# George is reading a novel which has 303 pages. how many times does the digit 3 get printed in the page numbers? how many pages are there which has only one 3 present in the page number?

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by kavyakristi

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by kavyakristi

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Case 1:

numbers with 3 are like 3,13,23,......,303

they are in AP with a=3, d=10

let the total number of pages with the digit 3 be n.

303 = 3 + (n-1)10

⇒ n-1 = 300/10 = 30

⇒ n = 31

number of pages = 31

case 2:

number like 30,31,...39; 130,131,...139; 230,231,...239

from 30 to 39 - there are 10 numbers

from 130 to 139 - there are 10 numbers

from 230 to 239 - there are 10 numbers

total = 10+10+10 = 30

but we have already counted 33,133 and 233.

so number of pages = 30-3 = 27

case 3:

the number 300,301 and 302. (303 has already been counted)

so numbers = 3

So total number of pages = 31+27+3 = 61

every number having three has only one three except 33,133,233 and 303.

number of pages with 2 threes = 4

number of pages with only one three = 61 - 4 = 57

total number of printing of 3 = 57×1 +4×2 = 57+8 = 65.

**So the digit 3 has been printed 65 times and number of pages with only one 3 present is 57.**

numbers with 3 are like 3,13,23,......,303

they are in AP with a=3, d=10

let the total number of pages with the digit 3 be n.

303 = 3 + (n-1)10

⇒ n-1 = 300/10 = 30

⇒ n = 31

number of pages = 31

case 2:

number like 30,31,...39; 130,131,...139; 230,231,...239

from 30 to 39 - there are 10 numbers

from 130 to 139 - there are 10 numbers

from 230 to 239 - there are 10 numbers

total = 10+10+10 = 30

but we have already counted 33,133 and 233.

so number of pages = 30-3 = 27

case 3:

the number 300,301 and 302. (303 has already been counted)

so numbers = 3

So total number of pages = 31+27+3 = 61

every number having three has only one three except 33,133,233 and 303.

number of pages with 2 threes = 4

number of pages with only one three = 61 - 4 = 57

total number of printing of 3 = 57×1 +4×2 = 57+8 = 65.