One way is: (4 + 4 + 6 + 9)^(100/50) = 23^2 = 529
81 and 9 if you add them. 4 and 25 if you multiply them.
1, 4, 9, 16, 25, 36, 42, 64, 81, 100
100 ÷ 4 = 25 → 24 numbers between 0 and 100 exclusive are divisible by 4 100 ÷ 6 = 16 2/3 → 16 numbers between 0 and 100 are divisible by 6 lcm(4, 6) = 12 → 100 ÷ 12 = 8 1/3 → 8 numbers between 0 and 100 are divisible by 4 and 6.
106------------------------------------------------------------------------------------------------From Rafaelrz.You can make, 4P3 = 4!/(4-3)! = 24, different 3 digit numbers using 1, 2, 3 and 4.
With 4 different numbers, 24.
As a product of its prime factors: 2*2*5*5 = 100
As a product of its prime factors: 2*2*5*5 = 100
One way is: (4 + 4 + 6 + 9)^(100/50) = 23^2 = 529
As a product of its prime factors: 2 * 2 * 5 * 5 = 100
81 and 9 if you add them. 4 and 25 if you multiply them.
Since 100/4 = 25, there are 25 numbers between 0 and 100 divisible by 4.
1+23-4+56+7+8+9+100
1, 4, 9, 16, 25, 36, 42, 64, 81, 100
100 ÷ 4 = 25 → 24 numbers between 0 and 100 exclusive are divisible by 4 100 ÷ 6 = 16 2/3 → 16 numbers between 0 and 100 are divisible by 6 lcm(4, 6) = 12 → 100 ÷ 12 = 8 1/3 → 8 numbers between 0 and 100 are divisible by 4 and 6.
48
106------------------------------------------------------------------------------------------------From Rafaelrz.You can make, 4P3 = 4!/(4-3)! = 24, different 3 digit numbers using 1, 2, 3 and 4.