100
Using only 16, 17, 23, 24, 39 and 40 and addition, make 100 with the minimum number of numbers. e.g. 40+39+21=100 uses only 3 numbers - unfortunately 21 isn't available. You may use a number more than once e.g. 23+23+... is ok. What are the numbers?
Labels: mathemagic
posted by Chris at
9:08 PM
16 Comments:
16+16+17+17+17+17=100
it's wrong,i just wanted to be frist post.
I see Pequod's been helping you again. You got it right.
I hope Cam fancies explaining why that is correct, otherwise I'll have to finish working it out.
Most of the prolems posted by me tonight don't explain the answer, they just state it.
wow...i just took a guess.
Pequod's answer was way out there.
...yeah, yeah, yeah; if you say so Mr B.
I left a word out in my previous post. I meant my posted problem sources don't provide a explantion of how to derive the result.
January 17, 2010 10:24 PM
Chris,
It's fairly simple to prove using modulo arithmetic....
Cam
Hi Cam. I know, but it's still quite a lot of steps - unless you've seen something really slick.
Can eliminate 39 and 40 almost immediately. That leaves 16,17,23,24 as only numbers to bother with. Can't do it with 4 or less as 4*24 is less than 100.
So need at least 5 numbers.
100=0; 16=1, 17=2, 23=3, 24=4 (mod 5)
Must choose 5 numbers whose remainders add to 0 (mod 5). Choose candidates only by looking at the remainders.
5 of any number are automatically candidates (as 100 = 0 (mod 0)), but then need 5*20 so no good. A few seconds though shows can't be 4*one number. Trying 3*a given, easily see that 3*16 is no good, 3*17+16+23 is too small, 3*23+17+24 is too small, 3*24+16+17 is too large.
So 2*16+17+2*23 is too small. 2*17+2*23 is too small, 2*16+2*24 is too small, etc. (I'm bored)
That forces us to conclude that least at least 6 numbers.
100=4; 16=4,17=5,23=5,24=0 (mod 6)
6*a given automatically can't work as the remainder would be zero, bu also because we'd need 6*16.666... , so no good. But tells us that 16 must be present (an average cannot be less than the minimum value). So if had 1*16, would need 4 numbers whose remainders add to 0. 1*16+5*17 is 101, so only option (again useful to think about averages) is 2*16+4*17 = 100.
I'm sorry that's a bit sloppy. I had to rush it as I've go to go out.
Chris,
Try:
(16,17,23,24,39,40,100)mod16
(16,17,23,24,39,40,100)mod17
(16,17,23,24,39,40,100)mod23
(16,17,23,24,39,40,100)mod24
(16,17,23,24,39,40,100)mod39
(16,17,23,24,39,40,100)mod40
Look at the results and it becomes clear that only 16,17 can be used to match the mod of 100 and keep the total <100.
Cam
I can do it in 3 numbers..... The childish way - the 1 from either the 16 or 17 and I'll use the 0 from the 40, twice.... 100.
Hi Cam. I hadn't thought of that idea. I've now made that table, but can't say that I can see what to do with it as so many of the remainders are so large that I don't feel that the problem has been simplified.
I'll stare at it a bit more in a short while, and hope a penny drops. I usually am hopeless at mental arithmetic. Took me ages to learn how to factorise a quadratic, even with carefully selected ones that were easy. Come to that, I probably couldn't remember the 12 times tables until I was a teenager.
Here is a brief explanation of the method.
Create a table.
(16,17,23,24,39,40,100)mod16
(16,17,23,24,39,40,100)mod17
(16,17,23,24,39,40,100)mod23
(16,17,23,24,39,40,100)mod24
(16,17,23,24,39,40,100)mod39
(16,17,23,24,39,40,100)mod40
Look at results. Only need to look at digits right of 0s. (Looking at number to left of 0s is duplication of effort, and a good mathematician is a lazy one.)
You must create the mod of 100, by using multiples of the other numbers. Constrain result to multiples that total under 100.
i.e. <=6 16s,<=5 17s,<=4 23s,<=4 24s,<=2 39s,<=2 40s.
By inspection, you should be able to discard all but multiples of 1 to make 4,which is 17 mod 16. 4*1=4 so 4 multiples of 17. (100-4*17)/16=2 So 2 multiples of 16.
