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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