Difficult Problems

[nippon276]

Here are some problems I need help with. I have the answer key, yet I don't know how to do a lot of them. These are the first five:
1. If the temperature is more than 1^o Celsius but less than 74^o Celsius, find the sum of all possible distinct Fahrenheit temperatures such that the Fahrenheit temperature is a prime integer and that the equivalent Celsius temperature is any integer.

Obviously I could do this manually, but that would take too long. What's a quicker way?

2. How many distinct integers between 1000 and 2010, inclusive, are divisible by 3, 5, or 7?

Again, too long to do by hand. I tried dividing 2010 by 3, 5, and 7 (excluding the remainder) and got 670, 402, and 287. I did the same with 1000 and subtracted the respective values (2010 answers-1000 answers) and got 685 (the answer key says 550).

3. A book store sells 3 types of books: math books for $5 each, physics books for $1.50 each, and comic books for $0.25 each. Derby buys 100 books and pays exactly $50. If Derby buys at least 1 book of each type, find the ordered triple of the form (# math books, # physics books, # comic books) that Derby buys.

I tried coming up with a system of equations to solve, but there's three variables and I only had two equations (one for the price and the other for the number i.e. 5m+1.5p+.25p=50 and m+p+c=100), so that didn't work.

4. An original number is increased by k%, and then the result is decreased by w%. The final result is 85% of the original number. If k is a positive integer less than 173 and if w is a positive integer, find the sum of all possible distinct values of k.

Just a lot of crazy answers.

5. How many distinct positive integers less than 1000 have prime factorizations of the form p^aq^br^c with p , q , and r distinct positive primes and with a , b , and c positive integers such that p + q + r <13 ?

Not sure on how to do this one.

 

[Ishuda]

1. I don't really know a way other than manually but you can initially cut down the numbers by noting that, since
C = 5 (F-32) / 9 = 5 (F - 5)/9 - 15
and is greater than 1 and an integer, F must be greater than 32 and have a remainder of 5 when divided by 9, i.e. F = 9 k + 5, k>3. Since C is less than 74, we also have k < 18 and
C = 5 k -15; 3 < k < 18.
So what are the primes of the form
F = 9k + 5; 3 < k < 18?
Well k can't be divisible by 5 (or F would be) so that cuts it down a bit more (to 14).

2. Watch the end points. Another way: For 3, the first value is 1002 and the last number is 2010 so the form of the number is
n₃ = 999 + 3 k; k = 1, 2, 3, ..., 337 [ = (2010-999)/3 ]
For 5, the first number is ...
BTW: I'm not too sure the answer key is correct if the problem is as stated.

3. Let the triplet (x, y, z) represent the books bought minus 1, i.e. x+1 is the number of math books bought. The formula for how much money was spent is, as you pointed out,
50 = 5 (x+1) + 1.5 (y + 1) + 0.25 (z+1)
or, getting rid of the fractions by multiplying by 4 and expanding out the cost of the one book, we have
173 = 20 x + 6 y + z
The number of books equation becomes
97 = x + y + z
The constraints are
x \(\displaystyle \ge\) 0, y \(\displaystyle \ge\) 0, z \(\displaystyle \ge\) 0.
Let
z = 97 - x - y
then solutions are any non-negative integers x and y satisfying
76 = 19 x + 5 y
x+y < 98
What is y for x=0? x=1? ...

Have to go so that's it for now - maybe someone else will look at the rest.

EDIT: Fix dumb mistake (had forgot to subtract the 1 from the 20 and 6) and see below about multiples for 3, 5, 7 combined.

 

[doive]

2. How many distinct integers between 1000 and 2010, inclusive, are divisible by 3, 5, or 7?

Again, too long to do by hand. I tried dividing 2010 by 3, 5, and 7 (excluding the remainder) and got 670, 402, and 287. I did the same with 1000 and subtracted the respective values (2010 answers-1000 answers) and got 685 (the answer key says 550).

Interesting questions!
I'll have a more detailed look later but I think I see where 2 is coming from.

You have the right method for finding how many numbers are divisible by 3, 5, 7 -> (2010-1000)/n. but 1500 for example is divisible by both 5 and 3 - you need to take away the multiples of 15, 21, 35 and 105 which should get you back to your answer of 550

 

[Ishuda]

Interesting questions!
I'll have a more detailed look later but I think I see where 2 is coming from.

You have the right method for finding how many numbers are divisible by 3, 5, 7 -> (2010-1000)/n. but 1500 for example is divisible by both 5 and 3 - you need to take away the multiples of 15, 21, 35 and 105 which should get you back to your answer of 550

Of course. So also need multiples of 15, 21, and 35.

