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More ways to get Braingle...

A Very Good Year

Math brain teasers require computations to solve.

 

Puzzle ID:#38403
Fun:*** (2.74)
Difficulty:*** (2.82)
Category:Math
Submitted By:soccerguy0428Aus*!!!!!

 

 

 



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_jrca*
Sep 22, 2007

nice.. how did you think of this?
MellowWillowus*
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
avonmaAus*
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

a*1000+b*100+c*10+d = the year

b*9 = a*10 + c*9 + 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.
coolguy5678Aza*
Dec 21, 2007

I did exactly the same thing as Nomez
leftclickAau*
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));

leftclickAau*
Dec 21, 2007

Oh yeah, the output...

Found a match: 2307
bradon182001*us*
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.
dstarbirdAus*
Dec 21, 2007

Darn...I got 2417, which is the one after that. Owell...
leftclickAau*
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.
UlsterCharlotteAus*
Dec 21, 2007

Meh, I got 2747... I'm happy enough just to get a number that was right. Very cool puzzle, though.
phyllisa*us*
Dec 21, 2007

Intersting but I agree with tommysmo.
FatHeadAus*
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 +10*B +C must equal 10*A + B. Thus
9*A = 20 + 9*B + C. A, B and C MUST BE integers. Dividing by 9

A = 2 + B + (2+C)/9 and C
mumbojumbo1234A*
Dec 21, 2007

2031 ;x
mumbojumbo1234A*
Dec 21, 2007

Oh heh misread it as first and last together get middle.
mi2mo2tx*us*
Dec 21, 2007

ouch -- too tough for my brain. kudos to all who attempted and then got it!!
UptheHillA*
Dec 21, 2007

That was too hard, even if coffee was available!!!
dogg6pound9Aus*
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
dogg6pound9Aus*
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*us*
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.
auntiesisAus*
Dec 22, 2007

? Huh ?
mondayschild59Aus*
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*us*
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*us*
Jan 08, 2009

The first equation in my previous post should be

2B + CD = BC
javaguru*us*
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



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