25 replies to this topic

[Wolf O'Donnell]

I am really stumped. Lately, I've been attempting New Scientist's puzzles, which tend to be very time consuming. The past few have been reasonably okay, but now they've set one that seems rather daunting. The puzzle is as follows:



In the addition sum above, the digits have been replaced by letters and smiley faces. Different letters stand for different digits, the same letter stands for the same digit, a smiley face can be any digit, and leading digits cannot be zero.

If FIVE is divisible by 5 and ELEVEN is divisible by 11, what number is FIFTEEN?

Now, I've figured out that E must be five as any number disivible by five ends with a 5 or a 0. Since no leading digit can be a 0, E can only be 5 as ELEVEN cannot start with a 0. Now previous puzzles I've attempted managed to replace either all 10 digits or 9 digits, which made things a lot easier. However, in this puzzle, only 7 digits have been replaced. I'm kinda stumped. In previous puzzles, I did all sorts of possible combinations but this is a bit too drastic with this puzzle.

There are all sorts of combinations of 5L5V5N where L, V and N can be any of nine digits. FIVE is just as difficult to work out. All I know is that F cannot be 0.

What's the best way to attack this puzzle?

[Diogenes]

 

What's the best way to attack this puzzle?

With a very sharp object, or a paper shredder.


But seriously, no clue.  I'll take a look at it later when I'm not so bogged down and (hopefully) give you a more helpful response.

[SteveT]

If FIVE is divisible by 5, then E has to be 0 or 5. Since ELEVEN starts with E, and numbers can't start with 0, then E =/= 0. Therefore E = 5.

That's as much as I've thought about the question so far. Does that help you at all?

[Veteran]

A smiley face can be ANY digit, meaning they're not all the same one? Well surely that makes it impossible!

[Diogenes]

Any digit, but can't lead the number with 0

[Veteran]

What I mean is, since any one smiley can be 1 to 9, and one smiley can be a different number from another smiley, surely there are an infinite number of variables that make this impossible to solve bar guess work.

[Diogenes]

I had a feeling the smiley was ONE digit.  Like, a variable.

[Showsni]

E=5

11|ELEVEN
Therefore,
11|E-L+E-V+E-N
11|15-(L+V+N)

(L+V+N)= one of 4 15 26

F >=3

A bit of trial and error, simultaneous equations and scribblings on a bit of paper later and I cam up with

F=4 I=0 E=5 V=8 L=6 N=1 T=3

Then FIFTEEN is 4043551
FIVE = 4085 which is divisible by 5.
ELEVEN = 565851 which is divisible by 11.
And

******8
1294085
1340859
1408599
_______
4043551

for example.

Hm, I haven't ruled out any other possible solutions...

[Wolf O'Donnell]

E=5

11|ELEVEN
Therefore,
11|E-L+E-V+E-N
11|15-(L+V+N)

(L+V+N)= one of 4 15 26

F >=3

A bit of trial and error, simultaneous equations and scribblings on a bit of paper later and I cam up with

F=4 I=0 E=5 V=8 L=6 N=1 T=3

Then FIFTEEN is 4043551
FIVE = 4085 which is divisible by 5.
ELEVEN = 565851 which is divisible by 11.
And

******8
1294085
1340859
1408599
_______
4043551

for example.

Hm, I haven't ruled out any other possible solutions...

You'll have to tell me how you came up with the idea that L+V+N must equal 4, 15 or 26 and how F must be greater than or equal to 3, because I can't quite see it.

[Veteran]

I can see why F has to be 3 or more. Since the end smilies can't be 0, the lowest F can be is 1 + 1 + 1.

But the L+V+N thing totally baffles me as well.

[Egann]

I can see why F has to be 3 or more. Since the end smilies can't be 0, the lowest F can be is 1 + 1 + 1.

But the L+V+N thing totally baffles me as well.

