206k views
0 votes
easiest question everrr for 40 points math . Metal digits from 0 to 9 were ordered to number all houses (from 1 to 126) in a newly constructed complex. If the first 78 houses have already been numbered, then what part of the ordered amount of digits is left to be used ?

User Ylama
by
7.8k points

1 Answer

3 votes

Answer-

The part of the ordered amount of digits is left to be used is 41/90

Solution-

Metal digits from 0 to 9 were ordered to number all houses from 1 to 126.

The number of metal digits ordered = sum of all digits from 1 to 126

As, from 1 to 126 there are 9 single digit numbers (i.e 1 to 9), 90 two digit numbers (i.e 10 to 99) and 27 three digits number (i.e 100 to 126)


\text{So the number of metal digits ordered} = (9* 1)+(90* 2)+(27* 3)=9+180+81=270

The number of metal digits used = sum of all digits from 1 to 78

As, from 1 to 78 there are 9 single digit numbers (i.e 1 to 9), 69 two digit numbers (i.e 10 to 78)


\text{So the number of metal digits ordered} = (9* 1)+(69* 2)=9+138=147

The number of metsal digits left = 270 - 147 = 123

Then part of the ordered amount of digits is left to be used,


(123)/(270) =(41)/(90)


User Jords
by
8.0k points

Related questions

asked Nov 9, 2022 51.9k views
Hashchen asked Nov 9, 2022
by Hashchen
9.2k points
1 answer
4 votes
51.9k views
1 answer
21 votes
228k views
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.