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