# SOLUTION: I need help with this problem. The problem states: Two digits of this number were erased: 273*49*5. However we know that 9 and 11 divide the number. What is it? What I have d

Algebra ->  Algebra  -> Divisibility and Prime Numbers -> SOLUTION: I need help with this problem. The problem states: Two digits of this number were erased: 273*49*5. However we know that 9 and 11 divide the number. What is it? What I have d      Log On

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 Algebra: Divisibility and Prime Numbers Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Divisibility and Prime Numbers Question 86740This question is from textbook Mathematics for elementary teachers : I need help with this problem. The problem states: Two digits of this number were erased: 273*49*5. However we know that 9 and 11 divide the number. What is it? What I have done so far: 2+7+3+*+4+9+*+5 = 30+*+* The only number divisible by both 9 and 11 is 99. However, to make the sum of this number equal 99 the two digits would have to be 35 and 34. I am not sure if these numbers qualify? I tried other single numbers and haven't found a sum divided by both 9 and 11. Your help would be most appreciated. Thanks. Sincerely, R. FrankeThis question is from textbook Mathematics for elementary teachers Answer by scott8148(6628)   (Show Source): You can put this solution on YOUR website!let x equal the missing thousands digit and y equal the missing tens digit a number is divisible by 9 if the sum of the digits is divisible by 9 ... so 30+x+y=36 or 30+x+y=45 ... x+y=6 or x+y=15 a number is divisible by 11 if the difference between the sum of the odd numbered digits and the sum of the even numbered digits is divisible by 11 ... so 2+3+4+y-(7+x+9+5) or y-x-12=-11 ... y-x=1 ... this means an ODD sum x+y=15 ... y-x=1 ... 2y=16 ... y=8 ... x=7