Directions: Type in your answer in the input box provided below the question. Five positive integers have their mean, median and mode each equal to 10. If their range is equal to 15, how many possible values can the least of the five integers assume?

Let the integers be a, b, c, d, e in increasing order. Median = c = 10. Mean = (a+b+10+d+e)/5 = 10, so a+b+d+e = 40. Range = e - a = 15, so e = a + 15. Mode = 10, so at least two numbers must be 10. Since median is 10, either b=10 or d=10. If b=10, then a+10+10+d+e = 50, so a+d+e = 30. Since e=a+15, a+d+a+15 = 30, 2a+d = 15. Possible values for a: if a=1, d=13 (e=16, set: 1, 10, 10, 13, 16); if a=2, d=11 (e=17, set: 2, 10, 10, 11, 17); if a=3, d=9 (invalid, d must be >= 10); if a=4, d=7 (invalid). If d=10, then a+b+10+10+e = 50, a+b+e = 30. e=a+15, so 2a+b+15 = 30, 2a+b = 15. If a=1, b=13 (invalid, b must be <= 10); if a=2, b=11 (invalid); if a=3, b=9 (set: 3, 9, 10, 10, 18); if a=4, b=7 (set: 4, 7, 10, 10, 19); if a=5, b=5 (set: 5, 5, 10, 10, 20). Total 4 valid sets.