 

[Ishuda]

I must admit that I sometimes miss the obvious but seems to me a spread sheet program or some such is necessary for solving these in any sensible time frame. If someone has an 'easy' solution, I sure would appreciate having my horizons expanded.

4. An original number is increased by k%, and then the result is decreased by w%. The final result is 85% of the original number. If k is a positive integer less than 173 and if w is a positive integer, find the sum of all possible distinct values of k.
Let ' refer to fractional rather than percent, i.e. w' = w / 100
a * (1 + k') * (1 - w') = 0.85 * a
so
k' = - 1 + 0.85 / (1 - w')
or
k = 100 (w - 15) / (100 - w)
which cuts down on the number of numbers needing checking

5. How many distinct positive integers less than 1000 have prime factorizations of the form p^aq^br^c with p , q , and r distinct positive primes and with a , b , and c positive integers such that p + q + r <13 ?

Eligible p, q, and r are 2, 3, 5, and 7, i.e. including 11 violates the sum constraint. Order doesn't matter so maximum of ₄C₃ = 4 choices (that is, one must choose all three and not allow any to just be the trivial 1 since a, b, and c are greater than zero). Use constraint and reduces to 2, i.e. 2+5+7>13 and 3+5+7>13.

So question reduces to how many of form 2^a 3^b 7^c or 2^a 3^b 5^c ... which is a bit of help.

 

[doive]

I must admit that I sometimes miss the obvious but seems to me a spread sheet program or some such is necessary for solving these in any sensible time frame. If someone has an 'easy' solution, I sure would appreciate having my horizons expanded.

4. An original number is increased by k%, and then the result is decreased by w%. The final result is 85% of the original number. If k is a positive integer less than 173 and if w is a positive integer, find the sum of all possible distinct values of k.
Let ' refer to fractional rather than percent, i.e. w' = w / 100
a * (1 + k') * (1 - w') = 0.85 * a
so
k' = - 1 + 0.85 / (1 - w')
or
k = 100 (w - 15) / (100 - w)
which cuts down on the number of numbers needing checking

5. How many distinct positive integers less than 1000 have prime factorizations of the form p^aq^br^c with p , q , and r distinct positive primes and with a , b , and c positive integers such that p + q + r <13 ?

Eligible p, q, and r are 2, 3, 5, and 7, i.e. including 11 violates the sum constraint. Order doesn't matter so maximum of ₄C₃ = 4 choices (that is, one must choose all three and not allow any to just be the trivial 1 since a, b, and c are greater than zero). Use constraint and reduces to 2, i.e. 2+5+7>13 and 3+5+7>13.

So question reduces to how many of form 2^a 3^b 7^c or 2^a 3^b 5^c ... which is a bit of help.

4. (1 + k') * (1 - w') = 0.85
if you rearrange with to make w' the subject and multiply through by 100 for k and w
w = 100 - 8500/(100+k)
Given w and k are both integers \(\displaystyle \implies \frac{8500}{100+k} \) must also be an integer
100+k must then be an exact divisor of 8500
8500 has factors 2*2*5*5*5*17 so 100+k must be some subset of those i.e. 100<2^a*5^b*17^c<273


5.
2^a 3^b 7^c
a = 1 to 5
b = 1 to 3
c = 1 to 2
2^a 3^b 5^c
a = 1 to 6
b = 1 to 4
c = 1 to 2
That's a sufficiently small set to check manually

 

[doive]

Here are some problems I need help with. I have the answer key, yet I don't know how to do a lot of them. These are the first five:
1. If the temperature is more than 1^o Celsius but less than 74^o Celsius, find the sum of all possible distinct Fahrenheit temperatures such that the Fahrenheit temperature is a prime integer and that the equivalent Celsius temperature is any integer.

C = 1.8*F+32
if F and C are both integers then C/1.8 is an integer
How many numbers between 1-74 divide exactly by 1.8? -> 18,36,54,72
You can work back from the to get the Farenheit temps and test which is or is not prime and sum the remaining ones

 

[doive]

3. A book store sells 3 types of books: math books for $5 each, physics books for $1.50 each, and comic books for $0.25 each. Derby buys 100 books and pays exactly $50. If Derby buys at least 1 book of each type, find the ordered triple of the form (# math books, # physics books, # comic books) that Derby buys.

I tried coming up with a system of equations to solve, but there's three variables and I only had two equations (one for the price and the other for the number i.e. 5m+1.5p+.25p=50 and m+p+c=100), so that didn't work.

