Randomly scramble the digits 0, 1, 2, 3, 4, 5 and 6 to form a seven-digit number (or a six-digit number if the leading digit is 0). Rank the following events in order of likelihood—ties are possible.
a. The number is divisible by 2
b. The number is divisible by 3
c. The number is divisible by 4
d. The number is divisible by 5
e. The number is divisible by 6
f. The number is divisible by 9
g. The number is divisible by 10
h. The number is divisible by 12
i. The number is divisible by 100
j. The number is divisible by 150
There are quick tricks for determining divisibility by certain numbers. If you haven’t learned these before, feel free to look some up.
The answer is b, a/e (tie), c/d/h (tie), g, j, f/i (tie). Let’s go through each:
b. The number is divisible by 3. Probability: 100 percent.
A number is divisible by 3 if the sum of its digits is divisible by 3. Regardless of their order, the sum of 0 + 1 + 2 + 3 + 4 + 5 + 6 = 21, which is divisible by 3.
a/e (tie). The number is divisible by 2/The number is divisible by 6. Probability: 4/7.
A number is divisible by 2 if its last digit is even. Of the seven digits 0, 1, 2, 3, 4, 5 and 6, four of them are even.
A number is divisible by 6 if it is divisible by both 2 and 3. Because all of the outcomes are divisible by 3, the likelihood that the number is divisible by 6 is the same as the likelihood that it is divisible by 2.
c/d/h (three-way tie). The number is divisible by 4/The number is divisible by 5/The number is divisible by 12. Probability: 2/7.
A number is divisible by 4 if its last two digits form a number divisible by 4. For the numbers in our set, this happens when the last two digits are any of the following 12 combinations: 04, 12, 16, 20, 24, 32, 36, 40, 52, 56, 60 or 64. There are 42 total possible outcomes for the last two digits (seven possible outcomes for the last digit times six for the next-to-last, or 7 × 6 = 42), yielding 12/42 = 2/7.
A number is divisible by 5 if it ends with a 0 or a 5. Two out of the seven possible final digits result in a number divisible by 5, yielding 2/7
A number is divisible by 12 if it is divisible by 3 and 4. All outcomes are divisible by 3, so this has the same likelihood as divisibility by 4.
g. The number is divisible by 10. Probability: 1/7
A number is divisible by 10 if it ends in a 0. There is only one favorable ending digit out of the seven possible.
j. The number is divisible by 150. Probability: 1/42
A number is divisible by 150 if it is divisible by 3 and 50. All of the outcomes are divisible by 3. The number will be divisible by 50 only if it ends in 50. (It’s impossible for it to end in 00 because we only have one 0.) This gives one favorable two-digit ending out of 42 possible two-digit endings.
f/i (tie). The number is divisible by 9/The number is divisible by 100. Probability: 0 percent.
A number is divisible by 9 only if the sum of its digits is divisible by 9. The numbers 0 through 6 sum to 21, which is not divisible by 9.
A number is divisible by 100 only if it ends in 00, which is impossible because we only have one 0.
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