4 and 432 2 and 864 8 and 216
2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 = 1728
1958
To determine if 1728 is divisible by 6, we need to check if the sum of its digits is divisible by 3 and if it is an even number. The sum of the digits of 1728 is 1+7+2+8=18, which is divisible by 3. Additionally, the last digit of 1728 is 8, which is an even number. Therefore, 1728 is divisible by 6.
576
The prime factors of 1728 are 2 and 3
4 and 432 2 and 864 8 and 216
1728 = 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3 = 2^6 * 3^3 They are in exponents: 2^6 and 3^3
We know that the cube of 10 is 10^(3) = 1000 So try 11^(3) = 1331 (Too low) So try 12^(3) = 1728 THE ANSWER!!!! Go No futher. So the cube root (1728) = 12
As a product of its prime factors in exponents: 2^6 times 3^3 = 1728
2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 = 1728
1958
120% of 1728 = 1.2 x 1728 = 2073.6
12 x 144 = 1728...........144..........x 12========...........288.......+ 144========..........1728
To determine if 1728 is divisible by 6, we need to check if the sum of its digits is divisible by 3 and if it is an even number. The sum of the digits of 1728 is 1+7+2+8=18, which is divisible by 3. Additionally, the last digit of 1728 is 8, which is an even number. Therefore, 1728 is divisible by 6.
1728 - 1779 = -51
1728