### Brain Teasers

# A Very Good Year

Math
Math brain teasers require computations to solve.

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?

### Answer

The next year to have the same property will be 2307.23 + 07 = 30

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## Comments

nice.. how did you think of this?

Interesting.

boy!, you had me here with the calculator trying to add numbers...gave up at 1985

cute

cute

Dec 11, 2007

What about '2042'

Dec 11, 2007

Never mind the last comment, sorry

Sorry. I'm not very good at numbers...

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

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

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.

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.

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.

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.

I did exactly the same thing as Nomez

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));

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));

Oh yeah, the output...

Found a match: 2307

Found a match: 2307

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.

Darn...I got 2417, which is the one after that. Owell...

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

1978 was a great year, btw Full marks to this teaser

i didn't feel like doing the math this morning.

Meh, I got 2747... I'm happy enough just to get a number that was right. Very cool puzzle, though.

Intersting but I agree with tommysmo.

Wow, surintan and Nomez, I just used Guess and Check.

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

9*A = 20 + 9*B + C. A, B and C MUST BE integers. Dividing by 9

A = 2 + B + (2+C)/9 and C

2031 ;x

Oh heh misread it as first and last together get middle.

ouch -- too tough for my brain. kudos to all who attempted and then got it!!

That was too hard, even if coffee was available!!!

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

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???

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.

? Huh ?

I didn't even try. Math is not my thing. Trivia is my forte. Thanks for posting. I know the others enjoyed it.

Monday~

Monday~

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

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

The first equation in my previous post should be

2B + CD = BC

2B + CD = BC

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.

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.

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.

After reading the comments, I can see that my "brute-force" solution turned out to be very quick due to some luck...

hurray

hurray

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