Question 1133243
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The smallest possible sum of 10 (1 more than 9) distinct powers of 2 is

{{{2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9 = 1+2+4+8+16+32+64+128+256+512 = 1023}}}
We must omit 1 of those powers of 2 to get a sum of exactly 9 distinct ones.
1023 is a 4-digit number.  To get the largest 3-digit number, 999, we must
subtract 24 from 1023, and to get the smallest 3-digit number, 100, we must
subtract 923 from 1023.

So to get a 3-digit number from that sum, we must omit powers of 2 that
are between 24 and 923.  The smallest power of 2 between those is 32, which
is 2<sup>5</sup> and and the largest power of 2 between those is 512, which is 2<sup>9</sup>. 
So there are 5 powers of 2 that we can omit from the sum of the first
10 powers of 2, to get the sum of 9 powers of 2 that is a 3-digit number.  Answer: 5     
 
Here are all 5 ways to omit a power of 2 from the sum of the smallest 10 powers
of 2 that will give the sum of 9 powers of 2 that will be a 3-digit number: 

{{{2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8 = 1+2+4+8+16+32+64+128+256 = 511}}}
{{{2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^9 = 1+2+4+8+16+32+64+128+512 = 767}}}
{{{2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^8+2^9 = 1+2+4+8+16+32+64+256+512 = 895}}}
{{{2^0+2^1+2^2+2^3+2^4+2^5+2^7+2^8+2^9 = 1+2+4+8+16+32+128+256+512 = 959}}}
{{{2^0+2^1+2^2+2^3+2^4+2^6+2^7+2^8+2^9 = 1+2+4+8+16+64+128+256+512 = 991}}}

Edwin</pre>