It's the divisibility rule of 11. (You DID realize that 11 was prime and, therefore, has a divisibility rule, I hope.) I tried tinkering with it as well, but I couldn't get it to work. It's a rather complicated divisibility rule compared to 5, 2, or 3, so it tends to not get taught.

Test for divisibility by 11.

Subtract the last digit from the remaining leading truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary.
Example: 19151--> 1915-1 =1914 -->191-4=187 -->18-7=11, so yes, 19151 is divisible by 11.

Edited by Egann, 30 January 2010 - 09:51 AM.

[Wolf O'Donnell]

I can see why F has to be 3 or more. Since the end smilies can't be 0, the lowest F can be is 1 + 1 + 1.

But the L+V+N thing totally baffles me as well.

Of course, how stupid of me! By the way, Showsni, you could earn £15 by submitting that answer of yours.

£15 will be awarded to the sender of the first correct answer opened on Wednesday 24 February. The Editor's decision is final. Please send entries to Enigma 1579, New Scientist, Lacon House, 84 Theobald's Road, London WC1X 8NS, or to enigma@newscientist.com (please include your postal address).

I just do it for the challenge, because I know I'm not going to win. (After all, it's basically a lottery as to whether your answer is the first to be opened and I hate lotteries).

Edited by Wolf_ODonnell, 30 January 2010 - 09:50 AM.

[Veteran]

[quote name='Egann' date='30 January 2010 - 02:44 PM' timestamp='1264862665' post='504352']
[quote name='Veteran' date='30 January 2010 - 08:32 AM' timestamp='1264858337' post='504350']
I can see why F has to be 3 or more. Since the end smilies can't be 0, the lowest F can be is 1 + 1 + 1.

But the L+V+N thing totally baffles me as well.
[/quote]

It's the divisibility rule of 11. (You DID realize that 11 was prime and, therefore, has a divisibility rule, I hope.) I tried tinkering with it as well, but I couldn't get it to work. It's a rather complicated divisibility rule compared to 5, 2, or 3, so it tends to not get taught.

[quote='more info than you ever cared for']Test for divisibility by 11.

Subtract the last digit from the remaining leading truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary.
Example: 19151--> 1915-1 =1914 -->191-4=187 -->18-7=11, so yes, 19151 is divisible by 11. [/quote]

Nope, not heard of that rule before. So is it possible to write that rule algebraicly?

[Showsni]

If a number divides 11; in this case ELEVEN; then E+L-E+V-E+N divides 11... Or something like that. You only use that bit to work out what N can be, really, as the sum doesn't give any indication.

My working: first, I wrote a lot of simultaneous equations! One for each column of the sum. The (1), (2) etc. refer to the smiley faces; x, y, z, etc. are the tens digits of each column... so, substituting in E=5:

(1) + 5 + (7) + (10) = 10x + N
x + V + 5 + (9) = 10y + 5
y + I + V + 5 = 10z + 5
z + F + I + V = 10a + T
a + (4) + F + I = 10b + F
b + (3) + (6) + F = 10c + I
c + (2) + (3) + (8) = F

Each variable, being a digit, is between 0 and 9 inclusive. Then start by looking at x; it's at most 3, at least 0. Trial and error! Let's start by guessing x=3. Subbing into the second equation and simpifying:

3 + V + (9) = 10y

Then y is either 1 or 2. Trial and error! Let's guess 2. Then V is either 8 or 9. Looking at equation 3:

2 + I + V = 10z

If V=9, I=9. Error! So V=8, I=0, z=1.

F + 9 = 10a + T
a + (4) = 10b

F =/= 0, so so a>0. Then b=1, a=1, (4)=9 and F = T + 1

1 + (3) + (6) + F = 10c

L+V+N = L + N + 8
So, it can't be 4; if it were 26, L=N Error! So L + N = 7

Since we said x=3, N=0,1 or 2. I is already 0, and if N=2 L=5 which is E; so N=1 L=6. F and T are consecutive with F>3; with 5,6,8 already gone, we must have F=4 T=3.

