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

Step-by-step solution

Let us consider 10 successive five digit numbers aligned & a₁ a₂ a₃ a₄ 0 & a₁ a₂ a₃ a₄ 1 & a₁ a₂ a₃ a₄ 2 & . . & a₁ a₂ a₃ a₄ 9 aligned Where, a₁, a₂, a₃, a₄ are some digits. We see that half of these 10 numbers i.e. 5 have an even sum of digits. The first digit a₁ can takes 2,3,4,5 and each of the digits a₂, a₃, a₄ can takes 10 different values the units place digit can assume only 5 different values of which the sum of all digits is even. aligned So, value of K is & =4 10^3 5-1 & [ 20,000 will not include ] & =199