Sum Up to 999

Can you find a triple of three-digit numbers that sum up to 999 and collectively contain all digits from 1 to 9 exactly once? How many such triples are there? What if the sum was 1000?

There are exactly 180 such triples that sum up to 999 and none that sum up to 1000.

In order to see that, notice that the sum of the first digits of the numbers can be no more than 9. Therefore, the sum of the middle and the sum of the last digits cannot be less than 10. We then see that the sum of the last digits should be exactly 19 and the sum of the middle digits should be exactly 18. The sum of the first digits should be 45-19-18=8.

There are 2 ways to get 8 using unique digits from 1 to 9: 1+2+5 and 1+3+4.

  • If the first digits are {1, 2, 5}, the options for the middle digits are {3, 6, 9}, {3, 7, 8}, and {4, 6, 8}. The last digits end up {4, 7, 8}, {4, 6, 9}, and {3, 7, 9} respectively.
  • If the first digits are {1, 3, 4}, the options for the middle digits are {2, 7, 8} and {5, 6, 7}. The last digits end up {5, 6, 9} and {2, 8, 9} respectively.

Since the set of the first digits, the set of the middle digits, and the set of the last digits of the numbers can be permuted in 6 ways each, we get a total of 5×6×6×6=1080 solutions, or 180 up to permutation of the 3 three-digit numbers.

In order to see that we cannot get a sum of 1000, we note that since the sum of the digits from 1 to 9 is divisible by 9, then the sum of the 3 three-digit numbers should be divisible by 9 as well. Since 1000 is not divisible by 9, the statement follows.

FEATURED

A Broken Circle

There are N points on a circle. If we draw all the chords connecting these points and no three of them intersect at the same point, in how many parts will the interior of the circle get broken?

For example, when N is equal to 1, 2, 3, 4, and 5, we get 1, 2, 4, 8, and 16 parts respectively.

The answer, somewhat surprisingly, is not 2ᴺ⁻¹, but 1 + N(N-1)/2 + N(N-1)(N-2)(N-3)/24.

In order to see that, we start with a single sector, the interior of the circle, and keep successively drawing chords. Every time we draw a new chord, we increase the number of parts by 1 and then add 1 extra part for each intersection with previously drawn chords.

Therefore, the total number of parts at the end will be:

1 + the number of the chords + the number of the intersections of the chords

Each chord is determined by its 2 endpoints and therefore the number of chords is N(N-1)/2.

Each intersection is determined by the 4 endpoints of the two intersecting chords and therefore the number of intersections is N(N-1)(N-2)(N-3)/4!.

The Die Game

You pick a number between 1 and 6 and keep throwing a die until you get it. Does it matter which number you pick for maximizing the total sum of the numbers in the resulting sequence?

In the example below, the picked number is 6 and the total sum of the numbers in the resulting sequence is 35.

No matter what number you pick, the expected value of each throw is the average of the numbers from 1 to 6 which is 3.5. The choice of the number also does not affect the odds for the number of throws until the game ends, which is 6. Therefore, the total sum is always 3.5 × 6 = 21 on average, regardless of the chosen number.

Splitting Coins

You split 1000 coins into two piles and count the number of coins in each pile. If there are X coins in pile one and Y coins in pile two, you multiple the two numbers to get XY. Then you split both piles further, repeating the same counting and multiplication process, and adding the new multiplication results to the first one. The process ends when you end up with 1000 single-coin piles. Prove that you will always get the same final result, no matter how the piles have been divided during the splitting process.

For example, if you start with 5 coins and split them into a pile of 2 and a pile of 3, you get the number 2×3=6. Then, if you split the pile of 3 into a pile of 1 and a pile of 2, you will add 1×2=2 more to the 6 and get 8. Finally, if you split the two piles of 2 into single-coin piles, you will end up with 8+1+1=10.

Consider the sum of the squares of the numbers of coins in each pile, plus twice the sum of the products. On each step, if you split a pile of X+Y coins into a pile of X coins and a pile of Y coins, the sum of the squares will get reduced by 2XY, exactly the amount the sum of the products will increase by. Therefore, that number remains constant throughout the entire process and ends up exactly (1000²-1000)/2=499500.

