If v,w,x,y,z are non negative integer, each less than 11,then how many…
2023
If v,w,x,y,z are non negative integer, each less than 11,then how many distinct combinations(v,w,x,y,z) satisfy v(11^4)+w(11^3)+x(11^2)+y(11)+z=151001?
- A.
1
- B.
2
- C.
3
- D.
zero
Attempted by 28 students.
Show answer & explanation
Correct answer: A
Key idea: the left side v·11⁴ + w·11³ + x·11² + y·11 + z is exactly how a number is written in base 11, where v, w, x, y, z are the base-11 “digits”. Each digit is allowed to be 0 to 10, which is precisely the rule “each less than 11”. So the question is really asking: in how many ways can 151001 be written in base 11 using these five digit places?
Step-by-step conversion of 151001 to base 11:
11⁴ = 14641. Divide: 151001 ÷ 14641 gives v = 10, with remainder 151001 − 10×14641 = 4591.
11³ = 1331. Divide: 4591 ÷ 1331 gives w = 3, with remainder 4591 − 3×1331 = 598.
11² = 121. Divide: 598 ÷ 121 gives x = 4, with remainder 598 − 4×121 = 114.
11¹ = 11. Divide: 114 ÷ 11 gives y = 10, with remainder 114 − 10×11 = 4.
What is left is z = 4.
So (v, w, x, y, z) = (10, 3, 4, 10, 4), and every value is between 0 and 10, so it is a valid combination. Check: 10×14641 + 3×1331 + 4×121 + 10×11 + 4 = 151001. ✓
Why is the count exactly one? Because 11⁴ = 14641 ≤ 151001 < 161051 = 11⁵, the number needs exactly five base-11 places — no more, no fewer. And every whole number has only one base-11 representation, just as every number has only one ordinary base-10 form. So there is exactly one combination that works.
Answer: 1.