If a, b be two numbers prime to each other, a value may be found for n which will make (a^{n} − 1) a multiple of b.
Let k be a No. not containing 2 or 5 as a factor, i.e. let it be prime to 10. Then, if
(1)/ |
(k) |
(1)/ |
(k) |
This evidently holds good in any other scale of notation. Hence, if a be the radi× of the scale of notation, and b be a No. prime to a, a value may be found for n, which will make (a^{n} − 1) a multiple of b.
Q.E.D.
(1) With radix 10, find a value for n, which will make (10^{n} − 1) a multiple of 7.
1 |
7 |
142857 |
10^{6} − 1 |
Ans. n = 6.
(2) Let the two given Nos. be 8, 9.
Taking 8 as radix, we get
= .·0·7 =
1
9
7 |
8² − 1 |
Ans. n = 2.
Let the two given Nos. be 7, 13.
Taking 7 as radix, we get
= .·03524563142·1 =
1
13
35245631421 |
7^{12} − 1 |
Ans. n = 12.