unable to solve these 3 word problems

[strawberryicy]

I got these 3 word problems from the math club of our school , and unable to solve it. I am a fifth grader. Thank you!

1. For any whole number N, let #N be the number of letters it takes to write N as an English word. For example, #18=8, since the word "eighteen" has 8 letters. For how many whole numbers greater than 0 and less than 1000 does #N=N?


2.Between which of the following pairs of numbers does the greatest number of even numbers occur?

A) 25 and 141

B) 137 and 245

C) 183 and 297

D) 249 and 363

3. If I made a list of every seven-digit whole number greater than 1 million which has exactly six of its digits equal to 9, how many different numbers would be on my list?

 

[tkhunny]

1) This is a counting problem. Just count them.

2) Again, find a way to count them.

3) Counting, counting, counting.

There is nothing magic about any of these. Let's see your efforts.

 

[mmm4444bot]

unable to solve [them]

We would like to know where you are stuck, on each of these, before trying to figure out what guidance to provide. Please be specific.

Are there some words that you do not understand? Otherwise, what have you thought about or tried thus far?

The post titled, "Read Before Posting" provides information about how to ask for help on these boards.

Cheers

 

[strawberryicy]

Thank you all for your kind relpys.

1. Does it mean I need to count whole number from 1 to 1000? That means I need to check 1000 numbers. I did try this method the first time I saw this problem but I give up after counting 1 to 20. The teacher gave us 30 word problems to work on within 30 minutes, and this is one of them. I didn't have time to count all the numbers. Is there any other way to do this problem?

2. I got:
A) 25 and 141-->(140-26)/2+1=58
B) 137 and 245-->(244-136)/2+1=54
C) 183 and 297-->(296-184)/2+1=57
D) 249 and 363-->(362-250)/2+1=57

Our teacher didn't give us the answer. Am I correct?

3. I start from 1,000,000.
There are 7 digit.
I count: 1,111,114 and the digit of million goes from 1 to 9. There will be 9 numbers.
1, 111,123 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,111,213 and the digit of million goes from 1 to 9. There will be 9 numbers .
1,111,222 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,111,231 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,111,321 and the digit of million goes from 1 to 9. There will be 9 numbers.
and so on. Then the digit of hundred thousand, ten thousand, thousand, hundred, ten, one go from 1 to 9.
Again, I give up. Same as problem 1, The teacher gave us 30 word problems to work on within 30 minutes, and this is one of them. I didn't have time to count all the numbers. Is there any other way to do this problem?

Thank you!

 

[tkhunny]

1. Does it mean I need to count whole number from 1 to 1000? That means I need to check 1000 numbers. I did try this method the first time I saw this problem but I give up after counting 1 to 20. The teacher gave us 30 word problems to work on within 30 minutes, and this is one of them. I didn't have time to count all the numbers. Is there any other way to do this problem?

No, what would be the point of that. Develop a plan.

One - Three
Two - Three
Three - Five
Four - Four -- We have a winner.
Five - Four
Seven - Five
Eigth - Five
Nine - Four
Ten - Three
Eleven - Six
Twelve - Six
Thirteen - Eight
Fourteen - Eight
Fifteen - Seven

Jut thinking about it a bit, I doubt there are any more. The letters don't go up very quickly. The values do. I'm going with "1" Only.

By the way, should we count spaces? "Twenty Five" Is that 10 or 11?

 

[JeffM]

I got these 3 word problems from the math club of our school , and unable to solve it. I am a fifth grader. Thank you!

1. For any whole number N, let #N be the number of letters it takes to write N as an English word. For example, #18=8, since the word "eighteen" has 8 letters. For how many whole numbers greater than 0 and less than 1000 does #N=N?

These are all counting problems that require CLEVER counting.
The English names for numbers come in patterns after twelve.
What are the MAXIMUM number of letters in the name of any number that exceeds 99 and is exceeded by 1000?
I come up with 5 + 3 + 7 + 7 = 22. How do I get the elements of that sum?
What does that sum mean?
What is the MAXIMUM number of letters in the name any number that exceeds 19 and is exceeded by 100?
What is the maximum number of letter?
What is the MAXIMUM number of letters in the name of any number that exceeds twelve and is exceeded by 20?
Now can you answer the question for sure?

 

[strawberryicy]

I got these 3 word problems from the math club of our school , and unable to solve it. I am a fifth grader. Thank you!

1. For any whole number N, let #N be the number of letters it takes to write N as an English word. For example, #18=8, since the word "eighteen" has 8 letters. For how many whole numbers greater than 0 and less than 1000 does #N=N?

