🍁A∂vαทcє Maths🍁
Open in Telegram
Channel's :- join 👇👇👇 @success_on_motivations 🍁🍁🍁 @success_on_pdf 🍁🍁🍁 @Arithmetic_collection 🍁🍁🍁 Gk Group :- 👇👇👇join @gkgsforntpc 🍁🍁🍁
Show more1 087
Subscribers
No data24 hours
+47 days
+1030 days
Posts Archive
1 087
n²x² + 1/x² = n
n(nx² + 1/nx²) = n
(nx² + 1/nx²) = 1
So
(nx²)³ = - 1
n³x⁶ = - 1
n³x³ = - 1/x³
(nx)³ = - 1/x³
1/(nx)³ = - x³
Now
x³ + 1/(nx)³
x³ - x³
0
1 087
CD
+AB
1CE
It’s given that
C+A=C(mod10)
A=0(mod10)
So A=0 but this is not possible because then 1CE will not be possible
So it means B+D is a two digit number
Which means C+A+1=C(mod10)
A+1=0(mod10)
A=-1(mod10)
A=9(mod10)
So A=9
1 087
A,B,C,D,E can take max value of 9 as they are digits
10A+B+10C+D= 100+10C+E
10A+B+D=100+E
At Max value of D=B=9 and E=1
10A=83
A= 8.3
So min value of A should be 8.3 in worst case so A has to take 9
90+B+D= 100+E
B+D= 10+E
B,D,E should be distinct and none can be 9
8+7= 10+5 or any other combination
C can take any number from 1 to 4
So numbers will be
For example
98 and 17
115
Many such numbers are possible
But value of A is fixed at 9
Credit - Vivek Gupta
1 087
A,B,C,D,E can take max value of 9 as they are digits
10A+B+10C+D= 100+10C+E
10A+B+D=100+E
At Max value of D=B=9 and E=1
10A=83
A= 8.3
So min value of A should be 8.3 in worst case so A has to take 9
90+B+D= 100+E
B+D= 10+E
B,D,E should be distinct and none can be 9
8+7= 10+5
C can take any number from 1 to 4
So numbers will be
For example
98 and 17
115
Credit @viki600
