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Find the largest three-digit number (written in decimal notation) that is divisible by 22 and such that the sum of the units digit and the tens digit is 11.

 Dec 21, 2019
 #1
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I get 792

 

Let the number  be     a*10^2  + b*10  +  c

 

If the number is divisible by 22, it is also  divisible  by  11

 

This means (possibly)  that

 

a  -  b   +  c   =   0

 

And  b + c  =  11       subtract these

 

a - 2b  =  -11

 

2b - a   =  11

 

2b   =  11  +  a    

 

So....we want a  as large as possible    such that b  is a single digit....and  a  must be odd because 2b is even

 

The largest  value of a such that  b  is a single digit is when  a   = 7

 

So......this means  that  b  = 9

 

And

 

7 - 9 + 2  = 0

 

So...792 is the number

 

 

cool cool cool

 Dec 21, 2019
edited by CPhill  Dec 22, 2019

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