Take 9 from 6, 10 from 9, 50 from 40 and leave 6.
How Come ??
SIX - 9 (IX) = S
9 (IX) - 10 (X) = I
40 (XL) - 50 (L) = X
Henry has been caught stealing cattle, and is brought into town for justice. The judge is his ex-wife Gretchen, who wants to show him some sympathy, but the law clearly calls for two shots to be taken at Henry from close range. To make things a little better for Henry, Gretchen tells him she will place two bullets into a six-chambered revolver in successive order. She will spin the chamber, close it, and take one shot. If Henry is still alive, she will then either take another shot, or spin the chamber again before shooting.
Henry is a bit incredulous that his own ex-wife would carry out the punishment, and a bit sad that she was always such a rule follower. He steels himself as Gretchen loads the chambers, spins the revolver, and pulls the trigger. Whew! It was blank. Then Gretchen asks, 'Do you want me to pull the trigger again, or should I spin the chamber a second time before pulling the trigger?'
What should Henry choose?
Henry should have Gretchen pull the trigger again without spinning.
We know that the first chamber Gretchen fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Henry has Gretchen pull the trigger again, the probability that a bullet will be fired is 1/4.
If Gretchen spins the chamber again, the probability that she shoots Henry would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position.
A number with an interesting property:
When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 2.
When I divide it by 4, the remainder is 3.
When I divide it by 5, the remainder is 4.
When I divide it by 6, the remainder is 5.
When I divide it by 7, the remainder is 6.
When I divide it by 8, the remainder is 7.
When I divide it by 9, the remainder is 8.
When I divide it by 10, the remainder is 9.
It's not a small number, but it's not really big, either.
When I looked for a smaller number with this property I couldn't find one.
Can you find it?
The number has to end in 9.
Looked brute force for small numbers.
59 and 119 were promising, but no cigar.
Then looked for agreement among
39 + multiples of 40,
69 + multiples of 70 and
89 + multiples of 90
Smallest one was 2519.
Many people would think Friday the 13th will be an unlucky day. Is it possible that there is no Friday on 13th through the whole year? How many Fridays at 13th can we have in a year at most? Can you calculate it out?
We can calculate out how many days there will be for the 13th on each month if we count from the beginning of the year (January 1). Then we divide total days by 7 to get the remainders. We also need to consider the leap year. Through the whole year we had all kinds of remainders, from 0 to 6. The minimum of occurence for all the unique remainders was 1. It means that we have at least one Friday on 13th. In a regular year, the best chance you can get 3 Fridays on 13th, which are in February, March and December because the remainders of these 3 months are 2. In a leap year, the best chance you also can get 3 Fridays on 13th, which are in January, April and July because the remainders of these 3 months are 6.
Sonal asked the class to see if they could find the sum of the first 50 odd numbers. As everyone settled down to their addition, Lavesh ran to her and said, 'The sum is 2,500.' Ms. Sonal thought, 'Lucky guess,' and gave him the task of finding the sum of the first 75 odd numbers. Within 20 seconds, Lavesh was back with the correct answer
How does Lavesh find the sum so quickly and what is the answer ??
The following pattern holds: The sum is equal to n x n, when n is the number of consecutive odd numbers, starting with 1. For example, the sum of the first 3 odd numbers is equal to 3 x 3, or 9; the sum of the first 4 odd numbers is equal to 4 x 4, or 16; the sum of the first 5 odd numbers is equal to 5 x 5, or 25; and so on.
Try to find out a multi-digit number that if multiplied by the number 9 or any of its multiplications products (18, 27, 36, 45,..) will result in the multiplication factor repeated (n) number of times
The number is: 12345679
12345679 * 9 = 111111111
12345679 * 18 = 222222222
12345679 * 27 = 333333333
And so on...