Braingle: 'A Very Good Year' Brain Teaser

A Very Good Year

Math brain teasers require computations to solve.

Puzzle ID:	#38403
Fun:	(2.74)
Difficulty:	(2.82)
Category:	Math
Submitted By:	soccerguy0428

The year 1978 has an unusual property. When you add the 19 to the 78, the total is the same as the middle two digits (97). What will be the next year to have this same property?

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Comments

air_up_there_jr Sep 22, 2007	nice.. how did you think of this?
MellowWillow Sep 23, 2007	Interesting.
jasmin215 Oct 28, 2007	boy!, you had me here with the calculator trying to add numbers...gave up at 1985 cute
(user deleted) Dec 11, 2007	What about '2042'
(user deleted) Dec 11, 2007	Never mind the last comment, sorry
avonma Dec 21, 2007	Sorry. I'm not very good at numbers...
surintan Dec 21, 2007	After year 2307 Subsequently you can just increase the second and third digit by 1 each for the next 6 numbers, i.e. 2417 2527 2637 2747 2857 2967
Nomez Dec 21, 2007	a1000+b100+c10+d = the year b9 = a10 + c9 + d Any a,b,c and d that are greater than zero and satisfy this equation will fit this puzzle Of course, this alone won't satisfy the puzzle. Finding the lowest one takes a little bit of trial and error. I actually got 2417 the first time I tried it. That's because I forgot to account for 0's Cool puzzle.
gghali Dec 21, 2007	It's actually a bit more defining than that: You know that the number can be represented as abcd. We also know that ab+cd = bc So, there are 2 possibilites: b+d b+d = c a+c = b b+d >= 10 (we need to "carry the 1") b+d = 10+c a+c+1 = b In the former case, combining the formulas leads to b+d = b-a, which cannot work unless a=d=0, which doesn't apply since a>0 (197. So we know that it is the latter case. Again, combining formulas, we get a+c+1 = 10+c-d a+d = 9 so we know that years that work must be of the form 1xx8 2xx7 3xx6 ... Plugging this back into b=a+c+1, we see that in the case of a=1, the second digit must be 2 greater than the third digit. Clearly, the last of the years 2000. So we know it is of the form 2xx7, and that b=c+3. So the minimum values for b and c are set when c=0 -> b=3. Thus, 2307.
coolguy5678 Dec 21, 2007	I did exactly the same thing as Nomez
leftclick Dec 21, 2007	int i; for (i=1979; ((int)Math.floor((double)i / 100.0) + (i % 100)) != (int)Math.floor((double)(i % 1000) / 10.0); i++) { } System.out.println(String.format("Found a match: %d", i));
leftclick Dec 21, 2007	Oh yeah, the output... Found a match: 2307
bradon182001 Dec 21, 2007	Yeah right. I agree with all of the above since I have no clue. This one must have taken a great deal of thought. Good job Soccerguy.
dstarbird Dec 21, 2007	Darn...I got 2417, which is the one after that. Owell...
leftclick Dec 21, 2007	Oh yeah and a slightly modified program gives the full list: 1208, 1318, 1428, 1538, 1648, 1758, 1868, 1978, 2307, 2417, 2527, 2637, 2747, 2857, 2967, 3406, 3516, 3626, 3736, 3846, 3956, 4505, 4615, 4725, 4835, 4945, 5604, 5714, 5824, 5934, 6703, 6813, 6923, 7802, 7912, 8901 1978 was a great year, btw Full marks to this teaser
tommysmo Dec 21, 2007	i didn't feel like doing the math this morning.
UlsterCharlotte Dec 21, 2007	Meh, I got 2747... I'm happy enough just to get a number that was right. Very cool puzzle, though.
phyllisa Dec 21, 2007	Intersting but I agree with tommysmo.
FatHead Dec 21, 2007	Wow, surintan and Nomez, I just used Guess and Check.
lfishbach Dec 21, 2007	This is a classic Diaphantine? Problem. Let us find all the years starting with a 2 that fit the solution. The year can be written as 2ABC. Then 20+A +10B +C must equal 10A + B. Thus 9A = 20 + 9B + C. A, B and C MUST BE integers. Dividing by 9 A = 2 + B + (2+C)/9 and C
mumbojumbo1234 Dec 21, 2007	2031 ;x
mumbojumbo1234 Dec 21, 2007	Oh heh misread it as first and last together get middle.
mi2mo2tx Dec 21, 2007	ouch -- too tough for my brain. kudos to all who attempted and then got it!!
UptheHill Dec 21, 2007	That was too hard, even if coffee was available!!!
dogg6pound9 Dec 21, 2007	wow... i need to go back to school after reading nomez and suritans algebraic solutions... i came up with 2857 based on the "guess and check" theory, but obviously it doesnt come next... i couldnt find a logical pattern to solve it, or think of an equation to help... i highly commend those who know their algebra to solve this quickly and accurately
dogg6pound9 Dec 21, 2007	i forgot to comment about the teaser ... this is the best teaser i've seen so far... 5 *'s to soccerguy for thinking up this one... and i have to ask on behalf of myself and i'm sure everyone else... how did you come up with this???
nerdyiscool Dec 21, 2007	Didn't like it... but I'm a much more linguistically minded individual. Math frightens me, and I personally have several friends two years younger than me who could sooo kick my butt in maths. I'm sure it was great, just not my thing.
auntiesis Dec 22, 2007	? Huh ?
mondayschild59 Dec 22, 2007	I didn't even try. Math is not my thing. Trivia is my forte. Thanks for posting. I know the others enjoyed it. Monday~
javaguru Jan 08, 2009	I was certain there was a solution that started with 2, so 2A + CD = BC [(B + D) / 10] + 2 + C = B and (B + D) % 10 = C (where [x] is floor x--the integer portion of x--and % is the mod, or remainder, operator) [(B + D) / 10] is either 0 or 1. This gives: (B + D) % 10 = B - [(B + D) / 10] - 2 If B + D is less than 10, then B + D = B - 2, which is clearly not possible, so B + D >= 10. So now B + D - 10 = B - 3 D = 7 Also B >= 3. Plugging 3 in for B gives C = (3 + 7) - 10 = 0 2307
javaguru Jan 08, 2009	The first equation in my previous post should be 2B + CD = BC
javaguru Jan 08, 2009	By the way, why are so many people that don't like/aren't good at math commenting on a math teaser? It's not like not being good at math is something to wear with pride.
opqpop Sep 11, 2010	This wasn't really hard. We can quickly see 1979, 198X, 199X isn't going to work, so we go into the 2000's. The first 2 digit number is already 20, so we skip to 22XY (X and Y not necessarily distinct) with X > 2 because we need the middle two digit number to be > 22. Obviously, this isn't going to work for any X, so we skip to 23XY. Starting with X=0, we immediately see 2307 works and we're done.
opqpop Sep 11, 2010	After reading the comments, I can see that my "brute-force" solution turned out to be very quick due to some luck... hurray