Elitmus
Exam
Numerical Ability
Algebra
if v,w,x,y,z are non negative intergers each leass than 7,then how many distinct combinations are possible of(v,w,x,y,z) satisfy v(7^4) +w(7^3)+ x(7^2)+ y(7) +z=151001
Read Solution (Total 4)
-
- 151001=62*7^4+6*7^3+0*7^2+0*7^1+1
v=62
w=6
x=0
y=0
z=1
only these combinations can produce the result
so unique solution - 7 years agoHelpfull: Yes(1) No(3)
- Dividing 151001 by 11^4 give quotient 10 and remainder 4591
So
151001 = 10*11^4 + 4591
Dividing 4591 by 11^3 gives quotient 3 and remainder 598
So
151001 = 10*11^4 + 3*11^3 + 598
Dividing 598 by 11^2 gives quotient 4 and remainder 114
So
151001 = 10*11^4 + 3*11^3 + 4*11^2 + 114
Dividing 114 by 11 gives quotient 10 and remainder 4
So
151001 = 10*11^4 + 3*11^3 + 4*11^2 + 10*11 + 4
So v=10, w=3, x=4, y=10, z=4
is ONE solution. But is that the only one? We will
show that it indeed is the only solution: - 6 years agoHelpfull: Yes(1) No(1)
- In quest 11 s der not 7 n for 11 der exists only 1 solution
- 7 years agoHelpfull: Yes(0) No(1)
- 151001=62*7^7+6*7^3+1*7^2+0*7+32
v=62
w=6
x=1
y=0
z=32 - 6 years agoHelpfull: Yes(0) No(1)
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