Cornelius got a flat tire on a miserable, rainy day. As he was changing it on the side of the road, he placed the four lug nuts in the overturned hub cap. As our dear friend was moving back to put on the spare, he accidentally bumped the hub cap. He watched helplessly as all four lug nuts rolled into a deep sewer and were whisked away. Then our friend had a brilliant solution and within minutes was driving safely away.
As a result of temporary magical powers, you have made it to the Wimbledon finals and are playing Roger Federer for all the marbles. However, your powers cannot last the whole match. What score do you want it to be when they disappear, to maximize your chances of hanging on for a win?
It sounds obvious that you should ask to be ahead two sets to love (it takes 3 out of 5 sets to win the men’s), and in the third set, ahead 5-0 in games and 40-love in the sixth game. (Probably you want to be serving, but if your serve is like mine, you might prefer Roger to be serving the sixth game down 0-40 so that you can pray for a double fault.)
Not so fast! These solutions give you essentially 3 chances to get lucky and win, but you can get six chances—with three services by you and three by Roger. You still want to be up two sets to none, but let the game score be 6-6 in the third set and 6-0—in your favor, of course—in the tiebreaker.
A completely black dog was strolling down main street during a total blackout affecting the entire town. Not a single streetlight had been on for hours. As the dog crosses the center of the road a Buick Skylark with two broken headlights speeds towards it, but manages to swerve out of the way just in time. How could the driver see the dog to swerve in time?
Her birthday is December 31st. Today is January 1st so she was 7 two days ago, and turned 7 last year. Now she’s 8 and will turn 9 this year. And next year she’ll turn 10.
For example, if today is 01 Jan 2015:
30 Dec 2014 = age 7 (two days ago) 01 Jan 2015 = age 8 (today, she turned 8 on 31 Dec 2014, which was last year) 31 Dec 2015 = age 9 (her birthday this year) 31 Dec 2016 = age 10 (her birthday next year)
Find a six-digit number containing no zeros and no repeated digits that satisfies the following conditions:
1. The first and fourth digits sum to the last digit, as do the third and fifth digits. 2. The first and second digits when read as a two-digit number equal one quarter the fourth and fifth digits. 3. The last digit is four times the third digit.
If you call the number ABCDEF, then you get the following equations.
1. A + D = F and C + E = F 2. AB = DE / 4 3. F = 4 × C
The only numbers that work for C and E are 2 and 6 or 4 and 8, and in order to make F a single-digit number, we can deduce that C = 2, E = 6 and F = 8.
So far, our number is AB2D68.
We know A + D = 8 so A and D are both odd numbers. The only odd number less than 8 that we can use for D to make one-quarter of two-digit number D6 also be a two-digit number is 7, so D = 7 and A is 1. This makes the two-digit number AB 19.