DAY 442
THE FOUR LETTERS
Credit: GAMES Magazine
Nov/Dec 1981
Martin Gardner
A secretary types four letters to four people and addresses the four envelopes. If she inserts the letters at random, each in a different envelope, what is the probability that exactly three letters will go in the right envelopes?
Answer:
The probability is zero. If three letters match the envelopes, so will the fourth.
Thursday, 17 March 2016
Wednesday, 16 March 2016
DAY 441
CAN YOU FIGURE IT?
Credit: GAMES Magazine
Sept/Oct 1981
Michael Ecker
Quickly - Is 15% of 20 the same as 20% of 15?
Answer:
Yes.......the answer to both is 3.
Now, answer this question:
Which is larger?
94.1% of 23.25 or
23.25% of 94.1?
Answer:
They are equal - it is the same type of problem as the first example.
A% of B will always equal B% of A, since the former equals A/100 times B (or AB/100) and the latter is BA/100.
CAN YOU FIGURE IT?
Credit: GAMES Magazine
Sept/Oct 1981
Michael Ecker
Quickly - Is 15% of 20 the same as 20% of 15?
Answer:
Yes.......the answer to both is 3.
Now, answer this question:
Which is larger?
94.1% of 23.25 or
23.25% of 94.1?
Answer:
They are equal - it is the same type of problem as the first example.
A% of B will always equal B% of A, since the former equals A/100 times B (or AB/100) and the latter is BA/100.
DAY 440
TIME FLIES
Credit: GAMES Magazine
Sept/Oct 1981
D. P.
Two days ago I was 28, but next year I'll be 31. When's by birthday?
Answer:
December 31.......... He is speaking on January 1. Two days ago (Dec. 30) he was 28. Yesterday (Dec. 31 - his birthday) he was 29. Today (Jan. 1) he is (still) 29, but at the end of this year he will turn 30. So next year he will be 31.
TIME FLIES
Credit: GAMES Magazine
Sept/Oct 1981
D. P.
Two days ago I was 28, but next year I'll be 31. When's by birthday?
Answer:
December 31.......... He is speaking on January 1. Two days ago (Dec. 30) he was 28. Yesterday (Dec. 31 - his birthday) he was 29. Today (Jan. 1) he is (still) 29, but at the end of this year he will turn 30. So next year he will be 31.
DAY 439
TWO IN A MILLION
Credit: GAMES Magazine
Michael Ecker
There is only one pair of whole numbers that satisfies both these conditions: their product is one million, and no digit of either number is a zero. What are the two numbers?
Answer:
64 and 15,625
One million = 10 to the sixth power
10 to the sixth = 2 to the sixth x 5 to the sixth
2 to the sixth = 64
5 to the sixth = 15,625
64 x 15,625 = 1,000,000
TWO IN A MILLION
Credit: GAMES Magazine
Michael Ecker
There is only one pair of whole numbers that satisfies both these conditions: their product is one million, and no digit of either number is a zero. What are the two numbers?
Answer:
64 and 15,625
One million = 10 to the sixth power
10 to the sixth = 2 to the sixth x 5 to the sixth
2 to the sixth = 64
5 to the sixth = 15,625
64 x 15,625 = 1,000,000
Wednesday, 2 March 2016
DAY 438
EASY AS 1, 2, 3.....
Credit: GAMES Magazine
May, 2001
Martin Gardner
There are just two ways to insert four plus or minus signs between the digits 123456789 to yield a sum of 1:
1 + 23 - 45 - 67 + 89 = 1
1 - 23 + 45 + 67 - 89 = 1
Curiously, the solutions are the same except that the signs are reversed.
Now see if you can insert four plus or minus signs in the sequence 987654321 to make a sum of 1. There is only one solution.
Answer:
98 - 76 - 54 + 32 + 1 = 1
EASY AS 1, 2, 3.....
Credit: GAMES Magazine
May, 2001
Martin Gardner
There are just two ways to insert four plus or minus signs between the digits 123456789 to yield a sum of 1:
1 + 23 - 45 - 67 + 89 = 1
1 - 23 + 45 + 67 - 89 = 1
Curiously, the solutions are the same except that the signs are reversed.
Now see if you can insert four plus or minus signs in the sequence 987654321 to make a sum of 1. There is only one solution.
Answer:
98 - 76 - 54 + 32 + 1 = 1
Subscribe to:
Posts (Atom)