Find the smallest odd positive integer m such that the decimal expansion of 1/m has just 4 non-recurring digits and 2 repeating digits.

Question

Another maths problem i need help with any help is appreciated, Thanks!

Answer 1

I assume by non-recurring you mean the lag.

the non-recurring digits make up the lag that precedes the repetend

these digits occur when there are 2s or 5s in the denominator

the number 11 in the denominator will give you a 2-digit repetend

to get 4 digits in the lag you need 4 5s in the denominator; you can't use 4 2s because you want an odd integer for m

any 2 will make the integer m an even number

m=5*5*5*5*11=6875

1/6875=0.00014545454545...