the sum of N positive integers is 19. what is the maximum possible product of these N numbers?
thx.
Nice problem.
Let's look at some cases
1. N=2
clearly our only logical choices are 9,10 for a product of 90
It should be obvious that the numbers should be "centrally" positioned
e.g 2 and 17 add up to 19, but only have a product of 34
2. N=3
I considered 4,7,8 for a product of 224, 5,6,8 for a product of 240 and 2,8,9 for a product of only 144.
3. N=4
You should quickly realize that we can't start too high or we will go over 19
so 2,3,4,10 ---> product 240
2,3,5,9 ---> product 270
2,3,6,8 ---> product 288
3,4,5,7 ---> product 420 *!*!
4. N=5
We have to start with 1,2,3 or else we run over,
e.g. if we start with 2,3,4,5 we already have a sum of 14, so we need a 5, but we already used it.
so the only choices would be
1,2,3,4,9 ---> product 216
1,2,3,5,8 ---> product 240
1,2,3,6,7 ---> product 252
so it looks like the 4 numbers 3,4,5,7 which have a sum of 19 and a maximum product of 420 are it!
okay i get it!so you will just have to try out all the possible way!thanks for the help~
Do the numbers have to be different?
3+3+3+3+3+2+2=19,
3^5x2^2=972
this suks
this was poster on the day i was born HAHHA
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Find the product.
13 × (–5) × (–6)
To find the product, we simply multiply the three numbers:
13 × (–5) × (–6) = 390
Find the quotient.
–14 ÷ (–2)
To find the quotient, we simply divide -14 by -2:
-14 ÷ (-2) = 7
Therefore, -14 ÷ (-2) = 7.