Is that Divisible?

Question text

A number 32a5b¯¯¯¯¯¯¯¯¯¯¯¯

is divisible by 11. Its last two digits form a number 5b¯¯¯¯¯ that can be exactly divided by 3. What is the largest 5

-digit number with this with these properties?

Your Answer:

Solution

Since 5b¯¯¯¯¯

is divisible by 3, b should be either 1, 4 or 7. If 32a5b¯¯¯¯¯¯¯¯¯¯¯¯ is to be divisible by 11, it must be the case that

3−2+a−5+b=a+b−4

is a multiple of 11. If b=1, then a=3; if b=4 then a=0; and if b=7, then a=8. The three 4-digit numbers that satisfy these properties are then 32351, 32054 and 32857. Thus, the largest 5-digit number with these properties is 32857

.

The correct answer is: 32857

 

My Solution (That above one was crappy)

 If 5b is divisible by 3, then b must be 1,4, or 7, because only for those values will 5+b be divisible by 3. If 32a5b is divisible by 11, we must have 3-2+a-5+b, or a+b-4, must be divisible by 11. Let's say b is 7. Then a has to be 8. So, the resulting number is 32857. But what if b is 4? Then a is 0, for the answer to be 32054. If b is 1, then a is 3, 32351. The largest is 32857.

Important:

Remember that abcdef is divisible by 11 ifandonlyif a-b+c-d+e is divisible by 11.

A Symmetrical Octagon

Problem 6 – Correct! – Score: 1 / 7 (27650)

Report an ErrorProblem:

Ellie wants to color two sides of a regular octagon green, two blue, two red, and two yellow such that each pair of opposite sides of the octagon have the same color. How many different patterns can she form? (Two patterns are considered identical if one can be rotated to form the other.)

Solution:

Suppose she first chooses a side to color green. Because two patterns are considered identical if one can be rotated to form the other, it doesn't matter which side she chooses. She can then spin her octagon so that this first colored side is on "top". Then, suppose she goes clockwise around the octagon to finish coloring. She can't use green because the other green side must be opposite the first side. So, she has  choices for the first side after the green side. She then has  colors remaining for the next side, and remaining for the side after that. At this point, she has four consecutive sides that are four different colors. The other four sides are then determined: each is the color of its opposite side. So, there are  possible colorings.

Hint(s):

Grab some crayons and try it!

Your Response(s):

• :( 2520

• :( 315

• :( 3

• :( 24

• :) 6
• Analysis: There is only 4 sides that matter, since the other four sides are mirrored. There are 4! ways to color the four sides, but we overcounted because of rotations. 4!/4 = 3! = 6.

On Many Different Cases to Consider

Problem 1 Part b – Correct! – Score: 3 / 7 (1923)

Report an ErrorProblem:

We call a number special if every digit in the number either is a 1 or borders a 1. For example, 11111, 13, 141, 1441, 515151, and 101 are all special, but 10001, 222, 122, and 1333 are not special.

How many positive 3-digit numbers are special?

Solution:

A 3-digit number is special if it has a 1 in the middle, or it both starts and ends in a 1. There are  of the first type (we have 9 choices for the first digit and 10 choices for the last digit). There are also 10 of the second type (we must choose the middle digit), but we already counted 111 in the first case, so we don't count it again; we get 9 new special numbers in the second case. Therefore, there are 90 + 9 = 99 special three-digit numbers.

Your Response(s):

• :( 81

• :( 82

• :) 99

Analysis: The formats 1_1, _1_, and 111 aren't enough - the numbers can be in the form of 11_ or _11 too!

On Casework and Different Things

Solution

In topic: Distinguishability (Counting & Probability).

 

Correct! Way to go!

Matt's four cousins are coming to visit. There are four identical rooms that they can stay in. If any number of the cousins can stay in one room, how many different ways are there to put the cousins in the rooms?

Your First Answer: 14

Your Second Answer: 15

Solution:

Just counting the number of cousins staying in each room, there are the following possibilities: (4,0,0,0), (3,1,0,0), (2,2,0,0), (2,1,1,0), (1,1,1,1).

(4,0,0,0): There is only  way to put all the cousins in the same room (since the rooms are identical).

(3,1,0,0): There are  ways to choose which cousin will be in a different room than the others.

(2,2,0,0): Let us consider one of the cousins in one of the rooms. There are  ways to choose which of the other cousins will also stay in that room, and then the other two are automatically in the other room.

(2,1,1,0): There are  ways to choose which cousins stay the same room.

(1,1,1,1): There is one way for all the cousins to each stay in a different room.

The total number of possible arrangements is .

Symmetry, Reflection, and When to Divide

  Level 25 Counting & Probability  +123 XP

Review

In topic: Counting with Symmetry (Counting & Probability).

 

Incorrect. Oops!

In how many ways can  people sit around a round table if no two of the  people Pierre, Rosa, and Thomas can sit next to each other?

Your First Answer: 384

Your Second Answer: 1008

Solution:

After Pierre sits, we can place Rosa either two seats from Pierre (that is, with one seat between them) or three seats from Pierre. We tackle these two cases separately:

Case 1: Rosa is two seats from Pierre. There are  such seats. For either of these, there are then four empty seats in a row, and one empty seat between Rosa and Pierre. Thomas can sit in either of the middle two of the four empty seats in a row. So, there are  ways to seat Rosa and Thomas in this case. There are then  seats left, which the others can take in  ways. So, there are seatings in this case.

Case 2: Rosa is three seats from Pierre (that is, there are  seats between them). There are  such seats. Thomas can't sit in either of the seats directly between them, but after Rosa sits, there are  empty seats in a row still, and Thomas can only sit in the middle seat of these three. Once again, there are  empty seats remaining, and the  remaining people can sit in them in  ways. So, we have seatings in this case.

Putting our two cases together gives a total of  seatings.

Distinguishability is Annoying

  Level 25 Counting & Probability  +125 XP

Review

In topic: Distinguishability (Counting & Probability).

Source: AoPS Staff

Incorrect. Oops!

Rachel has two identical basil plants and an aloe plant. She also has two identical white lamps and two identical red lamps she can put each plant under (she can put more than one plant under a lamp, but each plant is under exactly one lamp). How many ways are there for Rachel to put her plants under her lamps?

Your First Answer: 18

Your Second Answer: 12

Solution:

We can split this up into cases.

First, consider the case when all three plants are under the same color lamp. Either all three plants are under the same lamp, both basil plants are under one lamp and the aloe plant is under the other lamp, or the aloe plant and one basil plant are under one lamp and the other basil plant is under the other lamp. This case gives us three possibilities for each color of lamp, for a total of six possibilities.

Next, consider the case where the aloe plant is under a different color of lamp than the two basil plants. Since the two lamps of the same color the aloe plant can be under are identical, it doesn't matter which one the aloe plant is under. The basil plants can either both be under the same lamp, or each be under a different lamp. This case gives us two possibilities when the aloe is under a white lamp and two possibilities when the aloe is under a red lamp, for a total of four possibilities.

Last, consider the case where the basil plants are each under a different colored lamp. The aloe plant can be under the same white lamp as a basil plant, the same red lamp as a basil plant, a different white lamp from the basil plant, or a different red lamp from the basil plant, for a total of four possibilities. In all, there are  possibilities.