Question

Find all integer solutions for
[x/2] + [x/3] + [x/4] + [x/5] = 100

where [z] denotes the integer part of z
for example: [e]= 2 and [pi]=3

Answer 1

find the LCD for 2, 3 4 and 5
rewrite the fractions
solve for x

Answer 2

then? after that?

Answer 3

what about the above mentioned condition?
"where [z] denotes the integer part of z"

Answer 4

integer positive whole number, negative whole number

Answer 5

do not use fractions

Answer 6

look at the example pi is actually 3.142 but only write 3

Answer 7

I'm afraid that's not gonna work.
for example:
taking the lcd and finding x in the normal way will give me 77.6something.
then taking the integer part will give me 77.
but we all know 77 is not the answer

Answer 8

taking the integer part of some numbers and adding them is not equal to adding those numbers and taking the integer part of the final solution

Answer 9

@satellite73 , @zepdrix , @radar , @Luigi0210 , @tkhunny , @shamil98
Do you guys have any idea on how to approach? I was struggling with this for more than two days now

Answer 10

@mukushla @oldrin.bataku

Answer 11

I think there are no solutions. I came to that conclusion simply counting and calculating. But there should be a better way

Answer 12

Interesting question!

Thinking "out loud" (but on paper) (but on a computer keyboard):

What about adding up all of those fractions, ignoring the brackets [ ] for now, obtaining x and then rounding x up to the nearest integer?

Answer 13

well I tried that. but that's not an answer
what I got was 77.something. Then int 77 is not the answer

Answer 14

\[\frac{ 30x+20x+15x+12x\frac{ }{ ? } }{ 60 }=100\] (please ignore the typos)

Answer 15

so x = 7

Answer 16

^ how come?

Answer 17

"taking the integer part of some numbers and adding them is not equal to adding those numbers and taking the integer part of the final solution"

Answer 18

Then 77x/60 = 100;
77x=6000;

x = Floor (6000/77)

Answer 19

the instructions i gave was same as mathmale

Answer 20

so 6000/77 = 77.999....
I tried applying 77,78,79 and 80 to the equation again. But it didn't work

Answer 21

How about defining, in your own words, what characteristics solution x must have? Perhaps that would help.

Answer 22

1.) x should be an integer
2.) sum of four integers should equal to 100
3.) those four integers should obey the given formula

Answer 23

I'm still thinking about your statement, "Taking the integer parts of some numbers ... "

Is that your interpretation of what I was doing?

Actually, I was suggesting summing up 30x/60 + 20x/60 + 15x/60 + 12x/60 and setting the result = to 100. This is not the same thing as summing up the floors (integer parts) of the fractions x/2, x/3, x/4 and x/5, is it?

Now, back to your specs for x:
1) x must be an integer: agreed
2) the sum of the four integers [x/2], [x+3], etc., must be 100. Agreed. Hadn't thought of this situation in those words until now.
3) those four integers must obey the given formula: Isn't this statement the same as (2)?

Answer 24

Actually, I was suggesting summing up 30x/60 + 20x/60 + 15x/60 + 12x/60 and setting the result = to 100. This is not the same thing as summing up the floors (integer parts) of the fractions x/2, x/3, x/4 and x/5, is it?

yes they are not the same I think

Answer 25

I'm hitting my head against a stone wall at the moment. Is it conceivable that this problem has no solution if x must be an integer? Didn't you say you went through various integers manually to identify a solution by trial and error?

Answer 26

30x/60 + 20x/60 + 15x/60 + 12x/60 will give me 77x/60
x=77.999
then I tried 77, but it sums up to 97
then 78, but it sums up to 99
then 79, still to 99
then 80, sums up to more than 100

Answer 27

Doesn't that seem to imply that there's no integer value of x that is a solution?

Answer 28

My prof won't take trial and error as a valid way of proving this since it's higher level math problem solving course.
That's the concern that I have.
trial and error method doesn't seem suitable for the question :(

Answer 29

plus we are not allowed to use calculators

Answer 30

Supposing, in the equation you've presented, that you replace the = sign with <=. Wouldn't there be a fair number of integer solutions x? Were you to restrict the given expression further, by writing something like 75 < [x/2]+[x/3] (etc.) <= 100, you'd get a solution set looking something like {77,78,79}?

Answer 31

You have aptly eliminated the integer 80 as a possible solution, and demonstrated that 77, 78, 79, when subst. individually into the given equation, do not produce the final, required sum 100.

So, we've tried algebraic methods to determine x, and have, apparently, failed at that; and we've tried a couple of values {77, 78, 79}, all of which are integers, all of which produce sums just less than 100, but none of which produce the sum of exactly 100. What more could your prof ask? Try putting yourself in his shoes and see whether you could answer that.

Answer 32

okay, If I couldn't come up with something by tomorrow I'll do that. Any really appreciate your effort dude. That is one way of writing the solution in a proper manner. Thanks again.
I've got to go. This is only one question due tomorrow, I have 5 more to mess around with. Thank you for your time

Answer 33

@triciaal @mathmale thanks alot for you time guys

Answer 34

There is a number, x= 77.999 (which you yourself have found), which would be a solution were it not for the fact that x must be an integer.

My pleasure, really. Great discussion. Glad you recognize that it's high time to move on to those other problems.

Answer 35

is it 5?
77-60 = 17
17x/60 = 100
using only the fractions 100/17 =5 integer only