And it checks out on subbing back in. Nice! Got it on my first trial and error guess!

[Wolf O'Donnell]

I need a bit more help again.

This time the puzzle features some annotation I don't understand.

The equation is as follows:

(A.B + 1) divided by (A.B.C + A + C) = 0.138



I'm pretty sure A.B is some kind of decimal number, but what's A.B.C?

[Showsni]

A point like that is the symbol for multiplication, Wolf. A.B means A times B; A.B.C means A time B times C.

Oh, and the answer is:

Spoiler

A=17 B=4 C=7

That was actually a pretty easy one.

[SteveT]

Is that a British thing? I'm used to seeing an x, *, parentheses, or a point centered vertically of the line, but never at the bottom of the line. Not that my keyboard can do that.

So many ways to multiply....

Oh, and the inverse log of the sum of the logs of the two numbers. That should also do it.

Edited by SteveT, 21 February 2010 - 10:35 PM.

[Veteran]

A dot used to represent multiply is only used when letters are involved to avoid confusion. Numbers we'd just use the regular x. Do you guys also put a line through a 'z' to distinguish it from a '2'?

If the dot was in the middle of the line it would be a vector dot product.

[Showsni]

Usually you'd just miss out the dot and write ABC, but I guess they didn't want people to think it was a three digit number. Once you're past GCSEs, a x symbol is pretty much never used...

The way to get the answer for that puzzle is pretty neat. I liked it.

[SteveT]

Do you guys also put a line through a 'z' to distinguish it from a '2'?

Only people who do a lot of math. It's actually one of the easiest ways to tell, in my experience. I also loop my lower-case L to distinguish from a 1, since my handwriting is sans- serif.

If the dot was in the middle of the line it would be a vector dot product.

Ah, ok. We always put it in the middle, and trust everyone to keep track of what's a vector and what isn't. (Likewise with the x and a cross-product).

Once you're past GCSEs, a x symbol is pretty much never used...

Similar situation. Once you start doing algebra with unknowns, using an x for multiplication is a bad idea.

I always stick to parentheses, myself. Personal preference.

Edited by SteveT, 22 February 2010 - 07:27 PM.

[Veteran]

I was also taught to do a curly 'x' for the letter to distinguish it from multiply. This was before they broke out the dot and ripped up the rule book.



It was a pain in chemistry when I had to scribble zeta and xi all the damn time. The Greeks were on drugs when they designed those two letters.

[SteveT]

I've had classes where you'd use E, epsilon, and script E in the same equation.

And yes, having to draw zeta is cruel and unusual punishment.

And I meant draw. No one can write a zeta.

Edited by SteveT, 22 February 2010 - 07:48 PM.

[Wolf O'Donnell]

A point like that is the symbol for multiplication, Wolf. A.B means A times B; A.B.C means A time B times C.

Oh, and the answer is:

Spoiler

A=17 B=4 C=7

That was actually a pretty easy one.

Oh, I never knew that. When I was in GCSE, I always took ABC to mean A multiplied by B multiplied by C. Well, that makes the puzzle a lot easier to solve now that I know the dots mean multiply. Thanks, Showsni.

I've had classes where you'd use E, epsilon, and script E in the same equation.

And yes, having to draw zeta is cruel and unusual punishment.

And I meant draw. No one can write a zeta.

What about the Greeks?

Edited by Wolf O'Donnell, 24 February 2010 - 12:51 PM.

[SteveT]

Not even the Greeks. They made that letter up as a joke.

[nick]

I would like a good math puzzles website. Where is it you are getting these?

[Wolf O'Donnell]

I would like a good math puzzles website. Where is it you are getting these?

The New Scientist magazine. It's published in their weekly issues, which are available in the US as well as the UK and Australia. They even have a website, the URL of which is exactly what you might think it is...

http://www.newscientist.com/