The number of integers between 0 and 2002 that are relatively prime to 2002 are m , then (m/10) is equal to . Please explain
Answers
Answer:
m/10 = 72
Step-by-step explanation:
2002 = 2 * 7 * 11 * 13
All the number multiples of 2 * 7 * 11 * 13 will not be co-prime to 1002
Lets find which are not co prime
Multiple of 2 = All even numbers
= 2 , 4 , 6 ,...................................................2000
= 1000 Numbers
Multiple of 7 with odd numbers as Even numbers already considered
7 , 21 , 35 ........................................................... 7 * 285
= 143 numbers
Multiple of 11 with odd numbers as Even numbers already considered
11 , 33 , 55 ........................................................... 11 * 181
= 91 numbers
but Multiples of 77 are repeated 77 , 77 * 3 ..................77 * 25
= 13
91 - 13 = 78 Numbers
Multiple of 13 with odd numbers as Even numbers already considered
13 , 39 , .................. 13 * 153
Number = 77
multiples of 91 & Multiples of 143 to be reduced but 1001 would be common
91 * 1.................91*21 = 11 Numbers
143* 1 ...............143 * 13 = 7 Numbers
11 + 7 - 1 = 17 numbers to be reduce
77 - 17 = 60 Numbers
1000 + 143 + 78 + 60 = 1281
Total Numbers = 2001
m = 2001 - 1281
m = 720
m/10 = 72
Answer:
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