Question

In a book, 495 digits were used to number the pages.
How many pages were there?
How many 9s were used?​

Answer 1

Answer:

There are 9 pages that use a single digit (pages 1-9), leaving 495 digits - 9 pages × 1 digit/page = 486 digits

There are 90 pages that use 2 digits (pages 10-99), leaving 486 digits - 90 pages × 2 digits/page = 306 digits

There are 900 pages that use 3 digits (pages 100-999); this would be 2,700 digits, so the number of pages is somewhere in the hundreds.

306 digits ÷ 3 digits/page = 102 pages in the hundreds.

→ total number of pages = 102 + 90 + 9 = 201 pages.

Step-by-step explanation:

I hope it helps to you

please make me brainist please

Answer 2

Answer:

201 pages are used using 495 digits.

40 times the number 9 is used.

Step-by-step explanation:

Given 495 digits were used to number the pages.

The single digit numbers are 1 to 9.

So, 9 pages are numbered with single digit, 1 digit is used per page and hence the remaining pages are

$495 - 9\times 1 = 486 \mbox{pages}$

Next the 2 digit numbers are 10 to 99, with a total number of 90 two digit numbers are used.

Since 2 digits per paper will be used, the remaining pages are

$486 - 90 \times 2 = 486 - 180 =306 \mbox{pages}$

Then the 306 digits are used with 3 digits per each page.

Therefore,

$\dfrac{306}{3} =102\mbox{pages}$  = 102 pages in the hundreds.

And hence the total number of pages

$= 102 + 90 + 9 = 201 \mbox{pages}$

So, 201 pages are used using 495 digits.

And in total 40, the number 9 are used for 201 pages.

For more problems:

https://brainly.in/question/17939331

https://brainly.in/question/12452354

https://brainly.in/question/2499130