# SOLUTION: When 100 is divided by some positive integer x, the remainder is 10. What is the sum of the smallest and largest values of x?

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 Click here to see ALL problems on Divisibility and Prime Numbers Question 262678: When 100 is divided by some positive integer x, the remainder is 10. What is the sum of the smallest and largest values of x?Answer by Theo(3556)   (Show Source): You can put this solution on YOUR website!x is the divisor y is the result r is the remainder if the remainder is always 10, then the number of 100 - 10 = 90 must be divisible without a remainder. our equation becomes: 90/x = y with no remainder. multiply both sides of this equation by x to get: 90 = x*y find all the factors of x and you will be able to find the anser. 90 equals: 1*90 2*45 3*30 6*15 9*10 now take these factors and divide them into 100. 100/1 = 100 with no remainder 100/90 = 1 with a remainder of 10 100/2 = 50 with no remainder 100/45 = 2 with a remainder of 10 100/3 = 33 with a remainder of 1 100/30 = 3 with a remainder of 10 100/6 = 16 with a remainder of 4 100/15 = 6 with a remainder of 10 100/9 = 11 with a remainder of 1 100/10 = 10 with a remainder of 0. narrow it down to all those with a remainder of 10 and you get: 100/90 100/45 100/30 100/15 the divisor had to be greater than 10. the smallest divisor is 15. the largest divisor is 90. your answer is: the sum of the smallest and the largest divisor is 90 + 15 = 105.