Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.
First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.
Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:
1, 3, 5, 7, 9, 11, 13, 15 .
Number of items in the set . . . 8
Sum of all the items in the set . . . 64
Hmmm. That's interesting. 64 happens to be the square of 8 .
Do you think that might be all there is to it ?
Let's check it out:
Even-numbered lists of odd numbers:
1, 3 Items = 2, Sum = 4
1, 3, 5, 7 Items = 4, Sum = 16
1, 3, 5, 7, 9, 11 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .
Amazing ! The sum is always the square of the number of items in the set !
For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.
I slogged through it on my calculator, and it's true.
I never knew this before. It seems to be something valuable
to keep in my tool-box (and cherish always).
Answer:
I don't now why but by seeing this photo my mouth is watery
Answer:
f(5) = 4375
Step-by-step explanation:
Given f(n) = 5f(n - 1) and f(1) = 7
This allows us to find the next term in the sequence from the previous term
f(2) = 5f(1) = 5 × 7 = 35
f(3) = 5f(2) = 5 × 35 = 175
f(4) = 5f(3) = 5 × 175 = 875
f(5) = 5f(4) = 5 × 875 = 4375
Step-by-step explanation:
We are picking 6 numbers from the numbers 1,2,3,4,5,6,7,8,9,10. Since we care about numbers being next to each other, we might think of the 10 numbers as being distributed in 5 boxes (which you can think of as the holes):
| 1 2 | 3 4 | 5 6 | 7 8 | 9 10 |
So on the first box we have the numbers 1 and 2, on the second box we have the numbers 3 and 4, and so on. Since we are picking 6 numbers from those 10 numbers, that means we'll have to pick 6 boxes (and inside each box we pick a number), but we only have 5 available boxes, so by the pigeonhole principle, we'll have to pick 1 same box at least two times. Since on each picked box we'll need to pick a number, on this box which was picked two times, we will have to pick both of its numbers. And so those 2 numbers inside that box will be next to each other (meaning they're consecutive numbers).