These are all counting problems that require CLEVER counting.
The English names for numbers come in patterns after twelve.
What are the MAXIMUM number of letters in the name of any number that exceeds 99 and is exceeded by 1000?
I come up with 5 + 3 + 7 + 7 = 22. How do I get the elements of that sum?
What does that sum mean?
What is the MAXIMUM number of letters in the name any number that exceeds 19 and is exceeded by 100?
What is the maximum number of letter?
What is the MAXIMUM number of letters in the name of any number that exceeds twelve and is exceeded by 20?
Now can you answer the question for sure?

To JeffM,
Thank you for your reply!
I tried your questions, and I am confused.

If it is a 2-digit number, it would be 7+5 (Seventy+three/seven/eight) MAXIMUM. (I ignored the space between them.)
If it is a 3-digit number, it would be 5+7+7+5 (three/seven/eight+hundred+seventy+three/seven/eight) MAXIMUM. (I ignored the spaces.)

Am I lost?

 

[JeffM]

Am I lost?

It is always possible that I am lost.
If it is "a hundred and b-c," a can have at most 5 letters, "hundred" has seven letters, "and" has three letters, b can have at most seven letters ("seventy"), c can have at most 5 letters. So that is 5 + 7 + 3 + 7 + 5 = 27.

Yep, I was lost.

So the most letters that any English word for a three digit number can have is 27 so if N equals or exceeds 100 then the number of letters in the number's name is less than the number. So you do not have to count all the way to 1,000; you can stop at 99.

But wait. If the number is at least 20 and at most 99, its English name has the form b-c. We saw above that b has at most 7 letters and c at most 5. So the name has at most 12 letters, which is less than 20.

So you do not have to count all the way to 20. What about the teens?

Can you do the problem now?

 

[JeffM]

2.Between which of the following pairs of numbers does the greatest number of even numbers occur?

A) 25 and 141

B) 137 and 245

C) 183 and 297

D) 249 and 363

Let's try a simpler version.
How many even numbers are there between 3 and 5?
Between 13 and 11?
Between 3 and 7?
Between 25 and 29?
Between 3 and 9?
Between 105 and 111?
Do you see any pattern?

 

[mmm4444bot]

The teacher gave us 30 word problems to work on within 30 minutes

In the context of a fifth-grade class, I am very sorry to learn this.

A math club is supposed to be fun.

These types of word problems are brain-teasers. They provide an opportunity for fun, pattern recognition, discovery, and satisfaction. Yet, all of this is lost, when forcing butterflies into the students' gut.

I see no benefit from limiting club members to 60 seconds per exercise. Was a prize offered to the winner of this race?

 

[strawberryicy]

I got these 3 word problems from the math club of our school , and unable to solve it. I am a fifth grader. Thank you!

1. For any whole number N, let #N be the number of letters it takes to write N as an English word. For example, #18=8, since the word "eighteen" has 8 letters. For how many whole numbers greater than 0 and less than 1000 does #N=N?

These are all counting problems that require CLEVER counting.
The English names for numbers come in patterns after twelve.
What are the MAXIMUM number of letters in the name of any number that exceeds 99 and is exceeded by 1000?
I come up with 5 + 3 + 7 + 7 = 22. How do I get the elements of that sum?
What does that sum mean?
What is the MAXIMUM number of letters in the name any number that exceeds 19 and is exceeded by 100?
What is the maximum number of letter?
What is the MAXIMUM number of letters in the name of any number that exceeds twelve and is exceeded by 20?
Now can you answer the question for sure?

To JeffM,
Thank you for your reply!
I tried your questions, and I am confused.

If it is a 2-digit number, it would be 7+5 (Seventy+three/seven/eight) MAXIMUM. (I ignored the space between them.)
If it is a 3-digit number, it would be 5+7+7+5 (three/seven/eight+hundred+seventy+three/seven/eight) MAXIMUM. (I ignored the spaces.)

Am I lost?

To JeffM,

Thank you again for your hints. It helps me to clarify my thoughts a lot.

I re-read the problem again. It says #18=8 eighteen=8.
It doesn't say #18=18. Do we only consider the ones?

I come out:
Four=4 -----> #4=4
eighteen=8 --->#18=8
forty nine=9 ----> #48=9
fifty nine=9 --->#59=9
sixty nine=9 ---->#69=9
ninty nine=9 --->#99=9

and no more.

Am I right?

 

[strawberryicy]

Re:

The teacher gave us 30 word problems to work on within 30 minutes

In the context of a fifth-grade class, I am very sorry to learn this.

