Question

The number of zeros at the end of (2^123- 2^122 -2^121)(3^223-3^222-3^221) is

Answer 1

(2¹²³ - 2¹²² - 2¹²¹)( 3²²³ - 3²²² - 3²²¹)

= 2¹²¹(2² - 2¹ - 1)(3²²³ - 3²²² - 3²²¹ )

= 2¹²¹(4 - 2 - 1)(3²²¹)(3² - 3¹ - 1)

= 2¹²¹(4 - 3) 3²²¹ (9 - 3 - 1)

= 2¹²¹ × 3²²¹ × 5

= $2^{120}\times3^{221}$ (2 × 5)

= $2^{120}\times3^{221}$ × 10

hence, it is clear that, there is only zero contains at the end of (2¹²³ - 2¹²² - 2¹²¹)( 3²²³ - 3²²² - 3²²¹)

hence, answer is 1.

Answer 2

Step-by-step explanation:

(2¹²³ - 2¹²² - 2¹²¹)( 3²²³ - 3²²² - 3²²¹)

= 2¹²¹(2² - 2¹ - 1)(3²²³ - 3²²² - 3²²¹ )

= 2¹²¹(4 - 2 - 1)(3²²¹)(3² - 3¹ - 1)

= 2¹²¹(4 - 3) 3²²¹ (9 - 3 - 1)

= 2¹²¹ × 3²²¹ × 5

= 2^{120}\times3^{221}2

120

×3

221

(2 × 5)

= 2^{120}\times3^{221}2

120

×3

221

× 10

hence, it is clear that, there is only zero