No; it is divisible only by its factors (1, 2, 4, 5, 8, 10, 16, 20, 40, and 80).
The first number is 80
160 is divisible by all three numbers, 10, 2, and 5. 160 / 10 = 16 160 / 2 = 80 160 / 5 = 32
There are 80 such numbers.
A number is a multiple of 80 if the last 4 digits are a multiple of 80 The 1,000s digit is even and the last 3 digits are a multiple of 80 The 1,000s digit is odd and the last 3 digits are 40 times an odd number It also works if the number is a multiple of 5 and 16 at the same time because 80 = 2^4 x 5 = 16 x 5
80 is divisible by 2,4,5,and 6
80 is not divisible by 234569 but is divisible by 10.
No; it is divisible only by its factors (1, 2, 4, 5, 8, 10, 16, 20, 40, and 80).
Factors of 80 = 1, 2, 4, 5, 8, 10, 16, 20, 40, 80.
No, 80 is composite. it is divisible by 1, 2, 4, 5, 8, 10, 20, 40, and 80
101
84
80 is divisible by 5 and 4 and is greater than 75
1, 2, 4, 5, 8, 10, 16, 20, 40, 80
1, 2, 4, 5, 8, 10, 16, 20, 40, 80
80 is an even number, thus it is divisible by 2; 80/2 = 40. 40 is also even, so it too is divisible by 2; 40/2 = 20. The same applies to 20; 20/2 = 10, and to 10; 10/2 = 5. Since both 2 and 5 are prime numbers, the prime factorization of 80 is 24 X 5.
It could be 76 - not divisible by 5. It could be 75 - not divisible by 2.