DAY 449
GO FORTH AND MULTIPLY
Credit: GAMES/World of Puzzles
October, 2016
Raymond Love
Can you figure out what A must be and what B must be in the multiplication problem below to make it valid? The answer is unique.
AB
x A
BAA
Answer:
A = 8
B = 6
86
x 8
688
Tuesday, 1 November 2016
Monday, 26 September 2016
Saturday, 3 September 2016
Saturday, 4 June 2016
DAY 447
THREE COINS
Credit: NPR Weekend Edition Sunday Puzzle
July 3, 2005
Presented by Will Shortz
Puzzle by Eric Friedman
[Note from JA: I do not know how this problem was solved, as no explanation was given. This puzzle was used in the U.S. Puzzle Championship in 2005. There are two possible answers.]
A certain country mints coins in only three denominations. Each denomination is an integral number (1, 3, 4, etc.) The amounts 20, 23, and 29 can be made exactly by using three coins. What are the denominations of the coins?
Answers:
6, 7, 11 or
5, 9, 10
THREE COINS
Credit: NPR Weekend Edition Sunday Puzzle
July 3, 2005
Presented by Will Shortz
Puzzle by Eric Friedman
[Note from JA: I do not know how this problem was solved, as no explanation was given. This puzzle was used in the U.S. Puzzle Championship in 2005. There are two possible answers.]
A certain country mints coins in only three denominations. Each denomination is an integral number (1, 3, 4, etc.) The amounts 20, 23, and 29 can be made exactly by using three coins. What are the denominations of the coins?
Answers:
6, 7, 11 or
5, 9, 10
Friday, 29 April 2016
DAY 446
THE WOOD CHOPPING PROBLEM
Two men chopped wood all day long at exactly the same rate. The first man worked straight through without stopping to rest. At the end of the day he had a sizeable pile of logs. The second man would chop for fifty minutes and then take a ten-minute break. At the end of the day he had a much larger pile. He did not work any faster after his breaks, nor did the first man work any slower as the day wore on. How was the second man able to chop more wood?
HINT:
He did something else during his breaks other than resting.
Answer:
During each ten-minute break he sharpened his ax.
THE WOOD CHOPPING PROBLEM
Two men chopped wood all day long at exactly the same rate. The first man worked straight through without stopping to rest. At the end of the day he had a sizeable pile of logs. The second man would chop for fifty minutes and then take a ten-minute break. At the end of the day he had a much larger pile. He did not work any faster after his breaks, nor did the first man work any slower as the day wore on. How was the second man able to chop more wood?
HINT:
He did something else during his breaks other than resting.
Answer:
During each ten-minute break he sharpened his ax.
Friday, 22 April 2016
DAY 445
Credit: Nashville Banner
Red O-Donnell
In Reader's Digest
May, 1985
The minister was pleading for funds for the church. "The Lord has done so much for you," he reminded the congregation. "You should reciprocate. Each of you here today should give a tenth of your income."
One member was overcome by the sermon. "A tenth isn't enough," he shouted. "Each of us should give a twentieth!"
Credit: Nashville Banner
Red O-Donnell
In Reader's Digest
May, 1985
The minister was pleading for funds for the church. "The Lord has done so much for you," he reminded the congregation. "You should reciprocate. Each of you here today should give a tenth of your income."
One member was overcome by the sermon. "A tenth isn't enough," he shouted. "Each of us should give a twentieth!"
Monday, 4 April 2016
DAY 444
IT ALL ADDS UP
Credit: GAMES Magazine
March, 2013
Henry Dudeney
An old challenge is to get the numbers 1 through 9, in order, to form an expression that equals 100, using only basic arithmetic operations (+, -, x, div.) and parentheses. For example,
1 + 2 + 3 + 4 + 5 + 6 + 7 + (8 x 9) = 100
Can you find another solution that uses just three steps and no parentheses?
HINT:
#1: Only addition and subtraction are used.
Answer:
123 - 45 -67 + 89 = 100
IT ALL ADDS UP
Credit: GAMES Magazine
March, 2013
Henry Dudeney
An old challenge is to get the numbers 1 through 9, in order, to form an expression that equals 100, using only basic arithmetic operations (+, -, x, div.) and parentheses. For example,
1 + 2 + 3 + 4 + 5 + 6 + 7 + (8 x 9) = 100
Can you find another solution that uses just three steps and no parentheses?
HINT:
#1: Only addition and subtraction are used.
Answer:
123 - 45 -67 + 89 = 100
Sunday, 3 April 2016
Thursday, 17 March 2016
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.
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.
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
Thursday, 11 February 2016
DAY437
LEFTOVERS
Credit: GAMES Magazine
May, 1983
from Mathematical Games
by Marie Berrondo
What is the smallest number which, when divided by 2, 3, 4, 5, and 6 will give the numbers 1, 2, 3, 4, and 5 as remainders, respectively?
Answer:
59..............Let n = the unknown number. Since n divided by 2 leaves a remainder of 1, n + 1 must be divisible by 2. Since n divided by 3 leaves a remainder of 2, n + 1 must be divisible by 3. Similarly, n + 1 must be divisible by 4, 5, and 6. The smallest common multiple of 1, 2, 3, 4, 5, and 6 is 60. So n + 1 = 60, and n = 59.
