Q: How would you type 1-2-3 in base 2?

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This cannot be determined because the number 3 does not exist in base 3; only 1 and 2 do.

1*5^2 + 2*5^1 + 3*5^0

123/3 - 2, 123/3 and 123/3 + 2 ie 39, 41 and 43.

No. 123 is only divisible by: 1, 3, 41, 123.

−4a + 139 = 4a + 123 → −8a = −16 → a = 24a + 123 → 4(2) + 123 = 8 + 123 = 131°

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This cannot be determined because the number 3 does not exist in base 3; only 1 and 2 do.

1*5^2 + 2*5^1 + 3*5^0

2 km to 123 m = 2000 m to 123 m = 2000 to 1232 km to 123 m = 2000 m to 123 m = 2000 to 1232 km to 123 m = 2000 m to 123 m = 2000 to 1232 km to 123 m = 2000 m to 123 m = 2000 to 123

123 is not divisible by 2 it is divisible by 3

1252 + 123 = 125

123/3 - 2, 123/3 and 123/3 + 2 ie 39, 41 and 43.

No. 123 is only divisible by: 1, 3, 41, 123.

39, 41 and 43. or 123/3 -2, 123/3 and 123/3 +2

246 = 123 x 2

−4a + 139 = 4a + 123 → −8a = −16 → a = 24a + 123 → 4(2) + 123 = 8 + 123 = 131°

248

(1/2) x (123/5) = 123/10 = 12.3