Question Bank  Equations  Set 2

Q25) A leaf is torn from back of a book. The sum of remaining pages is 15000 . what are the page numbers on the torn leaf

Now, let the numbers of pages be n
Since , number of pages after a leaf is torn is 15000, then the sum of the numbers on the pages will be greater than 15000
Now, n(n+1)/2 > 15000
n (n+1) > 30000
(n+1)^2 > n (n+1) > 30000 > 173^2
So , 172 < n < 176
n could be 173, 174, 175
n = 173 then n(n+1)/2 = 173*174/2 = 15051
Sum of the numbers on torn pages.
= 15051 – 15000 = 51
And this would be y + ( y +1)
= 2y + 1 = 51
So page numbers are 25, 26
If n = 174 the n(n+1)/2 = 15225
Sum on torn pages
15225 – 15000 = 225
Page numbers on torn pages are 112, 113
When = 175, n(n+1)/2 = 15400
Sum on the torn pages = 400 which is even . So not possible .
Now , we have to know that the smaller number on the torn pages should be odd. So, the numbers on the torn pages are 25 and 26 and the number of pages is 173.

Q26) Find all positive integers for which n^2 + 96 is perfect square

We solve it by two approaches
Now suppose k is a positive integer such that
n^2+ 96 = k^2
k^2 –n^2 = 96
(kn) (k+n) = 96
Since (kn) , (k+n) must be both odd and even
But odd case is not possible
Only possible cases are..
kn = 2, k+n = 48
kn = 4, k+n = 24
kn = 6, k+n = 16
kn = 8 , k+n = 12
on solving these equations we get
pairs of k,n are… (25,23) (14,10) (11,5) (10,2)second approach
positive solutions = (no of factors of 96 after dividing by 4) /2
= 2^3*3=total factors are 8
So positive solutions = 8/2=4

Q27) Find all integral solutions of a^4 + b^4 + c^4 = 1991 + d^ 4.

In this type of question we would apply divisibility on L.H.S and R.H.S in most cases also these questions can be solved orally
Now, we will learn new concept.
Let n be an integer, When it is odd.
Let n^4 – 1 = ( n^2 1)(n^2 +1) = (n1)(n+1)(n^2+1)
It must be divisible by 16 for any odd n.
Also (n1) , (n+1) are consecutive even integers and one of them is divisible by 4.
If n is even n^4 must be divisible by 16.
Thus n^4 gives remainder 0 and 1 .
Possible remainders of a^4 + b^4 + c^4 – d^4 when divided by 16 are { 1, 0, 1, 2, 3}
But 1991 gives remainder 7 when divided by 16 so no integral solution exist because L.H.S & R.H.S have different remainders

Q28) Find all real solutions of x for the equation x^10  x^8 + 8x^6  24x^4 + 32x^2  48 = 0

x^8( x^2 1) +8 x^4( x^2 1) + 32 ( x^2 1) 16 x^4 16 = 0
(x^2 1){ x^8 + 8 x^4 +32 } 16( x^4 +1) = 0
(x^2 1 ){ (x^4 + 4)^2 +16 } = 16(x^4 +1)
(x^4 +4)^2+16 = 16( x^4+1)/(x^2 1)
Put x^4 = t
(t +4)^2+16 = 16(t+1)/(√t1)
Put t = 4
8^2 +16 = 16*5
64 + 16 = 80 ( satisfied)
x^4 = 4
=> x = +/ √2

Q29) Let f(x) be a fourth degree polynomial such that f(1) = 2, f(2) = 4, f(3) = 6 and f(4) = 12. Find f(5)

Q30) 1 + 1/3 + 1/6 + 1/10 + 1/15 +...... up to 9 terms = ?

1, 3, 6, 10, 15 .....
Let nth term = an^2 + bn + c
Put n = 1
a + b + c = 1
4a + 2b + c = 3
9a + 3b + c = 6
5a + b = 3
3a + b = 2
2a = 1
a= 1/2
b = 23/2= 1/2
C = 0
So (1/2)n^2 + 1/2(n)Nth term = 1/1/2(n^2+n)= 2/n(n+1)
=> 2 [ 1/n  1/(n+1)]
2 [ 1/11/2+1/21/3+.....+1/91/10]
2[ 11/10]
2 * 9/10 = 9/5