Question

If you write numbers from 3 to 3333 how many times would you have to write 3

Answer 1

Answer:

This is a good question ....

Hope it helps you ....

Answer 2

Answer:
The correct answer is 1336
Step-by-step explanation:


3*f(3) => for 0–999 and prepending 0,1, and 2, covering 0–2999

3*f(2) => prepending 30,31,32 to 0–99 (but NOT counting the leftmost 3), covering 3000 to 3299

3*f(1) => prepending 330,331,332 to 0–9, covering 3300 to 3329 (but NOT counting the leftmost and second left most 3)

1 => prepending 333 to 0 to 3 (but NOT counting the leftmost , second left most and third leftmost 3)

Now, including the 3 we ignored earlier:

Number of numbers between 3000 and 3333 = 333 + 1 = 334 (including the initial 3000 and the final 3333). They all have 3 as left most digit (ignored earlier)

Number of numbers between 3300 to 3333 = 33+1 = 34 (including the initial 3300 and the final 3333)They all have 3 as second leftmost digit (ignored earlier)

Number of numbers between 3330 and 3333 = 3 + 1 = 4 (including the initial 3330 and the final 3333) They all have 3 as third leftmost digit (ignored earlier).

Therefore:

Answer would be: 3*f(3) + 3*f(2) + 3*f(1) + 1 + 334 + 34 + 4

Now f(1) = 1

Therefore, f(2) = 10*f(1) + 10 = 20

Therefore, f(3)= 20*10 + 10^2 = 300

3*f(3) + 3*f(2) + 3*f(1) + 1 + 334 + 34 + 4

= 3*300 + 3*20 + 3*1 + 372 + 1

= 900 + 60 + 3 + 373

= 1336