One Hundred Rooms

There are 100 rooms in a row in a building and inside each room there is a lamp that is turned off. One person enters each room and switches the lamp inside. Then, a second person enters every second room (2, 4, 6, etc.) and switches the lamp inside. A third person switches the lamp in every third room and so on and so far, until person #100 switches the lamp in room 100. How many lamps are turned on at the end?

We can see that the only switches that have been switched an odd number of times are the ones in rooms with perfect square numbers.

Indeed, if person N has switched the switch in room M, then person M/N has done that as well. Since person N and M/N coincide only when M=N², the claim above follows.

We conclude that the number of lamps that are turned at the end is equal to the number of perfect squares less than or equal to 100; that is exactly 10 rooms.

Annihilating Matrix

The numbers 1, 2, … , 100 are arranged in a 10×10 table in increasing order, row by row and column by column, as shown below. The signs of 50 of these numbers are flipped, such that each row and each column have exactly 5 positive and 5 negative numbers. Prove that the sum of all numbers in the resulting table is equal to 0.

Represent the initial table as the sum of the following two tables:

Since the sum of the numbers in each row of the first table is equal to 0 and the sum of the numbers in each column of the second table is equal to 0, it follows that the sum of all numbers in both tables is equal to 0 as well.

Source:

Quantum Magazine, November-December 1991

Math Homework

A student was given math homework consisting of the following three problems. What is so special about this homework?

  1. What would the value of 190 in hexadecimal be?
  2. Twenty-nine is a prime example of what kind of number?
  3. At time t = 0, water begins pouring into an empty tank so that the volume of water is changing at a rate V’(t)=sec²(t). For time t = k, where 0 < k < π/2, determine the amount of water in the tank.

The answer to each of the problems in the homework is hidden within the question itself:

  1. What would the value of 190 in hexadecimal be?
    The answer is be.
  2. Twenty-nine is a prime example of what kind of number?
    The answer is prime.
  3. At time t = 0, water begins pouring into an empty tank so that the volume of water is changing at a rate V’(t)=sec²(t). For time t = k, where 0 < k < π / 2, determine the amount of water in the tank.
    The answer is tank.
Source:

Math Horizons, MAA

A Very Cool Number

Find a number containing every digit 0-9 exactly once, such that for every 1≤N≤10, the leftmost N digits comprise a number, divisible by N.

Let the number be ABCDEFGHIJ.

  • The digit J must be 0, so that the number is divisible by 10.
  • The digit E must be 5, so that the number comprised of the first 5 digits is divisible by 5.
  • The digits B, D, F, H, J must be even, and therefore the digits A, C, E, G, I must be odd.
  • The numbers CD and GH must be divisible by 4 and the number FGH must be divisible by 8. Since C and G are odd, D and H must be 2 and 6, or vice-versa.
  • Since ABC, ABCDEF, and ABCDEFGHI are all divisible by 3, we see that A+B+C, D+E+F, and G+H+I are divisible by 3 as well. Therefore, D+F+5 is divisible by 3 and since D and F are even, we have either D=2, H=6, F=8, B=4 or D=6, H=2, F=4, B=8.
    • D=2, H=6, F=8, B=4. Since FGH is divisible by 8 and G is odd, we have G=1 or G=9. Also, G+I+6 is divisible by 3 and since I is odd, we have G=1, I=3. The only possibilities for ABCDEFG are 7492581630 and 9472581630, but they are not divisible by 7.
    • D=6, H=2, F=4, B=8. Since FGH is divisible by 8 and G is odd, we have G=3 or G=7. Also, G+I+2 is divisible by 3 and since I is odd, we have G=3, I=1 or G=3, I=7 or G=7, I=3 or G=7, I=9. The only possibilities for ABCDEFG are 9876543, 7896543, 1896543, 9816543, 1896547, 9816547, 1836547, 3816547. Out of these, only 3816547 is divisible by 7.

We conclude that the only solution is 3816547290.

More Sisters on Average

Who have more sisters on average in a society: boys, girls, or is it equal?

Remark: Assume that each child is born a boy or a girl with equal probability, independent of its siblings.

The average number of sisters is roughly the same for both boys and girls. To see this, notice that every girl in every family contributes “one sister” to each of its siblings who are either a boy or a girl with equal probability. Therefore, every girl contributes on average the same number of sisters to the group of boys and to the group of girls. Since there is roughly the same number of boys and girls in the society, the average number of sisters for boys and girls is the same.