A math club is supposed to be fun.

These types of word problems are brain-teasers. They provide an opportunity for fun, pattern recognition, discovery, and satisfaction. Yet, all of this is lost, when forcing butterflies into the students' gut.

I see no benefit from limiting club members to 60 seconds per exercise. Was a prize offered to the winner of this race?

To mmm4444bot,

Thank you for your comment.

Actually, the math club is run by parent volunteers. We only have 30 minutes to run our math club during lunch break.
The teacher/coach gave us 30 word problems to do on our own first, then we discussed them.
Sometimes, the parent coach didn't know how to solve it, either.

Yes, I agree with you. These hard problems make me frustrated sometimes.
However, I feel even worse that I just put these hard problems away and don't know how to solve it the rest of my life.
I HAVE TO KNOW HOW TO SOLVE THESE PROBLEMS.
I hope someone (like you guys) can help me go through these problems. Just like JeffM did to me.
I asked my parents. They have no clues.

You are right. Hard problems need more time than 30 minutes.

 

[JeffM]

1. For any whole number N, let #N be the number of letters it takes to write N as an English word. For example, #18=8, since the word "eighteen" has 8 letters. For how many whole numbers greater than 0 and less than 1000 does #N=N?I re-read the problem again. It says #18=8 eighteen=8.
It doesn't say #18=18. Do we only consider the ones?

I come out:
Four=4 -----> #4=4
eighteen=8 --->#18=8
forty nine=9 ----> #48=9
fifty nine=9 --->#59=9
sixty nine=9 ---->#69=9
ninty nine=9 --->#99=9

and no more.

Am I right?

I think you have gotten a bit off track. #N stands for the number of letters in the English name of the whole number N from 1 through 999.
So, we decided that the most letters that there could be in the name of a number greater than 99 was 27. But 27 is less than 100 or any bigger number. That means that there is no number from 100 through 999 such that #N = N.
Do you see that?

We also decided that the most letters that there could be in a number from 20 through 99 was 12, and 12 is definitely less than 20. That means there is no number from 20 through 99 such that #N = N.

Now let's consider the numbers from 13 through 19. Their names all have the pattern ateen. So 17 has the most letters in its name, namely 9. So there is no number from 13 through 19 such that #N = N.

So we have rejected all the numbers from 13 through 999 by GENERAL reasoning rather than counting the letters in each name SPECIFICALLY. See how math can simplify your work. But it still leaves the first twelve numbers, whose names do not have a common pattern.

#1 = 3 letters in "one."
#2 = 3 letters in "two."
#3 = 5 letters in "three."
#4 = 4.
#5 = 4.
#6 = 3.
#7 = 5.
#8 = 5.
#9 = 4.
#10 = 3.
#11 = 6.
#12 = 6.

WOW Out of 999 numbers from 1 through 999, #N = N for only one number, namely four.

Now how did you do with my hint for the second question? Again it is a matter of counting cleverly. (These are HARD questions for fifth graders, but they are hard questions about counting, and you know how to do that.)

 

[strawberryicy]

I think you have gotten a bit off track. #N stands for the number of letters in the English name of the whole number N from 1 through 999.
So, we decided that the most letters that there could be in the name of a number greater than 99 was 27. But 27 is less than 100 or any bigger number. That means that there is no number from 100 through 999 such that #N = N.
Do you see that?

We also decided that the most letters that there could be in a number from 20 through 99 was 12, and 12 is definitely less than 20. That means there is no number from 20 through 99 such that #N = N.

Now let's consider the numbers from 13 through 19. Their names all have the pattern ateen. So 17 has the most letters in its name, namely 9. So there is no number from 13 through 19 such that #N = N.

So we have rejected all the numbers from 13 through 999 by GENERAL reasoning rather than counting the letters in each name SPECIFICALLY. See how math can simplify your work. But it still leaves the first twelve numbers, whose names do not have a common pattern.

#1 = 3 letters in "one."
#2 = 3 letters in "two."
#3 = 5 letters in "three."
#4 = 4.
#5 = 4.
#6 = 3.
#7 = 5.
#8 = 5.
#9 = 4.
#10 = 3.
#11 = 6.
#12 = 6.

WOW Out of 999 numbers from 1 through 999, #N = N for only one number, namely four.

Now how did you do with my hint for the second question? Again it is a matter of counting cleverly. (These are HARD questions for fifth graders, but they are hard questions about counting, and you know how to do that.)

To JeffM,

Oops! I misundertood the problem. I thought the example #18=8 is the rule. But actually it is only the example of how to count the letters.
Thank you for guiding me through this problem. It was fun!! I will work on Problem 2 after dinner.

