Question

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 ?

Answer 1

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)

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

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)

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

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

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

$\frac{123}{270} =\frac{41}{90}$