JEE Mains 2019 #2

Let $k \in \mathbb{N}$ satisfies the given condition. There are 4 cases:

Case 1: $1000 \leq k < 7000$

The 1st digit may take only 1 and 3, a total of 2 choices. For the remaining 3 digits, it may take all of them, with 5 choices.

$\Rightarrow$ there are $2×5×5×5=250$ possible numbers

Case 2: $100 \leq k \leq 999$

The 1st digit may take only 1, 3, 7 and 9, a total of 4 choices. For the remaining 2 digits, it may take all of them, with 5 choices.

$\Rightarrow$ there are $4×5×5=100$ possible numbers

Case 3: $10 \leq k \leq 99$

The 1st digit may take only 1, 3, 7 and 9, a total of 4 choices. For the remaining digit, it may take all of them, with 5 choices.

$\Rightarrow$ there are $4×5=20$ possible numbers

Case 4: $0 < k \leq 9$

These can only take 1, 3, 7 and 9, a total of 4 numbers.

$\therefore$ there are a total of $250+100+20+4=\boxed{374}$ natural numbers that satisfy the condition.

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Clarification: 0 is excluded because 0 is a whole number, but not a natural or a counting number. Natural numbers starts at 1, 2, 3, ...

Let the numbers be $\underline{A}$ , $\underline{B}$ , $\underline{C}$ , $\underline{D}$

A $\rightarrow$ can have 3 values [0,1,3] ( because if we use 7,9 number will be greater than 7000)

B $\rightarrow$ can have 5 values [0,1,3,7,9]

C $\rightarrow$ can have 5 values [0,1,3,7,9]

D $\rightarrow$ can have 5 values [0,1,3,7,9]

So total required numbers will be= 3×5×5×5 -1. ( subtracting 1 because 0000 is not a natural number)

So answer is $\boxed{374}$