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What is the remainder when 123^4 - 123^3 - 123^2 - 123 is divided by

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Re: What is the remainder when 123^4 - 123^3 - 123^2 - 123 is divided by [#permalink]
hi,

could you please explain this part: "Net remainder = 122 - 2 = 120"?
I got -2 from the original expression but i'm not sure how you get 120 from that
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What is the remainder when 123^4 - 123^3 - 123^2 - 123 is divided by [#permalink]
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meowmeowmeow22 wrote:
hi,

could you please explain this part: "Net remainder = 122 - 2 = 120"?
I got -2 from the original expression but i'm not sure how you get 120 from that

Hey meowmeowmeow22

Remainders can never be negative. So if you end up getting negative remainder the net (actual) remainder is equal to divisor + (the negative remainder)

Lets take an example to illustrate this

• Divisor = 7
• Dividend = 11

We know when 11 when divided by 7 results in a quotient of 1 and remainder of 4.

$$\frac{11}{7}$$ can also be represented as $$\frac{14 - 3 }{ 7}$$

14/7 leaves a remainder = 0

-3/7 leaves a remainder = -3

As the remainder is negative, the actual remainder is 7 - 3 = 4

Similarly, in the above question we obtained a (negative) remainder of -2. Hence the actual remainder is 122 - 2 = 120

Hope this helps.
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Re: What is the remainder when 123^4 - 123^3 - 123^2 - 123 is divided by [#permalink]
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Re: What is the remainder when 123^4 - 123^3 - 123^2 - 123 is divided by [#permalink]
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