Solve 23^5+23^4+23^3+23^2+2*3+2

$$2 { 3 }^{ 5 } +2 { 3 }^{ 4 } +2 { 3 }^{ 3 } +2 { 3 }^{ 2 } +2 \cdot 3+2$$

Evaluate

$$728$$

Solution Steps

Calculate 3 to the power of 5 and get 243.

$$2\times 243+2\times 3^{4}+2\times 3^{3}+2\times 3^{2}+2\times 3+2$$

Multiply 2 and 243 to get 486.

$$486+2\times 3^{4}+2\times 3^{3}+2\times 3^{2}+2\times 3+2$$

Calculate 3 to the power of 4 and get 81.

$$486+2\times 81+2\times 3^{3}+2\times 3^{2}+2\times 3+2$$

Multiply 2 and 81 to get 162.

$$486+162+2\times 3^{3}+2\times 3^{2}+2\times 3+2$$

Add 486 and 162 to get 648.

$$648+2\times 3^{3}+2\times 3^{2}+2\times 3+2$$

Calculate 3 to the power of 3 and get 27.

$$648+2\times 27+2\times 3^{2}+2\times 3+2$$

Multiply 2 and 27 to get 54.

$$648+54+2\times 3^{2}+2\times 3+2$$

Add 648 and 54 to get 702.

$$702+2\times 3^{2}+2\times 3+2$$

Calculate 3 to the power of 2 and get 9.

$$702+2\times 9+2\times 3+2$$

Multiply 2 and 9 to get 18.

$$702+18+2\times 3+2$$

Add 702 and 18 to get 720.

$$720+2\times 3+2$$

Multiply 2 and 3 to get 6.

$$720+6+2$$

Add 720 and 6 to get 726.

$$726+2$$

Add 726 and 2 to get 728.

$$728$$

Factor

$$2^{3}\times 7\times 13$$