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MATHMAA

1) When the tens digit of a three digit number abc is deleted, a two digit number is formed. How many numbers abc are there such that abc=9(ac)+4c ?

Solution:

Given :  abc is a number, where a is in hundreds, b is in tens and c is in units place and a,b, c are digits such that $$0\leq b, c\leq 9, 0< a\leq 9$$

We write the number abc as  100a+10b+c

If tens digit b is removed then we get the number ac and we write this as 10a+c

Given abc = 9(ac)+4c

100a+10b+c=9(10a+c)+4c

100a+10b+c-90a-9c-4c=0

10a + 10b -12c = 0

a + b = $$\frac{6}{5}c$$

from above we say that c is either 0 or 5 since a, b , c are digits and sum of the digits cannot be a fraction .

If c=0 then a+b =0 and a=-b which is not possible as 'a' cannot be a negative number.

c= 5 is solution.

a+b =6

All possible ways :

Hence the possible numbers are 155, 245, 335, 425, 515, 605 .

Therefore 6 numbers are possible which satisfies the given conditions.

2) Let  P(x) = x² + b x + c, where b and c are integers. If P(x) is a factor of both

x^4 + 6 x^2 + 25 and 3 x^4 + 4 x^2 + 28 x + 5 , find the value of P(1).

Solution:

P(x) is a factor of x^4 + 6 x^2 + 25 and 3 x^4 + 4 x^2 + 28 x + 5.

P(x) is a factor , remainder =0 .

b(c-6+c-b^2)=0 and 25-c(6-c+b^2)=0

b=0 (or) 2c-6-b^2=0 and c(6-c+b^2)=25

Case i :

b=0 then c(6-c+0)=25

6c-c^2=25 $$\rightarrow c^{2}-6c+25=0$$ , for which c is not real number .

Hence b $$\neq$$ 0 .

Case ii :

2c-b^2-6=0 then b^2=2c-6 $$\rightarrow$$ c(6-c+2c-6)=25

$$\rightarrow$$ c^2 = 25 then c= 5 (since c cannot be negative as b^2=2c-6 cannot be negative ) .

b^2=4 then b=-2, or +2 .

then P(x)= x^2 + 2 x + 5 or P(x) = x^2 - 2 x + 5 .

P(x) is also a factor of 3 x^4 + 4 x^2 + 28 x + 5 .

P(x) = x^2 - 2 x + 5   divides 3 x^4 + 4 x^2 + 28 x + 5 with remainder 0.

Therefor P(x) =x^2 - 2 x + 5

Hence P(1) = 1-2+5 = 4 .

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