Find the smallest integer such that the last six digits of coincide with those of , in the same order.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
It is required that 1 0 6 = 5 6 ∗ 2 6 divides n 2 − n = n ( n − 1 ) . Now 5 6 = 1 5 6 2 5 must divide one of the numbers n and n − 1 , while 2 6 = 6 4 must divide the other. Thus one of the numbers n and n − 1 must be of the form 1 5 6 2 5 k , and it must be congruent to 1 or 63 modulo 64. Now 1 5 6 2 5 ≡ 9 ( m o d 6 4 ) , so that n − 1 = 7 ∗ 1 5 6 2 5 leads to the smallest number with the required property, n = 7 ∗ 1 5 6 2 5 + 1 = 1 0 9 3 7 6