



 
Maryland, I hope it is clear that it is not 8 and 7 that are multiplied together, because it is impossible to get a result of 56 by subtracting two numbers from 1 to 6. The result of the multiplication has to be small, probably 1 times a small number. How about 1 times 2? then we would need to split the remaining numbers, 3, 4, 5, 6, 7, 8 in pairs with a difference of 2. So 3 has to go with 5 (5  3 = 2) and then 4 has to go with 6 (6  4 = 2), but then we are stuck with 7 and 8 which don't have a difference of 2. Let's try 1x3 = 3 for the multiplication: Then 2 has to go with 5 (5  3 = 2) and 4 has to go with 7. Then we are left with 6 and 8. 8  6 = 2, not 3, so again it does not work. I tried again with 1x4, and 1x5, and it never seems to work. So now I think about it, and I see it cannot work: If the result of the multiplication were odd, then the three subtraction groupings would be pairs of numbers with different parities (odd  even, or even  odd, to get an odd result), so we'd be left with an even and an odd number to multiply, giving an even result. So the result of the multiplication has to be even. This means that the subtraction groupings are of the form even  even or odd  odd. So the multiplication was also the form odd times odd or even times even. It cannot have been odd times odd because the answer is even, and the smallest even times even multiplication is 2x4 = 8, and you cannot get a difference of 8 by subtracting numbers from 1 to 8. Are you sure the problem asks for one multiplication and three subtractions? Claude
 


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 