SOLUTION: If v,w,x,y,z are non negative integers,each less than 11,then how many distinct combinations of v,w,x,y,z satisfy v(11^4)+w(11^3)+x(11^2)+y(11)+z=151001

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:

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So we show that that is the ONLY solution.

v cannot be larger than 10.
Suppose that v could be less than 10, then the largest

v(11^4)+w(11^3)+x(11^2)+y(11)+z 

could possibly be would be when v=9, w=10, x=10, y=10, z=10

9(11^4)+10(11^3)+10(11^2)+10(11)+10 = 146409

But that is less than 151001. So v can only be 10.

So we know that:

10(11^4)+w(11^3)+x(11^2)+y(11)+z=151001

  146410+w(11^3)+x(11^2)+y(11)+z=151001

         w(11^3)+x(11^2)+y(11)+z=4591

If w were more than 3, the smallest

         w(11^3)+x(11^2)+y(11)+z

could be is when w=4,x=0,y=0,z=0

         4(11^3)+0(11^2)+0(11)+0 = 5324

But that is larger than 4591, so w is not greater than 4.

On the other hand, 

If w were less than 3, the largest

         w(11^3)+x(11^2)+y(11)+z

could be is when w=2,x=10,y=10,z=10

         2(11^3)+10(11^2)+10(11)+10 = 3992

But that is smaller than 4591, so w is not less than 3 either.

Therefore w = 3 is the only possibility, and

         3(11^3)+x(11^2)+y(11)+z = 4591
                 x(11^2)+y(11)+z = 598
                 

If x were more than 4, the smallest

                 x(11^2)+y(11)+z

could be is when x=5,y=0,z=0

                 5(11^2)+0(11)+0 = 605

But that is larger than 598, so w is not greater than 4.

On the other hand, 

If x were less than 4, the largest

                 x(11^2)+y(11)+z

could be is when x=3,y=10,z=10

                 3(11^2)+10(11)+10 = 483

But that is smaller than 598, so x is not less than 4 either.

Therefore x = 4 is the only possibility, and

                 4(11^2)+y(11)+z = 598
                         y(11)+z = 114  

Y cannot be more than 10.
If y were less than 10, the largest

                         y(11)+z

could be is when y=9,z=10

                        9(11)+10 = 109

But that is smaller than 114, so x = 10.


So we have proved that v=10, w=3, x=4, y=10, z=4

is the only solution.

Edwin