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How many integers m are there such that 0 < m < 100 and gcd(m, 1001) is a single-digit number?

 Jan 26, 2021
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Hi Guest!

 

1001 prime factorzation is 7*11*13.

GCD that are possible and 2 digit is 11, 13, 77, and 91. 

So 11 has the numbers: 11, 22, 33, 44, 55, 66, 77, 88, and 99: or 9 numbers. 

And 13 has the numbers: 13, 26, 39, 52, 65, 78, and 91: or 7 numbers. 

 

 

Total there are 99 numbers, and 9+7+1+1 have double digit gcds. 

So the answer is 99 - (9+7) = 83. 

 

So the answer is 83. 

 

I hope this helped. :))

 

=^._.^=

 Jan 26, 2021
edited by catmg  Jan 26, 2021

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