 

[strawberryicy]

2.Between which of the following pairs of numbers does the greatest number of even numbers occur?

A) 25 and 141

B) 137 and 245

C) 183 and 297

D) 249 and 363

Let's try a simpler version.
How many even numbers are there between 3 and 5?
Between 13 and 11?
Between 3 and 7?
Between 25 and 29?
Between 3 and 9?
Between 105 and 111?
Do you see any pattern?

To JeffM,

I come out:
Between 13 and 11? ---->12
Between 3 and 7? ------>4, 6---> (6-4)/2+1=2
Between 25 and 29? ---->26, 28 --->(28-26)/2+1=2
Between 3 and 9? ---->4, 6, 8----> (8-4)/2+1=3
Between 105 and 111? --->106, 108, 110 -->(110-106)/2+1=3


Therefore,
A) 25 and 141-->(140-26)/2+1=58
B) 137 and 245-->(244-138)/2+1=54
C) 183 and 297-->(296-184)/2+1=57
D) 249 and 363-->(362-250)/2+1=57

The answer is A) 25 and 141

 

[JeffM]

I come out:
Between 13 and 11? ---->12
Between 3 and 7? ------>4, 6---> (6-4)/2+1=2
Between 25 and 29? ---->26, 28 --->(28-26)/2+1=2
Between 3 and 9? ---->4, 6, 8----> (8-4)/2+1=3
Between 105 and 111? --->106, 108, 110 -->(110-106)/2+1=3


Therefore,
A) 25 and 141-->(140-26)/2+1=58
B) 137 and 245-->(244-138)/2+1=54
C) 183 and 297-->(296-184)/2+1=57
D) 249 and 363-->(362-250)/2+1=57

The answer is A) 25 and 141

Very nicely done. You came up with a formula that allowed you to SKIP all that counting.

So what are your thoughts, not your answer, on the third question?

 

[strawberryicy]

For Problem 3:

I am not sure if I fully understand the question or not.

I start from 1,000,000.
There are 7 digit.
I count: 1,111,114 and the digit of million goes from 1 to 9. There will be 9 numbers.
1, 111,123 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,111,213 and the digit of million goes from 1 to 9. There will be 9 numbers .
1,111,222 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,111,231 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,111,321 and the digit of million goes from 1 to 9. There will be 9 numbers.
Then:
1,114,111 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,123,111 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,213,111 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,222,111 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,231,111 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,321,111 and the digit of million goes from 1 to 9. There will be 9 numbers.
Then
1,111,141 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,111,231 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,112,131 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,112,221 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,112,311 and the digit of million goes from 1 to 9. There will be 9 numbers.
1,113,211 and the digit of million goes from 1 to 9. There will be 9 numbers.
and so on.

Then the digit of hundred thousand, ten thousand, thousand, hundred, ten, one go from 1 to 9.

I don't see any pattern.

 

[Denis]

3. If I made a list of every seven-digit whole number greater than 1 million which has exactly six of its digits equal to 9, how many different numbers would be on my list?

I'm 99% sure this means simply that each number contains six 9's; like these examples:
1,999,999 ... 8,999,999
9,909,999 ... 9,989,999
Get my drift?
Lowest: 1,999,999
Highest: 9,999,998

 

[strawberryicy]

3. If I made a list of every seven-digit whole number greater than 1 million which has exactly six of its digits equal to 9, how many different numbers would be on my list?

I'm 99% sure this means simply that each number contains six 9's; like these examples:
1,999,999 ... 8,999,999
9,909,999 ... 9,989,999
Get my drift?
Lowest: 1,999,999
Highest: 9,999,998

To Denis,

Thank you for your reply!

Oops!
I misunderstood the question again!!
No wonder my mom said I have to improve my reading.
I thought it means the sum of six of its digits equal to 9.

Then,
1,999,999 to 8,999,999 --->8 numbers
9,999,990 to 9,999,998 --->9 numbers
9,999,909 to 9,999,989 ---->9 numbers
9,999,099 to 9,999,899 ---->9 numbers
9,990,999 to 9,998,999 --->9 numbers
9,909,999 to 9,989,999 --->9 numbers
9,099,999 to 9,899,999 --->9 numbers

The formula: 6 digits x9+ the greatest digit x8=63

Thank you all for your time on these problems.

PS: I start worrying that if I couldn't understand the question, how could I solve the word problem?

 

[strawberryicy]

Oops!
It should be:
The formula: 6 digits x9+ the greatest digit x8=62