SOLUTION: 1. What’s the smallest number that you can write with 1’s and 0’s which is divisible by 225? 2. The divisors of 216,000 add up to 792,480. What do their reciprocals add up to?

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Question 86891: 1. What’s the smallest number that you can write with 1’s and 0’s which is divisible by 225?

2. The divisors of 216,000 add up to 792,480. What do their reciprocals add up to?
3. Find the product: 1·2·4·8·16· . . . · 2(to the 399th power)

4. All natural numbers from 1 to 101 are written in a row. How can the signs “+” and “-” be placed between them so that the value of the resulting expression is 0?


Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
1. What’s the smallest number that you can write with 1’s and 0’s which is divisible by 225?


The multiples of 225 end in 25, 50, 75, or 00. Only those ending in 00 could
contain only the digits 1's and 0's. Those are the multiples of 4×225 or 900.
So we need a multiple of 900.


The multiples of 900 are just the multiples of 9 with two 0's annexed.


Theorem: A positive integer is divisible by 9 if and only if the sum of its digits is a multiple of 9.


So the smallest integer which contains only 1's and 0's and has sum of digits a multiple of 9 is 111111111 (9 1's). Then the smallest multiple of 900 that has only 1's and 0's is found by annexing two 0's to that. So the answer is 11111111100.




2. The divisors of 216,000 add up to 792,480. What do their reciprocals add up to?


Theorem: The sum of the reciprocals of positive integer n =
the sum of the divisors of n divided by n.


Let's do it with a smaller number like 12, so you can see why.


The divisors of 12 are 1,2,3,4,6,12. Their sum is 28.


Look at the sum of the reciprocals:


1%2B1%2F2%2B1%2F3%2B1%2F4%2B1%2F6%2B1%2F12


Obviously 12 is the LCD, so when we get the LCD of 12, we have


12%2F12%2B6%2F12%2B4%2F12%2B3%2F12%2B2%2F13%2B1%2F12


or


%2812%2B6%2B4%2B3%2B2%2B1%29%2F12


The numerator is just the sum of the divisors of 12, and
thus equals to the fraction 28%2F12 which reduces to 7/3


Let the set of divisors of 216,000 be S =


{1,2,3,4,5,6,8,9,10,12,15,16,18,20,...,216000}.


if x e S then 216000/x e S.


Look at this sum:




If we get the LCD of 216000, we have


216000%2F216000%2B108000%2F216000+...+1%2F216000


or the fraction


216000+108000+...+1
-------------------
`````216000


and the numerator is just the sum of the divisiors of 216000
which we are given to be 7982480.


So the answer is 792480%2F216000. Both numerator and
denominator can be divided by 480 to reduce that fraction to


1651%2F450




3. Find the product: 1·2·4·8·16· . . . · 2(to the 399th power)


2%5E0·2%5E1·2%5E2·2%5E3·2%5E4·...·2%5E399
The sum of those exponents are the sum of the first 399 positive
integers.


The formula for the first n positive integers is n%28n%2B1%29%29%2F2


So the sum of the exponents is %28399%28400%29%29%2F2 = 79800,


making the answer 2%5E79800


I'll come back to the last one. I haven't time just now.


Edwin