Caden’s class wants to win a prize for the closest estimate to the number of grains of rice in a \(1\) kg bag of rice.
They have been given three clues:
Clue (1): The number of grains is a \(5\)-digit even number.
Clue (2): The sum of the five digits in the number is \(15\).
Clue (3): The number, rounded to the nearest hundred, is \(51\,200\).
Help Caden’s class win the prize. Use the given clues to determine all possible estimates of the number of grains of rice in the bag.
Kendala, one of Caden’s classmates, found that the mass of a single grain of rice depends on the type, ranging from about \(20\) to \(45\) milligrams. When she used this to calculate how many grains were in the \(1\) kg bag, did her answer suggest the rice was of the lighter or heavier type?
From Clues (1) and (3), we know that the number is an even number between \(51\,150\) and \(51\,249\).
Therefore, the number is of the form \(51\,1XY\) or \(51\,2XY\), where \(X\) and \(Y\) are digits.
Suppose the number is of the form \(51\,1XY\). We know that \(Y\) is even, and \(X\geq 5\). Since \(5+1+1 = 7\), then from Clue (2), \(X+Y = 8\). Therefore, we could have \(X=6\) and \(Y=2\), or \(X=8\) and \(Y=0\).
Suppose the number is of the form \(51\,2XY\). We know that \(Y\) is even and \(X\leq 4\). Since \(5+1+2 = 8\), then from Clue (2), \(X+Y = 7\). Since \(Y\) is an even digit, we could have \(X=3\) and \(Y=4\), or \(X=1\) and \(Y=6\).
Therefore, the estimates that satisfy all three clues are \(51\,162\), \(51\,180\), \(51\,234\), and \(51\,216\).
Since \(1\text{ kg}=1000\text{ g}=1\,000\,000\text{ mg}\), for the lighter rice, Kendala would calculate \(1\,000\,000\div 20=50\,000\) grains of rice. For the heavier rice, she would calculate \(1\,000\,000\div 45\approx 22\,222\) grains of rice. Thus, her answers indicate that the class’s rice was of the lighter type, since that answer is closest to their estimate.