Hungluu92vn wrote:

If the remainder is 5 when x is divided by 101, what is the remainder when x^3+x^2+x is divided by 101?

A. 5

B. 6

C. 31

D. 54

E. 55

When positive integer x is divided by positive divisor D, the remainder is R.Given these conditions, the smallest possible case will always be as follows:

x = remainder R.

The remainder is 5 when x is divided by 101.In accordance with the rule above, the smallest possible case is x=5.

What is the remainder when \(x^3+x^2+x\) is divided by 101?Since the smallest possible case is x=5, plug x=5 into \(x^3+x^2+x\) and divide by 101:

\(5^3+5^2+5 = 125 + 25 + 5 = 155\)

Dividing 155 by 101 yields a remainder of 54.

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