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
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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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