AP EAMCET · Maths · Permutation Combination
Let \(m\) be a natural number such that \(20000 < m < 60000\) and let \(k\) be the sum of all the digits in \(m\). Then the number of numbers \(m\) for which \(k\) is even, is
- A 19909
- B 19989
- C 18999
- D 19999
Answer & Solution
Correct Answer
(D) 19999
Step-by-step Solution
Detailed explanation
Let us consider 10 successive five digit numbers \[ \begin{aligned} & a_1 a_2 a_3 a_4 0 \\ & a_1 a_2 a_3 a_4 1 \\ & a_1 a_2 a_3 a_4 2 \\ & \ldots \ldots \ldots \ldots \ldots . . \\ & a_1 a_2 a_3 a_4 9 \end{aligned} \] Where, \(a_1, a_2, a_3, a_4\) are some digits. We see that…
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