I believe the answer is A and B. Both cannot be expressed as a sum of 2 digit numbers whose digits are the same but interchanged.
Let M= AB(A.. B = 0â€¦9) and N = BA where A and B are the digits 0-9.
- One deducted rule is the last digit of the sum should be the sum of A+B . To illustrate further take for eg 44 the last digit is 4 which can be expressed as a sum of 2+2 , 1+3,0+4. (which means First digit is A and the second B) = 22+22 = 44, Similarly 13+31 = 44, 04+40 = 44. Therefore 44 can be expressed as per above rule. Applying the same rule for 99 = 9 can be expressed as sum of 1+8, 9+0,7+2, 5+4,6+3 all of which can yield 99. Therefore 99 is eliminated. For numbers less than 100 the sum of the last 2 digits cannot exceed 9.
- For numbers greater than 100 the last digit can be expressed as a sum of numbers from 0-9 with one of the numbers > 5. To illustrate this let us analyze 121. Last digit is 1. Since the number is > than 100 we need to express it as a sum of single digit number whose total is 11 therefore it can be 9+2,8+3,7+4,6+5, if we apply any of this we find 92+29 = 121, 83+38 = 121, 74+47+121, 65+56=121. Therefore 121 is eliminated.
164 = 4 can be expressed as 8+6, 9+5,7+7, = 86+68=154,95+59=154,77+77=154. Therefore 164 is one of the choices
Let us look at 181 = 1 can be expressed as 9+2,7+4,8+3,6+5, none of which will yield 181 . Therefore according to me there are 2 answers A or B.
I do not know whether this is the correct way. If someone can elaborate further that will be great.
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