## Four number problem

Posted by Chris on March 17, 2011 – 4:51 pm

There are four different positive integers that sum to less than 18. To determine those numbers you need to know their product and the smallest number. What are the numbers?

March 17th, 2011 at 5:32 pm

It seems to me that if you need to know both the product AND the smallest number then there must be two different sets of numbers with the same product. One set would have 1 as the smallest number and the other set would have 2 as the smallest number. 3 cannot be the smallest number as the sum of the numbers would not be less than 18.

Just off the top of my head, 120 could be the product. If you then know that 1 is the smallest number the answer would be 1,4,5,6. If 2 is the smallest number then 2,3,4,5 is the answer.

There may be other pairs of solutions. More likely I’ve misunderstood the question.

March 17th, 2011 at 5:50 pm

Hi Wiz, you gave up too quickly.

March 17th, 2011 at 6:00 pm

1,4,5,6 or 2,3,4,5

March 17th, 2011 at 6:04 pm

There is a unique solution.

March 17th, 2011 at 7:13 pm

hi Wiz- 1,3,5,8 has a product of 120 as well

March 17th, 2011 at 9:54 pm

Thanks, Knightmare, so a 120 product starting with a 1 has two solutions.

So, let’s check out the other possibilities. I count just seven combinations beginning with a 2 that sum to less than 18.

Replace the 2 with a 1 and double one of the other numbers to retain the same product. Then check to see if the resulting sets are unique for tht product, do not result in two numbers being repeated, and sum to less than 18. I can’t find any.

So, I don’t know . . .

I’m sure Chris will tell me to keep trying.

March 17th, 2011 at 10:43 pm

i count 33 sets of 4 integers that = less tham 18

if we were told that you just need the product, this would be too easy. there is a reason we need to know the lowest number. (one out of three)

March 18th, 2011 at 6:20 am

Hi Wiz. You don’t need my kicks, Knightmare is doing a wonderful job for me

March 18th, 2011 at 6:28 am

We need to know the product and the lowest number because this is the only way we get a unique answer. A product of 120 gives too many possibilities.

4 unique numbers which sum < 18

1234 = 24

1235 = 30

1236 = 36

1237 = 42

1238 = 48

1239 = 54

1345 = 60

1346 = 72

1347 = 84

1348 = 96

1349 = 108

1356 = 90

1357 = 105

1358 = 120

1367 = 126

1456 = 120

1457 = 140

2345 = 120

2346 = 144

2347 = 168

2348 = 192

2349 = 216

2456 = 180

We can eliminate all the duplicates with the same lowest integer, in this case the 120s with lowest integer 1.

1358 = 120

1456 = 120

120 with lowest integer of 2 is unique so it still qualifies.

Now if I know the lowest integer and product I can provide you with a unique response.

March 18th, 2011 at 7:40 am

I agree with John24. All except for 2,3,4,9. That adds to 18, not less than 18.

It doesn’t appear that there is a single combination that can be picked out simply from the stated rules. Only combinations that can be eliminated (like 1,3,5,8 and 1,4,5,6)

March 18th, 2011 at 7:46 am

Hi John 24. You’ve done all the hard work, you’ve now just got to realise one more thing. DP has (unknowingly?) given you a big hint.

March 18th, 2011 at 7:48 am

I agree that the integers are 2,3,4,5.

By looking at the seven scenarios with a 2 as the smallest, all but one has a unique product from the entire set.

A product of 120 is the only one that isn’t unique (one with 2 as the smallest and two with 1 as the smallest).

This tells me we need both pieces of information (lowest and product) to identify 2,3,4,5.

March 18th, 2011 at 7:55 am

Yes, you are correct. I totally missed that. If you wanted any combination with lowest number 1, all you need is the product. If I told you “the product is 42″, you would tell me the combination is 1,2,3,7. I wouldn’t need to tell you the lowest number, since it is implied by the solution.

120 as the product is the only one where you would also need the lowest number.

If I told you “The lowest number is 1 and the product is 120″, you could not come up with the answer because there are 2 solutions.

The only solution can be a product of 120 with a low number of 2. Combination 2,3,4,5.

Thank you Chris for pointing out the hint I gave myself, and thank you cazayoux for the solution!

March 18th, 2011 at 7:59 am

my next to last paragraph was not worded very well. I was a little excited about seeing the solution, so I tried to quickly post. I should have said:

The only solution must have a product of 120, and the low number must be 2. The winning combination is 2,3,4,5.

March 18th, 2011 at 8:18 am

LOL. 2,3,4,5 it is. I’m slightly baffled; cazayoux said he agreed with that; but nobody had actually/clearly said that was the answer.

John 24 was within an atomic radius, having travelled a league, of getting there.

March 18th, 2011 at 9:29 am

I also missed a few of the combinations.

1245 = 40

1246 = 48

1247 = 56

1248 = 64

1249 = 72

1256 = 60

1257 = 70

1258 = 80

1259 = 90

1267 = 84

1268 = 96

And had one additional value in my list caught by DP

2349

33 combinations as stated by Knightmare. But only one requires both the product and the sum to determine the set of 4 unique numbers.

March 18th, 2011 at 9:54 am

what about 2,3,5,6 and 2,3,5,7 ? I get that there are 35 combinations with a sum of less than 18.

March 18th, 2011 at 10:05 am

According to the source site (old ToM ) there are 38 possibles. I didn’t mention that as the problem was already tedious enough.

March 18th, 2011 at 10:40 am

Well I guess someone else will have to come up with the missing 3, because I am just not seeing it.

It can’t be anything other than a combination with 1 or 2 (or both) in it. We cannot use a low value of 3 as staed by Wiz, since the lowest combination of different integers would be 3+4+5+6 = 18.

What am I not seeing?!?!?

March 18th, 2011 at 11:03 am

Hi DP. It’s possible that the source was in error. But whatever; only 2,3,4,5 does the job.

March 18th, 2011 at 4:02 pm

I guess I was misled by the wording of the question. (I am easily misled). My reading of it was that there had to be an answer whichever smallest number was given which is why I was looking for pairs of solutions in posts 1 and 6. The question to me implied that there would be an answer if 1 was given as the smallest number as well as if 2 were given.

However, I can claim to be the first with the right answer in post 1, even if it was for completely the wrong reasons!

April 5th, 2011 at 5:21 am

how abt 1,3,5,7?? product:105 least value:1

April 5th, 2011 at 5:44 am

srinu. 1357 gives the unique product 105. So if you knew the product was 105, you wouldn’t need to know the least value as well.

April 22nd, 2011 at 9:51 am

2,3,5,6