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-

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)

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)

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

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

step-by-step explanation:

one day the were 5 ducks, and their king told them to multiply their kingdom. so each went their own seperate way and met 4 other ducks on their journey.

when the ducks returned to their kingdom, it was discovered that there were 5 groups of 5 ducks. when the kind counted the number of ducks in total (including the ones he sent out earlier) he counted 25.

therefore 5 times 5 is 25.

the end. : )

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