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- December 12th 2012, 03:09 PM #1

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- December 12th 2012, 05:33 PM #2

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## Re: The Lonley 8

the first (left-most) digit of the number being divided must be 1, for if it were 2 or more, then the first subtraction would consist of 4 X's. if the problem is posed fairly (that is if the divisor is truly a 3-digit number), than it is between 100 and 999.

note that we brought down 2 digits after the first subtraction, so the 8 must correspond to the 2nd subtraction. since 8*quotient is a 3-digit number, our divisor is between 100 and 124. the digit after the 8 in the quotient must be a 0, as is the digit in the quotient before the 8.

this means the last digit of the quotient is 9. now the greatest number that the first number we could be subtracting from the dividend is 992, and the first 4 digits of our dividend must be no more than 99 greater than 992. this tells us the 2nd left-most digit of the dividend is 0. also 7*124 = 868 which is more than 100 less than 1000, so the first digit of the quotient is 8. so the quotient is 80809.

the first digit of the first number we subtract must be 9, so we have 8*quotient > 900, so the divisor is between 113 and 124 (inclusive).

the first digit of the first remainder must be 1 (by a similar analysis of the first digit of the original dividend) and the second digit must likewise be 0.

this means that 10+(8*quotient) > 1000, so 8*quotient > 990. this means the quotient must be 124, since 8*123 = 984.

so the last subtraction is 124*9 = 1116, so the last 2 digits of the first 4-digit remainder are 03 (992+11 = 1003), and the first 4 digit of the dividend are 1002, giving:

10020316/124 = 80809

- December 12th 2012, 08:14 PM #3

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