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# 11*101*10001*100000001*10000000000000001*100000000000000000000000000000001...

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1200
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11*101*10001*100000001*10000000000000001*100000000000000000000000000000001...

If factor number m has n zeros, then factor number m+1 will have 2n+1 zeros.

11*101=  1111

11*101*10001= 11111111

11*101*10001*100000001= 1111111111111111

11*101*10001*100000001*10000000000000001= 11111111111111111111111111111111

Will the product always be repuit (natural numbers where all the digits are 1)?

If it is the case, then what's the reason for that?

Sep 27, 2014

#3
+17768
+5

11 = 10 + 1 = 10^1 + 1

101 = 10^2 + 1

11 x 101 = (10^1 + 1)(10^2 + 1)= 10^3 + 10^2 + 10^1 + 1

(The problem starts with 11 -- which is 10^1 + 1 -- Now mutiplying by 10^2 results in new terms -- 10^3 + 10^2 -- which are just larger than the old terms (which are saved by multiplying by 1) )

10001 = 10^4 + 1

11 x 101 x 1001 = (10^3 + 10^2 + 10^1 + 1) x (10^4 + 1) = 10^7 + 10^6 + 10^5 + 10^4 + 10^3 + 10^2 + 10^1 + 1

(The old answer was 1111 = 10^3 + 10^2 + 10^1 + 1 -- Now multiplying by 10^4 results in the new terms -- 10^7 + 10^6 + 10^5 + 10^4 -- which are just larger than the old terms 10^3 + 10^2 + 10^1 + 1 (which are saved by multiplying by 1) )

So, if we multiply by a number -- a power of ten that results in sufficiently new terms -- we will get an answer that begins with only 1s and if we then append to that the old terms by multiplying the old terms by 1, we will get an answer that has only 1s.

The necessary power of ten in the new factor is detemined by your requirement for the number of zeroes.

Sep 27, 2014

#1
+109
0

Yes, the product will always be repuit; I don't know the answer, but there must be one because anomalies do not occur very often in science/math. I am curious what the answer is now.

Sep 27, 2014
#2
+3

Thx for the answer. I guess there are two of us now :)

Sep 27, 2014
#3
+17768
+5

11 = 10 + 1 = 10^1 + 1

101 = 10^2 + 1

11 x 101 = (10^1 + 1)(10^2 + 1)= 10^3 + 10^2 + 10^1 + 1

(The problem starts with 11 -- which is 10^1 + 1 -- Now mutiplying by 10^2 results in new terms -- 10^3 + 10^2 -- which are just larger than the old terms (which are saved by multiplying by 1) )

10001 = 10^4 + 1

11 x 101 x 1001 = (10^3 + 10^2 + 10^1 + 1) x (10^4 + 1) = 10^7 + 10^6 + 10^5 + 10^4 + 10^3 + 10^2 + 10^1 + 1

(The old answer was 1111 = 10^3 + 10^2 + 10^1 + 1 -- Now multiplying by 10^4 results in the new terms -- 10^7 + 10^6 + 10^5 + 10^4 -- which are just larger than the old terms 10^3 + 10^2 + 10^1 + 1 (which are saved by multiplying by 1) )

So, if we multiply by a number -- a power of ten that results in sufficiently new terms -- we will get an answer that begins with only 1s and if we then append to that the old terms by multiplying the old terms by 1, we will get an answer that has only 1s.

The necessary power of ten in the new factor is detemined by your requirement for the number of zeroes.

geno3141 Sep 27, 2014
#4
0

Thx for the reply geno3141 :)

I knew someone would get the answer :)

Sep 27, 2014