Another Math Problem

yerayeah

CA+CB=CD

CxD/B is odd by even so it's an even number.
Therefore CA must be even as only two evens can give an even answer.
Therefore A must be even as for CA to be even, A must be even as C is odd.

C(A+B)=C(D)
A+B=D

Since they're all different numbers, only 2+4=6 fits the bill, so D=6 and you don't know what the rest are. My take on it anyway!

bowsie010

Your child gets some faarce hard homework.

fergalr

Sometimes searching problems like this are best solved on computer.
Computers are good at trying a lot of combinations of something fast.

• C is greater than 1.
• C is an odd number.
• B and D are even Numbers.
  
  
• CA + CB=DC

Here are some numbers that meet those criteria, assuming CA doesn't mean C*A, but instead A concatenated onto C:
1,0,1,2
1,0,21,4
1,2,3,6
1,2,63,12
1,4,5,10
1,6,7,14
1,8,9,18
1,10,1,12
1,12,3,34
1,14,5,56
1,16,7,78
1,20,31,34
1,22,53,58
1,24,75,82
1,30,1,14
.
.
.
3,0,3,6

Single Digit:

1,0,1,2
1,2,3,6
3,0,3,6

different Single Digit

1,2,3,6


Numbers in the range -9 <-> 0 don't satisfy it either.
Have to say, its actually quite robustly specified for a problem like this!

x in the city

fergalr wrote: »

Sometimes searching problems like this are best solved on computer.
Computers are good at trying a lot of combinations of something fast.



Here are some numbers that meet those criteria, assuming CA doesn't mean C*A, but instead A concatenated onto C:
1,0,1,2
1,0,21,4
1,2,3,6
1,2,63,12
1,4,5,10
1,6,7,14
1,8,9,18
1,10,1,12
1,12,3,34
1,14,5,56
1,16,7,78
1,20,31,34
1,22,53,58
1,24,75,82
1,30,1,14
.
.
.
3,0,3,6



1,0,1,2
1,2,3,6
3,0,3,6



1,2,3,6


Numbers in the range -9 <-> 0 don't satisfy it either.
Have to say, its actually quite robustly specified for a problem like this!

surprised of the intelligence of (some) of the ah motley crew tbh...:P

Doc

I was always told that if you have two numbers represented by letters such as A and B and you show them in a problem like this:

AB + B = C

The AB means that you multiply A by B.

So that if A=4 and B=2 the above problem would be…

(4x2) + 2 = C => C = 10

The way you have solved OP's question this AB would be 42.

So...

42 + 2 = C => C = 44

Which I believe is incorrect.

Am I wrong???

mojesius

Thank jeebaz I don't have to do maths anymore.

D-Generate

Wow some people went to such lengths to find out!

C > 1 and C is odd leaving 3, 5 ,7, 9. Since we end up with a 2 digit number at the end after adding CA + CB then it can't be greater than 3.

3A + 3B = D3

Now to get 3 we have to add 1,2 or else we have a carry or we are using a 3 again in the case of 3,0.

So A = 1, B = 2, C = 3 and D = 6.

However this is supplementary to the answer as the question only asks for D so answer is D is 6.

I am gutted that i figured it out in no time at all and was excited to be first to have the answer but alas was stripped of this accolade.

D-Generate

Doc wrote: »

I was always told that if you have two numbers represented by letters such as A and B and you show them in a problem like this:

AB + B = C

The AB means that you multiply A by B.

So that if A=4 and B=2 the above problem would be…

(4x2) + 2 = C => C = 10

The way you have solved OP's question this AB would be 42.

So...

42 + 2 = C => C = 44

Which I believe is incorrect.

Am I wrong???

Well if you did it that way then wouldn't it be
C(A + = C(D)
A + B = D

4, 2, 6
6, 2, 8
2, 4, 6
2, 6, 8

So those are the valid solutions in that instance that satisfy criteria.

Doc

D-Generate wrote: »

Well if you did it that way then wouldn't it be
C(A + = C(D)
A + B = D

4, 2, 6
6, 2, 8
2, 4, 6
2, 6, 8

So those are the valid solutions in that instance that satisfy criteria.

That’s my point you took CA + CB=DC to mean (C is the first part of the number with A as the second) plus (C as the first part of the number with B as the second) is equal to (D as the first part of the letter with C as the second) then you can come up with a definitive answer for the value of D.

yerayeah's way of answering is correct. You got the right answer but the wrong way

Adyx

D-Generate wrote: »

Wow some people went to such lengths to find out!

C > 1 and C is odd leaving 3, 5 ,7, 9. Since we end up with a 2 digit number at the end after adding CA + CB then it can't be greater than 3.

3A + 3B = D3

Now to get 3 we have to add 1,2 or else we have a carry or we are using a 3 again in the case of 3,0.

So A = 1, B = 2, C = 3 and D = 6.

However this is supplementary to the answer as the question only asks for D so answer is D is 6.

I am gutted that i figured it out in no time at all and was excited to be first to have the answer but alas was stripped of this accolade.

If A = 1, B = 2, C = 3 and D = 6, that implies 3*1 + 3*2 = 3*6. Since 9 ≠ 18, the only solution which satisfies the equation and the clues is A=2, B=4, C=3, D=6 which was pointed out several times already. You can check it on WolframAlpha.

DOC09UNAM

Adyx wrote: »

If A = 1, B = 2, C = 3 and D = 6, that implies 3*1 + 3*2 = 3*6. Since 9 ≠ 18, the only solution which satisfies the equation and the clues is A=2, B=4, C=3, D=6 which was pointed out several times already. You can check it on WolframAlpha.

Your not multiplying them at all. You are told the letters stand for digits, not numbers.