Take any two-digit number. Reverse the digits, and subtract your answer from your original number. What do you notice?
(10a + b)-(10b + a), ((Ten lots of a)+ one lot of b)- ((Ten lots of b) + one lot of a). This equals 9a - 9b, you can take out a factor of 9, 9(a-b) this means the answer is always a multiple of 9. 64 - 46 = 18 (2x9) 87- 78 = 9 (1x9) 10 - 01 = 9.
Take any two-digit number. Add its digits, and subtract your answer from your original number. What do you notice?
TU
ab can be written as 10 a + b. The digits are a and b (a + b) 10a + b - a + b = 9a + 0b = 9a (this is always a multiple of 9)
Take any three-digit number. Reverse the digits, and subtract your answer from your original number. What do you notice?
abc - cba (100a + 10b + c)- (100c + 10b + a) 100a - a = 99a 10b - 10b = 0b c - 100c = -99c so it is 99a - 99c, you can take out a factor of 99, 99(a - c)It is always a multiple of 99 and there for 11 and 9. Check, on first sum I got 529-925 to be -404 but then when I divided it by 11 I found it wasn't divisible by it. I checked the sum on my calculator to get -396, when I did the sum 5 - 9 I put it down as -4 but in the process made the whole thing minus when it was not. It is actually - 400 + 4!
Take any five-digit number. Reverse the digits, and subtract your answer from your original number. What do you notice?
abcde - edcba (10,000a + 1000b + 100c + 10d + 1e) - ( 1a + 10b + 100c + 1000d + 10,000e) = 9999a + 990b + 0 - 990d - 9999e worked out this equals 99(101a + 10b - 10d - 101e) this is always a multiple of 99.