Date of the Problem
October 26, 2020

As November (the 11th month) gets underway, it’s the perfect time to focus on 11. Eleven is the fourth prime number, and there is a fun divisibility rule for 11. For any integer, insert alternating “–” and “+” signs between the consecutive pairs of digits, starting with a “–” sign between the left-most pair of digits. For example, for the number 91,828 we would have 9 – 1 + 8 – 2 + 8. (Notice that the first minus went between the left-most pair of numbers, 9 and 1, and then we alternated with “+” and “–” signs.)  Now, simplify the expression. For our example, we have 9 – 1 + 8 – 2 + 8 = 22. Since this value, 22, is divisible by 11, the original number is divisible by 11. Using this rule, if the five-digit integer 76,9a2 is a multiple of 11, what is the value of a?

We can set up the expression 7 – 6 + 9 – a + 2 and simplify it to 12 – a. We now know that in order for 76,9a2 to be a multiple of 11, the expression 12 – a must be a multiple of 11. The multiples of 11 are 0, 11, 22, 33, etc. Since a must be a single digit, the only possibility is 12 – a = 11, and therefore, a = 1. We can check with a calculator to see that 76,912 is in fact divisible by 11. 

When playing many games, players must roll a pair of dice and find the sum of the two numbers rolled. With two dice, there are 11 possible sums ranging from 2 through 12. What is the probability that a player will roll a sum of 11 on his first roll of two dice? Express your answer as a common fraction.

Though there are 11 possible sums, it is much "harder" to roll a sum of 12 than a sum of 6. The 11 possible sums are not all equally likely. We could list out the different outcomes of a roll of two dice. (It might be easier to think of this as having a green die and a red die, so that we can clearly see that a roll of green-1 and red-3 is different than green-3 and red-1.) If we list out all of the outcomes, we see that there are 36 different possible roll outcomes resulting in sums from 2 to 12. (Alternatively, there are 6 possible rolls on each die, so rolling two dice could result in 6 × 6 = 36 different outcomes.) Of these 36 outcomes, only two of them result in a sum of 11. They are green-5 and red-6 and green-6 and red-5. Therefore, the probability of rolling a sum of 11 with two dice is 2/36 = 1/18.

Eleven is a palindrome. A palindrome is an integer that reads the same backward and forward. The integer 12,321 is a palindrome since writing the numbers in reverse order is also 12,321. What is the greatest possible four-digit palindrome that is a multiple of 6?

The greatest four-digit palindrome is 9999. We need the four-digit palindrome to be a multiple of 6, which means that the units digit (and therefore the thousands digit) must be even. To make the number as large as possible, we need to start with 8_ _8. To be a multiple of 6, the number must be even (a multiple of 2), which we already have, and the number must also be a multiple of 3. A multiple of three has a digit-sum that is divisible by 3. Obviously, we'd like to have 8998, since that's the largest number we can now make. The digit-sum is 8 + 9 + 9 + 8 = 34, which is not divisible by 3, so 8998 is not our answer. We can test 8888, but this also does not work since the digit-sum is 32. Notice that our digit-sum is decreasing by 2 when we decrease each of the missing digits by 1. When we get to 8778, we see that the digit-sum is 30, which is divisible by 3. Therefore, 8778 is the largest four-digit palindrome divisible by 6.

Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.

Page 2 contains ONLY PROBLEMS. ♦

CCSS (Common Core State Standard)
PDF Download