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renw언30ム구淑.no-uShuchanU |
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Answer» 100 is the answer Here's the explanation:he lowest number greater than a hundred and divisible by 9 is 108...the largest number divisible by 9 and less than a thousand is 999... we use this formula... T = A + (n-1)*d T is the nth term in an arithmetic sequenceA is the first term in an arithmetic sequencen is the number of termsd is the common difference you are asking for "the number of 3 digit numbers divisible by 9" or numbers that are greater than 100 but less than 1000 that are divisible by 9 so... we are looking for the number of terms given first term(A) = 108last term(T) = 999common difference(d) = 9 we subtitute the given values999 = 108 + ( n - 1 ) * 9 we then solve...999 = 108 + ( n - 1 ) * 9999 - 108 = ( n - 1 ) * 9891 = ( n - 1 ) * 9( 891 ) / 9 = [ ( n - 1 ) * 9 ] / 999 = n - 199 + 1 = n100 = n therefore...there are 100 3-digit numbers divisible by 9 |
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