Elitmus
Exam
Numerical Ability
Algebra
If v,w,x,y,z are non negative interger each less than 7.then how many distinct combination of (v,w,x,y,z) satisfy
v(7^4) + w(7^3)+ x(7^2) + y(7) + z = 151001
Read Solution (Total 1)
-
- 151001
11
4
151001114 = 10 with remainder = 4591
4591
11
3
4591113 = 3 with remainder = 598
598
11
2
598112 = 4 with remainder = 114
114
11
11411 = 10 with remainder = 4
Therefore
151001
=
10
(
11
4
)
+
3
(
11
3
)
+
4
(
11
2
)
+
10
(
11
)
+
4
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
(
11
4
)
+
10
(
11
3
)
+
10
(
11
2
)
+
10
(
11
)
+
10
=
146409
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