How many integers m are there such that 0 < m < 100 and gcd(m, 1001) is a single-digit number?
+2407 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. :))
=^._.^=