LEFTOVERS
Credit: GAMES Magazine
May, 1983
from Mathematical Games
by Marie Berrondo
What is the smallest number which, when divided by 2, 3, 4, 5, and 6 will give the numbers 1, 2, 3, 4, and 5 as remainders, respectively?
Answer:
59..............Let n = the unknown number. Since n divided by 2 leaves a remainder of 1, n + 1 must be divisible by 2. Since n divided by 3 leaves a remainder of 2, n + 1 must be divisible by 3. Similarly, n + 1 must be divisible by 4, 5, and 6. The smallest common multiple of 1, 2, 3, 4, 5, and 6 is 60. So n + 1 = 60, and n = 59.
Tuesday, 9 February 2016
DAY 436
MONEYBAGS
Credit: GAMES Magzaine
February, 1983
Neal M. Cohen
You have $1,000 in one-dollar bills and ten empty sacks. How can you distribute the bills among the sacks so that you can provide any whole number of dollars from $1 to $1,000 merely by combining sacks? Each sack must contain at least one bill.
Answer:
There are several solutions. One is :
$1
$2
$4
$8
$16
$32
$64
$128
$256
$489
MONEYBAGS
Credit: GAMES Magzaine
February, 1983
Neal M. Cohen
You have $1,000 in one-dollar bills and ten empty sacks. How can you distribute the bills among the sacks so that you can provide any whole number of dollars from $1 to $1,000 merely by combining sacks? Each sack must contain at least one bill.
Answer:
There are several solutions. One is :
$1
$2
$4
$8
$16
$32
$64
$128
$256
$489
Friday, 22 January 2016
DAY 435
A LENGTHY QUIZ
Credit: GAMES Magazine
May, 1985
Doug Putnam
Put these units of measurement in order from shortest to longest.
1. Inch
2. Mile
3. Centimeter
4. Fathom
5. Rod
6. Nautical mile
7. Furlong
8. Millimeter
9. League
10. Hand
Answers:
Millimeter (4/100 inch)
Centimeter (37/100 inch)
Inch
Hand (4 inches)
Fathom ( 6 feet)
Rod (16 1/2 feet)
Furlong ( 660 feet)
Mile (5,280 feet)
Nautical mile (6,076 feet)
League (3 miles)
A LENGTHY QUIZ
Credit: GAMES Magazine
May, 1985
Doug Putnam
Put these units of measurement in order from shortest to longest.
1. Inch
2. Mile
3. Centimeter
4. Fathom
5. Rod
6. Nautical mile
7. Furlong
8. Millimeter
9. League
10. Hand
Answers:
Millimeter (4/100 inch)
Centimeter (37/100 inch)
Inch
Hand (4 inches)
Fathom ( 6 feet)
Rod (16 1/2 feet)
Furlong ( 660 feet)
Mile (5,280 feet)
Nautical mile (6,076 feet)
League (3 miles)
Tuesday, 19 January 2016
DAY 434
PUZZLES FROM THE POLE VAULT
From a Polish puzzle magazine called Sam na Sam
Published in GAMES Magazine
August, 1986
When Adam reached the finish line of a 50-kilometer race, Eve was two kilometers behind him. The next day they decided to race again. To even up the contest, this time Adam started two kilometers behind Eve, while Eve began at the starting line as usual. Assuming they cycled at the same speeds as the day before, which cyclist won the second day's race - or was it a tie?
Answer:
Adam won again. Since Adam can cycle 50 kilometers in the time Eve can cycle 48, the two will be side by side 2 kilometers before the finish of their second race. Since Adam is the faster cyclist, he will go on to win.
His margin of victory, in case you're interested, will be
2 x (1 - 48/50) or .08 kilometers.
PUZZLES FROM THE POLE VAULT
From a Polish puzzle magazine called Sam na Sam
Published in GAMES Magazine
August, 1986
When Adam reached the finish line of a 50-kilometer race, Eve was two kilometers behind him. The next day they decided to race again. To even up the contest, this time Adam started two kilometers behind Eve, while Eve began at the starting line as usual. Assuming they cycled at the same speeds as the day before, which cyclist won the second day's race - or was it a tie?
Answer:
Adam won again. Since Adam can cycle 50 kilometers in the time Eve can cycle 48, the two will be side by side 2 kilometers before the finish of their second race. Since Adam is the faster cyclist, he will go on to win.
His margin of victory, in case you're interested, will be
2 x (1 - 48/50) or .08 kilometers.
Thursday, 14 January 2016
Saturday, 9 January 2016
DAY 432
GO FORTH AND MULTIPLY
Credit: GAMES Magazine
May/June 1980
Using each of the ten digits once, can you find two five-digit numbers with the largest possible product?
HINT:
The intuitive answer 98,765
x 43,210 is not correct.
Answer:
Use two principles, namely, the largest digits go to the left; and, the product of two numbers will be maximized if their difference is as small as possible.
The solution is 96,420
x 87,531
GO FORTH AND MULTIPLY
Credit: GAMES Magazine
May/June 1980
Using each of the ten digits once, can you find two five-digit numbers with the largest possible product?
HINT:
The intuitive answer 98,765
x 43,210 is not correct.
Answer:
Use two principles, namely, the largest digits go to the left; and, the product of two numbers will be maximized if their difference is as small as possible.
The solution is 96,420
x 87,531
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