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 ->  Divisibility and Prime Numbers  -> Lessons -> 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


   



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. Franke
This question is from textbook Mathematics for elementary teachers

Answer by scott8148(6628) About Me  (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