Once again integers is a big constraint and you can quickly put on some upper bounds based on price: m<10, p<33 and number c<100.
You can further constrain m<6 since you can't make the remaning 90+ books for greater than that total (6m+p+93c < 50)

For a given m only 1 pair of p and c is possible which means you now just need to solve for the cases m=1 to 5 which has reduced to problem to five 2-variable simultaneous equations which is quite solvable.
You can create constraints since c must be even and p+2c must be divisible by 4 (to make an exact number of dollars) and probably further constraints you could consider; but at this point I just solved the four equations to determine that there is only 1 valid solution m = 5, p = 1 c = 94

It is interesting to observe that you can only vary the price by a minimum of $0.25 and that the "minimum incremental price" is $1.25 (i.e. adding $1.5 and removing one lot of $0.25 to keep the total number of books the same) then for the more general problem where the total cost is $50*n there are solutions for 1 in every 5 m e.g. for cost = $200 there are solutions for m=5,10,15,20
- NB this is a bit rambly but does make sense - honest!

 

[Ishuda]

C = 1.8*F+32
if F and C are both integers then C/1.8 is an integer
How many numbers between 1-74 divide exactly by 1.8? -> 18,36,54,72
You can work back from the to get the Farenheit temps and test which is or is not prime and sum the remaining ones

You have that turned around
F = 1.8 C + 32
not the other way around.

 

[nippon276]

Let ' refer to fractional rather than percent, i.e. w' = w / 100
a * (1 + k') * (1 - w') = 0.85 * a
so
k' = - 1 + 0.85 / (1 - w')
or
k = 100 (w - 15) / (100 - w)
which cuts down on the number of numbers needing checking

Wait… So how did you get k=100 (w-15) / (100-w)? I checked by looking at a table of k and w values and that suitable k values are 25, 70, and 150 or 245 (the answer).

 

[nippon276]

5.
2^a 3^b 7^c
a = 1 to 5
b = 1 to 3
c = 1 to 2
2^a 3^b 5^c
a = 1 to 6
b = 1 to 4
c = 1 to 2
That's a sufficiently small set to check manually

How exactly did you find these sets? I got the whole p-q-r-thing though I'm still a bit confused on the rest of the problem (the a-b-c stuff and checking).

 

[nippon276]

Thank you everyone for your help so far! I'm understanding more and more.

 

[Ishuda]

Wait… So how did you get k=100 (w-15) / (100-w)? I checked by looking at a table of k and w values and that suitable k values are 25, 70, and 150 or 245 (the answer).

Working from
k' = - 1 + 0.85 / (1 - w')
and note that
k = 100 k' and w = 100 w', so multiply through by 100 and simplify. Then multiply through left hand side by 100 top and bottom.

 

[nippon276]

Working from
k' = - 1 + 0.85 / (1 - w')
and note that
k = 100 k' and w = 100 w', so multiply through by 100 and simplify. Then multiply through left hand side by 100 top and bottom.

Ok, thank you for your help. I'm almost done with these five problems (#5 still confusing).

 

[Ishuda]

So 2^1 * 3^1 * 5^1 = 30 would be the minimum integer?

Yep, at least the way I read it.

 

[nippon276]

Yep, at least the way I read it.

Do you possibly have an idea on what to do?

 

[nippon276]

So 2^1 * 3^1 * 5^1 = 30 would be the minimum integer?

Any ideas?

 

[Ishuda]

Any ideas?

To keep track of what you do, you can 'count': first (100's) digit is exponent of 2, second is exponent of 3, third is exponent of 5.
111: 2 * 3 * 5 = 30
112: 2 * 3 * 5² = 150
113:2 * 3 * 5³ = 750
121: 2 * 3² * 5 = 90
122: 2 * 3² * 5² = 450
131: 2 * 3³ * 5 = 270
132: 2 * 3⁴ * 5 = 810
211: 2² * 3 * 5 = 60
...

You will also have to do that for 2, 3, and 7; 2, 5, and 7; and 3, 5, and 7.

 

[nippon276]

To keep track of what you do, you can 'count': first (100's) digit is exponent of 2, second is exponent of 3, third is exponent of 5.
111: 2 * 3 * 5 = 30
112: 2 * 3 * 5² = 150
113:2 * 3 * 5³ = 750
121: 2 * 3² * 5 = 90
122: 2 * 3² * 5² = 450
131: 2 * 3³ * 5 = 270
132: 2 * 3⁴ * 5 = 810
211: 2² * 3 * 5 = 60
...

You will also have to do that for 2, 3, and 7; 2, 5, and 7; and 3, 5, and 7.

I see what you're thinking. I might just ask my teacher though, else it would take a lot of time.