Math Puzzle

[MindBlownMagicIllusion]

I’m seeing a lot of answers with possible solutions that could work. I’m looking for a single equation that proves the correct answer.

[prathmeshbojalwar]

All a, b, c, and d are made up of boxes so all of them are positive integers.
Now, we only have 3 equations and 4 variables, so there will be no trivial solution, but rather a series of solutions.
Let's see the range of each variable
a is made up of 2 boxes so a > 1
b is made up of 1 box so b > 0
c is made up of 1 box so c > 0
d is made up of 3 boxes d > 2
Let's narrow down the solution with the given equations

a+b = 13 -> b = 13 - a -> a < 13 (as b is positive whole no.)
similarly, b < 13

b+c=7 -> b = 7 - c -> c < 7 and also, b < 7 (same logic as above)

c+d = 18 -> c < 18 and d < 18 (same logic as above)

now put b < 7 in a+b = 13, will give us
6<a<13

Similarly, from c < 7 and c+d = 18, we get
11<d<18

The final range will be
6<a<13 (Possible Values : 7, 8, 9, 10, 11, 12)
0<b<7 (Possible Values : 1, 2, 3, 4, 5, 6)
0<c<7 (Possible Values : 1, 2, 3, 4, 5, 6)
11<d<18 (Possible Values : 12, 13, 14, 15, 16, 17)

So there will be only six solutions as follows :
7, 6, 1, 17
8, 5, 2, 16
9, 4, 3, 15
10, 3, 4, 14
11, 2, 5, 13
12, 1, 6, 12

Now if you consider the boxes in different variables are of different sizes but the boxes making any individual variable are of the same size, it makes sense to group same size boxes. This gives us

a = 2w
b = x
c = y
d = 3z

as d is divisible by 3 it narrows down our 6 solutions to 2. Because in the possible values of d, only 12 and 15 are divisible by 3. so the narrowed-down solutions will be

12, 1, 6, 12
9, 4, 3, 15

again we know a is divisible by 2, so we can discard solution 2, so we get the answer as

12, 1, 6, 12

[jpsell]

A is 12, B is 1, C is 6, D is 12.

It works in every scenario with substitution.

[rileyburdick]

A=11 B=2 C=5 D=13

11+2=13 which is A+B
2+5=7 which is B+C
5+13=18 which is C+D
11+13=24 which is A+D
11+2+5+13=31 which is A+B+C+D

[kchapi]

Many options. I chose 8, 5, 2, 16 for no reason other than it was my first calculation and it adds up, but it’s not the only one.

A+B 8+5=13
B+C 5+2=7
C+D 2+16=18
A+D 8+16=24
A+B+C+D=31

Or, to answer your actual question: what is the value of any single letter? A=8 based on my above math.

[artinteymourian]

You have 3 equations and 4 unknowns. That means there can be multiple solutions that are all valid. As you can see, several comments have already presented different answers that all work.

[romain]

there are a lot of solution that work if it isnt related to the number of boxes concerned

[daniworsley]

It is not stated but it seems logical to assume that each box MUST be a different number from all other boxes.
Since we know the following to be correct:

A+D is 24
A+B is 13
B+C is 7
C+D is 18
And the total sum of all letters is 31.

This tells us that one of the letters is odd.
Now we know A+D and C+D give us an EVEN number so B MUST be an odd number.
A+D and C+D sums are off by a factor of 6 so the difference between A and C must be 6.
Since B+C is only 7 then the options for B are limited to 1, 3 or 5 since these are the only odd numbers under 7 that work. And we already know that A is a difference of 6 from C so A is either 12, 10 or 8 respectively.
We KNOW that the total of all numbers is 31 so let's work backwards from there.
IF we subtract the possible options for B from that total we get 30, 28 or 26.
From here we need to focus on A+D.
If the value of A is 12 then the value of D is also 12 but we also know that it can't be equal to A since each box can only be used once and we can't find a sum from differing digits that follows the rules. So D>A. A can't be 12 which means C can't be 6.
BUT if C is 4 then A can be 10, which makes D 14 and B is 3.

