I need to find the answer of 41 adding or subtracting those 5 numbers, I can only use those numbers, & use them only once! ,
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Let's simplify the problem by dividing all the numbers by 9 and noting only the remainder (the 9s remainder):

2, 0, 4, 8, 3. We can only use + or - as the arithmetic operators, and these will be inserted between the numbers. We can drop the zero for now and just look at 2, 4, 8, 3. So we use only three operators giving us a total of 8 possibilities. The 9s remainder of 41 is 5. We can apply the complete set of operators and determine the 9s remainder of each result. If the result is negative, we simply add 9. If the result is 9 the 9s remainder is 0:

2+4+8+3=8, 2+4+8-3=2, 2+4-8+3=1, 2+4-8-3=4,

2-4+8+3=0, 2-4+8-3=3, 2-4-8+3=2, 2-4-8-3=5.

The 9s remainder of 41 is 5 which is the same as -4, so we are looking for results whose absolute value is 4 or 5. The only such results are: 2+4-8-3=4 and 2-4-8-3=5. If we negate both sides of the first of these candidates we get: -2-4+8+3=5. We can apply the combinations of signs to the original numbers (leaving out 27 for the time being):

-56-04+17+93=50 and 56-04-17-93=-58. We can reverse the signs in the second equation:

-56+04+17+93=58. If we add and subtract 27 from these results we get:

77 and 23, and 31 and 85, none of which are the desired result of 41. So it's not possible, using only addition and subtraction, to get the result 41 from the five given numbers. 

We now need to consider combining two of the given numbers to make a 3- or 4-digit number. The smallest is 0417 or 417. Then we need to subtract the sum of the remaining numbers: 56+27+93=176. But 417-176=241, which is the smallest possible difference between the longer number and the sum of the remaining 3 numbers. And 241≠41, so there's no solution here.

Finally we need to create two 4-digit numbers from two pairs of the given numbers. Their difference has to be the remaining number. But the smallest difference is 2704-1793=911≠56.

by Top Rated User (1.2m points)

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