## 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?
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.
Source:

## 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.

## Circular Racetrack

Suppose you are on a one-way circular racetrack. There are 100 gas cans randomly placed on different locations of the track and the total amount of gas in the cans is enough for your car to complete an entire circle. Assume that your car has no gas initially, but you can place it at any location on the track, then pick up the gas cans along the way and use them to fill the tank. Show that you can always choose a starting position so that you can complete an entire circle.

Imagine you put your car at any location, but instead with an empty tank, you start with enough gas to complete the circle. Then, simply track the amount of gas you have and locate the point on the track where it is the lowest. If you choose that location as a starting point, you will be able to complete the track.

## Send More Money

Replace every letter with a different digit between 0 and 9, such that you get a correct calculation.

The answer is 9567 + 1085 = 10652.

You can first see that M = 1. Then S = 8 or 9 and O = 0. MORE is at most 1098, and if S = 8, then E = 9, N is either 9 or 0, but this is impossible. Therefore S = 9, N = E + 1. You can check easily the 6 possibilities for (N, E) and then find the answer.

FEATURED

## Sum Up to 15

Tango and Cash are playing the following game: Each of them chooses a number between 1 and 9 without replacement. The first one to get 3 numbers that sum up to 15 wins. Does any of them have a winning strategy?

Place the numbers from 1 to 9 in a 3×3 grid so that they form a magic square. Now the game comes down to a standard TIC-TAC-TOE, and it is well-known that it always leads to a draw when both players use optimal strategies.

FEATURED

## David Copperfield

David Copperfield and his assistant perform the following magic trick. The assistant offers a person from the audience to pick 5 arbitrary cards from a regular deck and then hands them back to him. After the assistant sees the cards, he returns one of them to the audience member and gives the rest one by one to David Copperfield. After the magician receives the fourth card, he correctly guesses what card the audience member holds in his hand. How did they perform the trick?

Out of the five cards, there will be (at least) two of the same suit; assume they are clubs. Now imagine all clubs are arranged in a circle in a cyclic manner – A, 2, 3, … J, Q, K (clock-wise), and locate the two chosen ones on it. There are two arks on the circle which are connecting them and exactly one of them will contain X cards, with X between 0 and 5. Now the assistant will pass to David Copperfield first the clubs card which is located on the left end of this ark, will return to the audience member the clubs card which is located on the right end of it and, with the remaining three cards, will encode the number X. In order to do this, he will arrange the three extra cards in increasing order – first clubs A-K, then diamonds A-K, then hearts A-K and finally spades A-K. Let us call the smallest card in this order “1”, the middle one “2” and the largest one “3”. Now, depending on the value of X, the assistant will pass the cards “1”, “2” and “3” in the following order:

X=0 ⇾ 1, 2, 3
X=1 ⇾ 1, 3, 2
X=2 ⇾ 2, 1, 3
X=3 ⇾ 2, 3, 1
X=4 ⇾ 3, 1, 2
X=5 ⇾ 3, 2, 1

In this way David Copperfield will know the suit of the audience member’s card and also with what number he should increase the card he received first in order to get value as well. Therefore, he will be able to guess correctly.

FEATURED

## Population

In certain society all parents stop having children right after they get their first boy. After 1000 years, approximately what will be the percentage of the women in the society?

The answer is 50%. The reason is that no matter what birth plan the society comes up with, the chance for having a boy or a girl during every birth is 50/50.

FEATURED

## Shark Attack

A man stands in the center of a circular field which is encompassed by a narrow ring of water. In the water there is a shark which is swimming four times as fast as the man is running. Can the man escape the field and get past the water to safety?

Yes, he can. Let the radius of the field is R and its center I. First the man should start running along a circle with center I and radius slightly less than R/4. His angular speed will be larger than the angular speed of the shark, so he can keep running until gets opposite to it with respect to I. Then, the man should dash away (in a straight line) towards the water. Since he will need to cover approximately 3R/4 distance and the shark will have to cover approximately 3.14R distance, the man will have enough time to escape.