If one is still struggling and it is too hard to see from inspection that this is true then one could create 4 more tables
[2*(16,17,23,24,39,40)]mod16
[2*(16,17,23,24,39,40)]mod17
[2*(16,17,23,24,39,40)]mod23
[2*(16,17,23,24,39,40)]mod24
[2*(16,17,23,24,39,40)]mod39
[2*(16,17,23,24,39,40)]mod40
[3*(16,17,23,24,39,40)]mod16
[3*(16,17,23,24,39,40)]mod17
[3*(16,17,23,24,39,40)]mod23
[3*(16,17,23,24,39,40)]mod24
[3*(16,17,23,24,39,40)]mod39
[3*(16,17,23,24,39,40)]mod40
[4*(16,17,23,24,39,40)]mod16
[4*(16,17,23,24,39,40)]mod17
[4*(16,17,23,24,39,40)]mod23
[4*(16,17,23,24,39,40)]mod24
[4*(16,17,23,24,39,40)]mod39
[4*(16,17,23,24,39,40)]mod40
[5*(16,17,23,24,39,40)]mod16
[5*(16,17,23,24,39,40)]mod17
[5*(16,17,23,24,39,40)]mod23
[5*(16,17,23,24,39,40)]mod24
[5*(16,17,23,24,39,40)]mod39
[5*(16,17,23,24,39,40)]mod40
Simply pick the mod that matches the 100 mod 16,17,23,24,39,40 number. Apply constraints for max multiples
e.g. in the 3x table 3*39 mod 17=15=100 mod 17, but it is over 2 multiples, so discard.
in the 4x table 4*17 mod 16=4=100 mod 16 and 4 multiples is <5 so 4 multiples of 17 works.
Once the mod matches the 100 mod x then fill in the remainder with the number with mod=0
e.g. (100-4*17)/16=2
Anyhow, the answer is 2 multiples of 16 and 4 multiples of 17.
Cam
Hi Cam, hmmm. I shall look at your enhanced answer, with the hope that it all comes together.
But I'm still concerned that the remainders are so large that it hasn't simplified the problem. It seems that you may as well just do trial and error from scratch.
I also don't understand why you only need to look to the right of the 0's. The given 6 numbers seem to be unrelated to each other (i.e. they seem to be random), so I don't understand what work would be duplicated. Am I being Mr. Thickie?
Although I charged through my version, I personally felt that the small remainders were where the labour savings came in - i.e. I didn't need a calculator for any part of my solution. But it was still quite a bit of work.
I was a bit naughty in getting rid of the 39 and 40 without having demonstrated fully why that was easy. (I'm a very lazy mathematician. LOL).
Hi Cam. One penny dropped, one(?) to go. I understand why you only need to look to the right or left of the 0s. Thanks for that. Prolly back soon.
Hi Cam. I am about to put this one to bed. That's mainly because I
don't think there's a really super duper slick solution.
However, I made this table (whilst playing)
(16,17,23,24,39,40;100)modN
( 0, 1, 1, 0, 1, 0; 0)mod2
( 1, 2, 2, 0, 0, 1; 1)mod3
( 0, 1, 3, 0, 3, 0; 0)mod4
( 1, 2, 3, 4, 4, 0; 0)mod5
( 4, 5, 5, 0, 3, 4; 4)mod6
( 2, 3, 2, 3, 4, 4; 2)mod7
I had implicitly done this in my ^solution^. I did it under some
strange delusion that I had to use mod n if I was choosing n
numbers at a time.
I now realise (and rather suspect that you have been aware of
it all along) that the mod can be to any base that you wish.
Any set of the n numbers must add up to 100 in every base.
I mean we require Sum = 100 (mod m), for all m (m > 1).
So it must simultaneously satisfy all residue tables like above.
So pick and mix if you like.
It is clear that 6 is the maximum we need to look at because
7*16 > 100.
If use 40: Can't use 3 times as 3*40 > 100. If use twice, then
need another 20 and that's not possible. If use once, then 60 to
go. If use 39 then have 21 to go and that's not possible. If use
24, then have 36 to go and that's not possible. If use 23, then
have 37 to go and that's not possible. If use 17, then have 43
to go, and that's not possible; the same for 16. Bye-bye 40.
Pretty much the same for 39. So 39 is out.
In fact can dispose of 24 quite quickly. 4*24 leaves 4 to go,
so no go. 3*24 leaves 28 to go, so no go. 2*24 leaves 52 to go,
so need an average of 26 in two of 15, 17 and 23, so no go.
1*24 leaves 76. Try 4*23 => 92, too big. 3*23 => 69 so 7 to go,
so no go. 2*23 = 46, leaves 30 to go, so no go. 1*23 leaves
53 to go, using 16 and 17. 2*17 leaves 19 to go, so no go. 1*17
leaves 36 to go, so no go. So back to 76 to go using only 16,
and that can't be done. Bye-bye 24.
Now only 16, 17 and 23 left. I can't be bothered to write any
more. I didn't really use the table at all. It might be able to
save a few steps, but at the expense of having to juggle with
it as well.
How strange - it seemed like a good idea at the time! Thanks
for keeping up the posting. I'm sorry that I've made the
explanation overly concise.
Haha you ppl have no lives!!
Dear Anonymous,
I may have no life, in your opinion, but I work as a physicist and I find these mathematical conundrums ease the day. I also get paid a lot of money to do my job, and I am a leader in my field, and in my poor miserable life, I am about to go home from work for a 22 mile drive through the countryside in an Aston Martin DBS (the older 2008 model).
Do I want to trade places.....?
Karl, I think your awesome. Also, very nice comeback, you threw that into his/her face quite nicely. (a different anonymous)
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