Latest Elitmus Aptitude Question SOLUTION: if v,w,x,y,z are non negative intergers each leass than 11,then how many distinct combinations are possible of(v,w,x,y,z) satisfy v(11^4) +w(11^3)+ x(11^2)+ y(11) +z=151001

change151001 to base 11 number .
it will be
10 3 4 10 4

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

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 only ONE solution.

Only 1 possible combination.

=10, w=3,x=4, y=10, z=4

11(11^3*v + 11^2*w + 11*x + y) + z = 11*13727 +4
so z=4
11(11^2*v + 11*w + x ) + y = 11*1247 +10
so y=10
11(11*v + w ) + x = 11*113 +4
so x=4
11*v + w = 11*10 + 3
so w=3
v=10

if v,w,x,y,z are non negative intergers each leass than 11,then how many distinct combinations are possible of(v,w,x,y,z) satisfy v(11^4) +w(11^3)+ x(11^2)+ y(11) +z=151001
a)0
b)1
c)2
d)3

i think its asking 4 distinct combination of v,w,x,y,z so the answer will be
0

11^4v+ 11^3w + 11^2x +11^y +z = 151001
11(11^3v+11^2w+11^x+y) + z = 11*13727 + 4 so, z = 4
now,
11(11^3v+11^2w+11^x+y)=11*13727
11(11^2v + 11^w +x) + y =11*1247 + 10 y = 10

11(11^v + w ) + x = 11*113 + 4 x = 4

11^v + w = 11*10 + 3 v=10 , w=3

how u can said only one combinatio is possible

0 Will be the answer as it is asking for distinct combinations.There are 2 10s in the solution provided.

First of all, I think you should type the given equation as
33 w + 32 x + 3y + z= 34,
Since w & x are non-negative integers,
both w and x cannot be greater than 0.
In other words, one of w and x must be 0.
When w =0, we have 32 x + 3 y + z = 34.
If x = 0,then 3y+z = 34.
But both y & z are less than 3, so 3y + z

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:

---------------------------------------------------

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.


Source : https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.857977.html

Taking 11 common from LHS separating z and breaking 151001 as a factor of 11 with some remainder in RHS
11(113v+112w+11x+y) + z = 11x13727 + 4 [hence; z = 4]
Equating z =4 and cancelling 11 from each side we get the following simplified equation and again repeating the above procedure;
11(112v + 11w +x) + y =11x1247 + 10 [y = 10]
11(11v + w ) + x = 11x113 + 4 [x = 4]
11v + w = 11x10 + 3 [v=10 , w=3]
So we get the unique solution for the above equaton.
1 Answer

151001/11^4 = 10 with remainder = 4591

4591/11^3 = 3 with remainder = 598

598/11^2 = 4 with remainder = 114

114/11 = 10 with remainder = 4

Therefore 151001=10(114)+3(113)+4(112)+10(11)+4

i.e., (v,w,x,y,z) is (10,3,4,10,4). These values satisfies given conditions and is one valid option.

Let's analyze if further option available satisfying the given condition. v,w,x,y,z should be less than 11 and greater than zero. The solution we got have v as 10. Suppose v=9. With this, the maximum value possible under the given constraints is
9(114)+10(113)+10(112)+10(11)+10=146409

answer is 1 because it is asking how many distinct combination. It is not asking for distinct values.