Mathematics Times – July 2019

Therefore there are four distinct pairs, that

satisfy the simultaneous equations.

6.Sol: In general ,we obtain , for n n^3   ( 1) ...^3

 2 1^33 and n even , the formula is

(^2) 3 3 3 3
1 1
( 1) (2 ) (2 1)
m m
n n
n n n
 

   

2

1

12 6 1

m

n

n n



  

2

1

12 12 6 1

m

n

n n n



   

124 2 6 1 1

m

n

n n



   

     

   



1 1

24 6

3 2

m m

m

     
    
   
m m[4( 1)( 1) 3( 1) 1]    m m
m m^2 (4 3)
which is divisible by m^2 .Since 2m = 2014, we
get m =1007. Therefore summation is divisible
by 1007^2.
7.Sol: Given

(^) 2 4 8 328a  b c
We know 328 8 41  , which is RHS-of the
equation can be factored so LHS. That is
2 (1 2a  ^2 b a 2 ) 8 41^3 c a  
which yields a = 3. Now we have
1 2  ^2 b a 23 c a 41
i.e., 2 2 3b  2 3 3c 40
Again we can see RHS- of the equation is
8 5 , so LHS can be factored.
i.e., 2 (1 23 3c  2 3 3 3)b  c  8 5
Which yields c = 2. Finally we have
1 2 2 3 6 3b    5
i.e., 2 2 6b  4
which yields b = 4. Now we have
2 3 3 8 6 17
3 4 2 24
a b c
abc
   
 
 
8.Sol: We know sum of interior angles of a
polygon of n - sides is( 2)n . Therefore the
sum of interior angles of 10 sides polygon is
8 . Since it has 8 180 , in which we have
7 are greater than 180 . Therefore, maximum
possible value of k is 7.
9.Sol: It happens, for the choices of the arguments
of the radicals in this problem, that we can
make this a bit less of a headache to think
about by noting that setting u x  1 i.e.,
u x^2   1 reduces the original equation to
6.Sol:
7.Sol:
8.Sol:
9.Sol:
10.Sol:
( 4) 4u^2   u ( 9) 6 1u^2   u
u    2 u 3 1
Since these terms must be positive or zero,
we can choose, say, u  2 a and u 3 1
a, with 0  a 1. Two of the possible
equations, u  2 a and 3   u 1 a produce
u  2 a, while using 2  u a implies that
u  2 a or u  3 1 a implies that
u  4 a are not mutually consistent results.
So we have the single interval
0  a 1
implies that
2    u 2 a 3
From this, we have
4    u x^2 1 9
implies that
5  x 10
10.Sol: Rewrite the given equation as
log log
log log
b a
c
a b
 

Since, a b, 1 so log 0a. Put

log

log

b

t

a

 and

rewriting the above equation as

1

t c

t

 

but we have

(^21)
2
t
t

. i.e., c 2. Therefore