There are five children under 10 years old. Their ages are listed as whole numbers, such as 0, 1, 2, etcetera. None have a decimal age such as 2.5; their ages increase by 1 on their birthday. The mean, median, mode and range of their ages are all equal. One of the children has a birthday and becomes a year older. The median and mode of the group remain equal, but both change value. What are the children’s ages before and after the birthday?
Definition reminders:
Mean = the sum of the numbers in the set divided by the total amount of numbers
Median = the middle number when the set is ordered from smallest to biggest
Mode = the most frequently occurring number in the set. (Note: if there are multiple most frequently occurring numbers in a dataset, the mode is not unique. All modes in this puzzle are unique)
Range = the difference between the highest and lowest numbers in the set
The children’s ages must be {2, 5, 5, 6, 7} before the birthday and {2, 5, 6, 6, 7} after.
Call the ages, in ascending order, a ≤ b ≤ c≤d ≤e. The median is the middle number, c. Because the mode equals the median and the mode is unique, the value c must appear at least twice. This means that at least b or d must equal c (remember: the letters are in ascending numerical order). We’ll show that d cannot equal c and therefore conclude that b must be equal to it.
The only way for the old unique mode to become a new unique mode by adding 1 to a single number is to change a number from the current mode, c, to a new mode of c + 1. If d = c, this transformation looks like {a, b, c, c, c + 1} → {a, b, c, c + 1, c + 1} regardless of whether b also equals c. But this doesn’t change the median. Instead we must have b = c, so the transformation looks like {a, c, c, c + 1, e} → {a, c, c + 1, c + 1, e}.
The mean and range both equal c, giving the system of equations:
(a + c + c + (c+1) + e) / 5 = c (mean)
e – a = c (range)
Algebraically combining these equations yields 2a + 1 = c. From this, we can find that e = c + a = 3a + 1. This allows us to write all the ages in terms of a:
{a, 2a + 1, 2a + 1, 2a + 2, 3a + 1}
Plugging in small values for a, we see that a = 2 is the only one that works:
a = 0: {0, 1, 1, 2, 1}is invalid because the last age in the set must be the largest
a = 1: {1, 3, 3, 4, 4} does not meet the rule that the mode must be unique
a = 2: {2, 5, 5, 6, 7} works; it is the correct solution
a = 3: {3, 7, 7, 8, 10} breaks the rule that the children are under 10 (making a larger only makes this problem worse)
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