Find the sum of all three -digit numbers which are multiples of 9
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567
Step-by-step explanation:
Let three consecutive multiple of 9 are x,x+1,x+2
then , (x+x+1+x+2)×9=567
x=20
1st= 180,2nd = 189,3rd= 198
THEREFORE SUM OF ALL DIGIT =180+189+198
THEREFORE ANSWER =567
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