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How many positive integers n have the property that the remainder of dividing 2003 by n is equal to 23?

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posted May 15 by Pushpak Chauhan

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1 Solution

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(2003 / 30)
(2003 / 33)
(2003 / 36)
(2003 / 44)
(2003 / 45)
(2003 / 55)
(2003 / 60)
(2003 / 66)
(2003 / 90)
(2003 / 99)
(2003 / 110)
(2003 / 132)
(2003 / 165)
(2003 / 180)
(2003 / 198)
(2003 / 220)
(2003 / 330)
(2003 / 396)
(2003 / 495)
(2003 / 660)
(2003 / 990)
(2003 / 1980)
So I could find 22 of such numbers giving a remainder of 23 with the number 2003.

solution May 16 by Tejas Naik



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