That checks out but what are the box values?

B and C are determined but what makes 10 and 14 that doesn't use 3 and 4?

D boxes CANNOT include 9 or 8. But ot can be eirher 7, 5 and 2 OR 7, 6 and 1.

A is either 9 and 1 or 8 and 2 respectively.

So the answer is:
A = 10 (either 9+1) or (8+2)
B = 3
C = 4
D = 14 (either 2, 5 and 7) or (1, 6 and 7)

Have a great day!

[notmee]

Do we know that there is only 1 solution?
I don’t think there is enough data to solve. If we take the CD block and fill in 0, 7 or 1, 6, then solve for the rest using those numbers, there are multiple possibilities and I see no way to isolate a single variable. So my solution is that the puzzle has no single solution

[andyanderson]

A = 6, B = 7, C = 0, D = 18. Without knowing the operations required in each set of boxes, theres not enough information to continue.

[AceAlandor]

I think the solution is 12, 1, 6, 12. Assuming all blocks and numbers are positive integers

The block hint (meaning a is composed of 2 blocks and D of 3) I believe means D can only divide by 3 and A by 2

Your A+D= 24 is a big hint because this means the D block is lower than 8 (-: 3* d block = 24). Further, because A is even (as is composed of 2 blocks), D is also even. This means the D block can be either 2, 4, 6 and D (3* D block) 6, 12, 18.

If D is 6 A is too big and B negative. If D is 18 C will be 0. Thus D is 12 and the rest follows

[Crazykiller-onPSN]

It’s Sunday man I ain’t doing math today 😊

[damarcuscreed]

A=6
B=7
C=0
D=18
6+7=13
7+0=7
0+18=18
6+7+0+18=31
⁶+18=24

[Rawr-qv]

I think a = 9.
A + B = 13
(9) + B = 13
B =13-9
B = 4
So plug it in.
9 + 4 = 13

[Jimtimpop]

B=7, C=0, A=6 (though the two boxes of A would be 3 and 3), and D =18 (with the three boxes being 6, 6, and 6)

[russellmitchell]

If all boxes must be a whole number, positive, and >0, and letters can have the same value, then (and this is the same as the person who said the total for D is 12 and work from there), then the value of each box with the letter is:

A box=6
B box=1
C box=6
D box=4

A+B=13 then A (2 boxes of 6 each) is 12 and B is 1, 12+1=13

B+C=7 then with B as 1 C must be 6, 1+6=7

C+D=18, C is 6, so the 3 boxes of D total 12, so each D box is 4, 6+12=18

A+B+C+D=31, 12+1+6+12=31

A+D=24, A is 6 per box or 12 and D is 4 per box or 12, so 12+12=24

[stevemarschner]

A=7
B=6
C=1
D=17
A can be divided into any numbers that add to 7
D can be divided into any numbers that equal 17

[pinkmind]

A=12, B=1, C=6, D=4. Because D is a multiple of 3 (#3boxes), A is an even number(#2 boxes). Therefore B is an odd number and therefore C is an even number. So c =6 (it cant be a 3 or 9 or 12 or 15) and D=4.

[Laurencetw]

this was published in Ivan Moscovich's masterpiece 'the little book of big brain games' (p. 358). the solution is achieved via pure logic and nothing more:

Two Russian mathematicians meet on a plane.

'If I remember correctly, you have three sons, ' says Ivan. 'What are their ages today?'

'The product of their ages is 36, ' says Igor. 'And the sum of their ages is exactly today's date.'

'I'm sorry, Igor, ' Ivan says after a minute, 'but that doesn't tell me the ages of your boys.'

'Oh, I forgot to tell you, my youngest son has red hair.'

'Ah, now it's clear, ' Ivan says. 'I now know exactly how old your three sons are.'

How did Ivan figure out the ages?

[CJONES]

Ouch! I love math, but damn this made my head hurt.