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# If the remainder is 5 when x is divided by 101,

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Intern
Joined: 10 Feb 2017
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Location: Viet Nam
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If the remainder is 5 when x is divided by 101,  [#permalink]

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15 Aug 2018, 02:14
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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
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Status: GMATINSIGHT Tutor
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Posts: 2959
Location: India
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If the remainder is 5 when x is divided by 101,  [#permalink]

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Updated on: 15 Aug 2018, 02:40
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1
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 x is divided by 101 then remainder is 5
i.e. When x^2 is divided by 101 then remainder is 5^2 = 25
i.e. When x^3 is divided by 101 then remainder is 5^3 = 125 but this is further divisible by 101 hence remainder = R(125/101) = 24

Remainder (x^3+x^2+x)/101 =5+25+24 = 54

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Originally posted by GMATinsight on 15 Aug 2018, 02:27.
Last edited by GMATinsight on 15 Aug 2018, 02:40, edited 1 time in total.
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If the remainder is 5 when x is divided by 101,  [#permalink]

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15 Aug 2018, 02:33
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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Re: If the remainder is 5 when x is divided by 101,  [#permalink]

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16 Aug 2018, 07:47
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

Remainder when $$x$$ is divided by 101 is 5

Given the expression $$x^3+x^2+x$$, we need remainder when this is divided by 101

Using remainder theorem, we get remainder as 5*5*5 + 5*5 + 5 = 125 + 25 + 5

Further division by 101 we get remainder as 24 + 25 + 5 = 54

Thanks,
GyM
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Posts: 7764
Re: If the remainder is 5 when x is divided by 101,  [#permalink]

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16 Aug 2018, 07:58
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

you can always add or multiply the remainders

$$x^3+x^2+x=5^3+5^2+5=125+25+5=155$$
when 155 is divided by 101, remainder will be 155-101=54

D
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Re: If the remainder is 5 when x is divided by 101,   [#permalink] 16 Aug 2018, 07:58
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# If the remainder is 5 when x is divided by 101,

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