What is the remainder when Z is divided by 122?

Question

Z = 123^4 - 123^3 + 123^2 - 123

A. 0
B. 1
C. 2
D. 3
E. 4

Elonmusk1857 · Answer

Reminder will be  Z = 123^4 - 123^3 + 123^2 - 123  =  1^4-1^3+1^2-1  =0

Option A

divyadna · Answer

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onlymalapink · Answer

i dont understand the solutions here can anybody help?

Bunuel · Answer

Z = 123^4 - 123^3 + 123^2 - 123

What is the remainder when Z is divided by 122?

A. 0
B. 1
C. 2
D. 3
E. 4

Z = 123^4 - 123^3 + 123^2 - 123=

= 123^3(123 - 1) + 123(123- 1)= 
 
  = 123^3*122 + 123 *122 = 
 
  = 122(123^3+123) 
 
Since the above is a multiple of 122, the remainder will be 0.

Answer: A.

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