Question 86891
1. What’s the smallest number that you can write with 1’s and 0’s which is divisible by 225?
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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.
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The multiples of 900 are just the multiples of 9 with two 0's annexed.  
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Theorem:  A positive integer is divisible by 9 if and only if the sum of its digits is a multiple of 9.  
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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.  
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2. The divisors of 216,000 add up to 792,480. What do their reciprocals add up to? 
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Theorem:  The sum of the reciprocals of positive integer n =
the sum of the divisors of n divided by n.
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Let's do it with a smaller number like 12, so you can see why.
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The divisors of 12 are 1,2,3,4,6,12.  Their sum is 28.
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Look at the sum of the reciprocals:
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{{{1+1/2+1/3+1/4+1/6+1/12}}}
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Obviously 12 is the LCD, so when we get the LCD of 12, we have
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{{{12/12+6/12+4/12+3/12+2/13+1/12}}}
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or
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{{{(12+6+4+3+2+1)/12}}}
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The numerator is just the sum of the divisors of 12, and
thus equals to the fraction {{{28/12}}} which reduces to 7/3
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Let the set of divisors of 216,000 be S = 
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{1,2,3,4,5,6,8,9,10,12,15,16,18,20,...,216000}.  
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if x e S then 216000/x e S. 
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Look at this sum:
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{{{1/1+1/2+1/3+1/4+1/5+1/6+1/8+1/9+1/10+1/12+1/15+1/16+1/18+1/20+...+1/216000}}}<br> 
If we get the LCD of 216000, we have
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{{{216000/216000+108000/216000}}}+...+{{{1/216000}}}
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or the fraction
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216000+108000+...+1
-------------------
`````216000
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and the numerator is just the sum of the divisiors of 216000
which we are given to be 7982480.
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So the answer is {{{792480/216000}}}. Both numerator and
denominator can be divided by 480 to reduce that fraction to
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{{{1651/450}}}
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3. Find the product: 1·2·4·8·16· . . . · 2(to the 399th power) 
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{{{2^0}}}·{{{2^1}}}·{{{2^2}}}·{{{2^3}}}·{{{2^4}}}·...·{{{2^399}}}
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The sum of those exponents are the sum of the first 399 positive 
integers.
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The formula for the first n positive integers is {{{n(n+1))/2}}}
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So the sum of the exponents is {{{(399(400))/2}}} = 79800,
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making the answer {{{2^79800}}}
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I'll come back to the last one. I haven't time just now.
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Edwin