000RM.dvi

416 Numbers with many repeating digits

14.1 A quick multiplication..................

What is the smallest positive integer with the property that when the digit

1 is appended to both ends, the new number is 99 times the original?

Suppose the numberXhasndigits. We require 10 n+10X+1 = 99X

or 89 X=10n+1. Note that 89 is prime. To find the smallestnwe divide

89 into the number 1 followed by a string of zeros, extended if necessary,

until a remainder 80 occurs. Then add 1 and obtain an integer quotient

which is the smallest possibleX.

The first time this occurs is at the 22nd zero. Thus, the smallest pos-

sible value ofXis

1022 +1

89

= 112359550561797752809.

For this value ofXwe have

99 ×112359550561797752809 = 11123595505617977528091.

See Appendix for the factorization of numbers of the form (^10) n− 11.