2400 = 3  * 800

243 =  3 * 3 * 3 * 3 * 3

343 = 7 * 7 * 7

Total Factors of 3  = 11

Total Factor of 7 =  6

Hence total Factors of 21  =  6

77 x 42 x 37 x 57 x 30 x 90 x 70 x 2400 x 243x343

is perfectly divisible by 21⁶

n = 6  

maximum value of n = 6

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