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# Recurring Decimal Or Not

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If N is an odd prime number > 10 is (N+(N+1)) /1100) always recurring ?

try :

(11+12)/1100

(13+14)/1100

(17+18)/1100

(19+20)/1100

(23+24)/1100 ...

and so on

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petrush;

Almost !

If p is a prime>11 then d is a repeating decimal.

d=p(p+1)/1100

d=[p/11][(p+1)/100]

The 1st term has no common divisors.

The 2nd term will have 2 decimal positions.

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petrush;

Almost !

If p is a prime>11 then d is a recurring decimal.

d=p(p+1)/1100

d=[p/11][(p+1)/100]

The 1st term has no common divisors.

The 2nd term will have 2 decimal positions.

Edited by sluggo
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• 6 months later...

I thought it was well known that any fraction has a repeating decimal expansion!  (I am thinking of terminating decimals as repeating "0"s.)

When you are dividing integer n by integer m,  you write n as n.00000000... Eventually you get to "bringing down" those "0"s.  Further, each time you divide you bring down a remainder smaller than m (if you bring down "0" it stops and you have a terminating decimal).  There are only a finite number of integers smaller than m so eventually you will have a remainder that you had before- then you will down a 0 like you did before and every thing will repeat!

For example to convert the fraction 3/7 to its decimal equivalent, you divide 7 into 3.000000...  7 divides into 3.0 .4 times with remainder 3.0- 2.8= 0.2.  7 divides into 0.20 0.02 times with remainder 0.20- 0.14= 0.06. 7 divides into 0.060 .008 times with remainder 0.060- 0.056= 0.004.  7 divides into 0.0040 0.0005 times with remainder 0.0040- 0.0035= 0.0005.  7 divides into 0.00050 0.00007 times with remainder 0.00050- 0.00049= 0.00001. 7 divides into 0.000010 0.000001 times with remainder 0.000010- 0.000007= 0.000003.

Notice that we have had remainders with last digits 2, 6, 4, 5, 1, 3.  Since any remainder must have last digit less than 6,  the next division must either give a remainder of 0, and so be a terminating decimal, or one of those 6  in which case everything repeats, we bring down the same "0", to get the same number again,, divide by 7 to get the same quotient and same remainder continuing the repetition.

Here that last remainder was 3, the same as the original numerator so 7 will, again, divide into 30 4 times with remainder 2, etc.

3/7= 0.428571428571428571.... with "428671" repeating.

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HallsofIvy;

With repeating values that alter the value of the expression as it is extended.

Repeating 0’s are redundant, the expression doesn’t change.

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