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What is the smallest possible difference between two different nine digit integers each of which includes all of the digits 1 to 9?
Wiki User
∙ 12y ago
Pick a and b as the last two digits (and we may assume a is greater than b). Make the remaining first seven digits the same as each other (any way will be fine). The last two digits will be ab for the first number and ba for the second number. The value of ab is 10a+b, the value of ba is 10b+a. The difference is (10a+b)-(10b+a)= 9a-9b = 9(a-b). To minimise this we just make a and b consecutive integers - there are eight ways of doing this. There are 7! ways of arranging the first seven digits. So 8! ways of choosing the numbers.
Tom 1107
Wiki User
∙ 12y agoOne number is 987654321 minus the othernumber, which is 987654312 and the answer is therefore 9
Tom 1107
Nine. Pick a and b as the last two digits (and we may assume a is greater than b). Make the remaining first seven digits the same as each other (any way will be fine). The last two digits will be ab for the first number and ba for the second number. The value of ab is 10a+b, the value of ba is 10b+a. The difference is (10a+b)-(10b+a)= 9a-9b = 9(a-b). To minimise this we just make a and b consecutive integers - there are eight ways of doing this, each of which will give the answer nine. There are 7! ways of arranging the first seven digits. So 8x7!=8! ways of choosing the numbers.
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