Conflicting Words

What unique feature do the following words share?

FRIEND, FEAST, THERE, THOROUGH, FLIGHT, WONDERFUL, RESIGN, ENDURING, PEST, COVERT

Each of these words contains its antonym as a sub-word:

FRIEND – FIEND, FEAST – FAST, THERE – HERE, THOROUGH – ROUGH, FLIGHT – FIGHT, WONDERFUL – WOEFUL, RESIGN – REIGN, ENDURING – ENDING, PEST – PET, COVERT – OVERT

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.