Answer to Question #4351 in Algebra for arun kumar vishwakarma

Question

you have to make 44 out of 6;7;8;and 9 each number has to exactly once and you are allowed only with the four basic operations (addition,substrction,multiplication,and division)you may use the same operators more than once

Answer 1

6,7,8,9 … 44
Consider sum 6+7+8+9=30<44. Hence, we have to use product to get 44.
44=2x2x11 and 6=2x3, 7=7, 8=2x2x2, 9=3x3. Hence, we can’t get 44 only with products, because we don’t have 11.
Also, the only way we can divide is 8x9/6=(2x2x2x3x3)/(2x3)=2x2x3=12. And we can’t make 44 from 7 and 12, because 7+12=19<44 and 7 and 12=2x2x3 doesn’t consist 11.
Consider all possible products:
6x7=42 and 8,9
6x8=48 and 7,9
6x9=54 and 7,8
7x8=56 and 6,9
7x9=63 and 6,8
8x9=72 and 6,7
As we prove, we can’t use products.
With 42, 8 and 9 we can get 43=42-8+9.
With 48, 7 and 9 we can get 46=48+7-9.
With 54, 7 and 8 we can get 39=54-7-8.
With 56, 6 and 9 we can get 41=56-6-9.
With 63, 6 and 8 we can get 49=63-6-8.
With 72, 6 and 7 we can’t get anything similar to 44.
All other operations (I mean + or -) won’t give us better value.
Hence, we can’t get 44 from 6,7,8,9 by using allowed operations.