Perfect Squares (Posted on 2003-04-05)

	Submitted by Ravi Raja
Rating: 4.0000 (3 votes)
Solution:	(Hide)
Let the given number be 444444 4444888888 8889, where the digit 4 appears ‘k’ times, the digit 8 appears ‘(k – 1)’ times and the digit 9 appears exactly once (and always in the unit’s place). Let N(k) = 444444 4444888888 8889, where the digit 4 appears ‘k’ times, the digit 8 appears ‘(k – 1)’ times and the digit 9 appears exactly once. N(k) – 1 = 444444 4444888888 8888, where both the digits 4 and 8 now appear ‘k’ times. Therefore, it is clear that {N(k) – 1} is divisible by the number 111111 1111, where there are ‘k’ 1’s and the quotient thus obtained will be 4000000 0008, where there is exactly one 4 (the leftmost digit) and exactly one 8 (the rightmost digit) separated by ‘(k – 1)’ zeros (0’s) in between. Also, we can see that the numbers 408, 4008, 40008, 400008, and similar numbers with exactly one 4 (the leftmost digit) and exactly one 8 (the rightmost digit) separated by any number of zeros (0’s) in between is an exact multiple of 6, giving quotients 68, 668, 6668, 66668, (with the same number of 6’s in the quotient as there are number of zeros in the dividend). Hence, the number 4000000 0008, where there is exactly one 4 (the leftmost digit) and exactly one 8 (the rightmost digit) separated by ‘(k – 1)’ zeros (0’s) in between is also a multiple of 6, the quotient being 666666 6668, where there are ‘(k – 1)’ 6’s and one 8. So, from the above results, the number {N(k) – 1} can be written as the product of two factors as follows: {N(k) – 1} = 444444 4444888888 8888, where both the digits 4 and 8 now appear ‘k’ times. = (666666 6666)x(666666 6668), where there are ‘(k)’ 6’s in the first factor and ‘(k – 1)’ 6’s and one 8 in the second factor. Again, the two factors can be rewritten in the form: (666666 6667 – 1)x(666666 6667 + 1), which is of the form (A – 1)(A + 1); and A = 666666 6667, where there are ‘(k – 1)’ 6’s and ONE 7. Thus, we now have the following form for N(k): {N(k) – 1} = (A – 1)(A + 1) or, N(k) – 1 = (A)^2 – 1 or, N(k) = (A)^2, where A = 666666 6667, where there are ‘(k – 1)’ 6’s and ONE 7. Thus proving that every number of the form 444444 4444888888 8889, [where the digit 4 appears ‘k’ times, the digit 8 appears ‘(k – 1)’ times and the digit 9 appears exactly once (and always in the unit’s place)], is always a PERFECT SQUARE.

	Subject	Author	Date
	Even simpler.	broll	2012-09-17 03:10:19
	Solution(simple)	Praneeth Yalavarthi	2007-07-10 13:06:21
	re(3): solution	Ravi Raja	2003-04-07 04:29:26
	Partial idea	Gamer	2003-04-06 05:03:37
	re(2): solution	Gamer	2003-04-06 04:44:16
	re: solution	Ravi Raja	2003-04-06 04:23:41
	re: solution	Ravi Raja	2003-04-06 04:18:56
	re: solution	Ravi Raja	2003-04-06 04:12:57
	solution	Charlie	2003-04-05 13:42:37
	Umm	DJ	2003-04-05 05:56:26