Japanese University Entrance Examination

James Wong

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Mar 19, 2011, 8:17:12 PM3/19/11

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Hello Mathematicians,

I was looking at a sample question from the aforementioned test, and found
this question.

Suppose the polynomial P(x) with integer coefficients satisfies the
following conditions.
1. If P(x) is divided by x^2 -4x + 3 the remainder is 65x-68
2. If P(x) is divided by x^2 + 6x -7, the remainder is -5x + a.

What is a?

I tried letting
P(x)/divisor1 = SomeQuotient1 + remainder1/P(x)
and
P(x)/divisor2 = SomeQuotient2 + remainder2/P(x)

but it doesn't look like this is the right track. Any hints?

Thanks,
James

A N Niel

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Mar 19, 2011, 8:29:29 PM3/19/11

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In article <mdCdnSRLyeYU2hjQ...@wavecable.com>, James Wong
<jw...@ocf.berkeley.edu> wrote:

Yours is, indeed, wrong. Should be:

P(x)/divisor1 = SomeQuotient1 + remainder1/divisor1
and
P(x)/divisor2 = SomeQuotient2 + remainder2/divisor2

Peter

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Mar 19, 2011, 8:34:35 PM3/19/11

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Actually your approach made sense.

From (1) we know that P(x) = (x^2 - 4x + 3)*Q(x) + 65x - 68 for some
polynomial Q and
from (2) we know that P(x) = (x^2 + 6x - 7)*R(x) - 5x + a for some
polynomial R

Because x^2 - 4x + 3 = (x - 1)*(x - 3) and x^2 + 6x - 7 = (x + 7)*(x -
1),
from (1) we get P(1) = -3 and from (2) we get P(1) = -5 + a

Combining the above two equalities, you find the value of a.

William Elliot

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Mar 19, 2011, 10:26:58 PM3/19/11

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On Sat, 19 Mar 2011, Peter wrote:

Would you explain why you introduced a
and removed the problem statement?

The Qurqirish Dragon

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Mar 20, 2011, 9:57:46 AM3/20/11

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"a" was in the OP, he didn't introduce it.

The Qurqirish Dragon

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Mar 20, 2011, 10:24:02 AM3/20/11

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Assume P(x) = b x^2 + cx + d, for unknown constants b, c, and d.
if you decrease b by 1, then the linear coefficient of the remainder
in statement (1) reduces by 4, and that in statement (2) increases by
6. By proper choice of b, you can get the difference in linear terms
to be anything.

Similarly, the constant term in (1) increases by 3, and that in (2)
decreases by 7. In particular, the change in the linear terms'
coefficients is the additive inverse of that in the constant terms.

Since the difference in the linear coefficients in the remainders is
-70, the difference in the constant terms must be +70, so a=2 is the
solution.

Check: assume b=0. then P(x) = cx + d. To get 65x - 68 as the
remainder in (1), we find c=65, d=-68. The difference in the linear
coefficients is 0. Since we need the difference to be -70, we need to
increase b by 7.
Thus, P(x) = 7x^2 + cx +d, where c and d have to be redetermined.
To get the proper remainder in (1), we need:
c - (7 * (-4)) = 65, and d - (7*3) = -86, so c=37 and d=-47
7x^2 + 37x - 47, divided by x^2 + 6x -7 is -5x + 2, which agrees with
the answer I derived above.

William Elliot

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Mar 20, 2011, 10:39:41 PM3/20/11

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On Sat, 19 Mar 2011, James Wong wrote:

> Hello Mathematicians,
>
> I was looking at a sample question from the aforementioned test, and
> found this question.
>
> Suppose the polynomial P(x) with integer coefficients satisfies the
> following conditions.
> 1. If P(x) is divided by x^2 -4x + 3 the remainder is 65x-68
> 2. If P(x) is divided by x^2 + 6x -7, the remainder is -5x + a.
>
> What is a?

p(x) = q(x)(x^2 - 4x + 3) + 65x - 68, for some polynomial q.
p(x) = r(x)(x^2 + 6x - 7) - 5x + a, for some polynomial r.

x^2 - 4x + 3 = (x - 1)(x - 3)
x^2 + 6x - 7 = (x - 1)(x + 7)

p(1) = -3
p(1) = -5 + a