Seven dierent playing cards, with values from ace to seven, are shued and
Seven dierent playing cards, with values from ace to seven, are shued and
placed in a row on a table to form a seven-digit number. What is the probability that this
seven-digit number is divisible by 11?
Note: Each of the possible seven-digit numbers is equally likely to occur.
Answer/Solution
4/35
Steps/Work
To be divisible by 11, the digits must be arranged so that the difference between
the sum of one set of alternate digits, and the sum of the other set of alternate digits, is either
0 or a multiple of 11. The sum of all seven digits is 28. It is easy to nd that 28 can be
partitioned in only two ways that meet the 11 test: 14j14, and 25j3. The 25j3 is ruled out
because no sum of three dierent digits can be as low as 3. Therefore, only the 14j14 partition
need to be considered.
There are 35 dierent combinations of three digits that can fall into the B positions in the
number ABABABA. Of those 35, only four sum to 14: 167, 257, 347, and 356. Therefore, the
probability that the number will be divisible by 11 is 4/35.
correct answer A
the sum of one set of alternate digits, and the sum of the other set of alternate digits, is either
0 or a multiple of 11. The sum of all seven digits is 28. It is easy to nd that 28 can be
partitioned in only two ways that meet the 11 test: 14j14, and 25j3. The 25j3 is ruled out
because no sum of three dierent digits can be as low as 3. Therefore, only the 14j14 partition
need to be considered.
There are 35 dierent combinations of three digits that can fall into the B positions in the
number ABABABA. Of those 35, only four sum to 14: 167, 257, 347, and 356. Therefore, the
probability that the number will be divisible by 11 is 4/35.
correct answer A