Another Math Problem

Paddycrumlinman

Brains of AH.... assistance is required.

Brucy bonus point to the person who can explain the answer in simple terms.

I had great success in the last one I posted.

Thanks in Advance

The letters A,B,C and D each stand for a different Single Digit. Use the clues below to find the digit. Use the clues to find the digits.

Clues

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

What number does D Stand for?

Scuid Mhór

does that mean we can deduce a is also an odd number?

flyton5

Could you not google it? Or copy it from someone in school tomorrow?

Scuid Mhór

flyton5 wrote: »

Could you not google it? Or copy it from someone in school tomorrow?

there is no fun in that, flyton5. to make the after hours crew, you gotta be able to conquer every challenge presented to you. i advise you start trying to solve it instead of sitting on your moral pedestal.

PanchoVilla

Paddycrumlinman wrote: »

Brains of AH.... assistance is required.

Brucy bonus point to the person who can explain the answer in simple terms.

I had great success in the last one I posted.

Thanks in Advance

The letters A,B,C and D each stand for a different Single Digit. Use the clues below to find the digit. Use the clues to find the digits.

Clues

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

What number does D Stand for?

D stands for do your own fúcking homework.

Paddycrumlinman

strobe wrote: »

D = 2

But it might not be the only answer.

Explanation if possible Sir?

strobe

Paddycrumlinman wrote: »

Explanation if possible Sir?

Blast my slow deletion skills. I was wrong.

irelandrover

A=2, B=4, C=3, D=6

Voodu Child

Eleventy seven

Scuid Mhór

irelandrover wrote: »

A=2, B=4, C=3, D=6

i FOR ONE welcome our new after hours crew homie.

flyton5

Scuid Mhór wrote: »

there is no fun in that, flyton5. to make the after hours crew, you gotta be able to conquer every challenge presented to you. i advise you start trying to solve it instead of sitting on your moral pedestal.

I prefer to stand on my moral pedestal. I left a high horse around here somewhere too...

cruizer101

C is greater than 1 and C is odd so that limits its to 3,5,7,9
Considering the answer is a two digit number C has to be 3.

Thirty something plus thirty something is sixty something so D is 6.

A + B has to be less than 10 so is 2 & 4.

W.Shakes-Beer

irelandrover wrote: »

A=2, B=4, C=3, D=6

fine im gonna be immature and just laugh at that.





C=3

mickdw

Paddycrumlinman wrote: »

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

What number does D Stand for?

D=6

enda1

irelandrover wrote: »

A=2, B=4, C=3, D=6

cruizer101 wrote: »

C is greater than 1 and C is odd so that limits its to 3,5,7,9
Considering the answer is a two digit number C has to be 3.

Thirty something plus thirty something is sixty something so D is 6.

A + B has to be less than 10 so is 2 & 4.

Paddycrumlinman wrote: »

Brains of AH.... assistance is required.

Brucy bonus point to the person who can explain the answer in simple terms.

I had great success in the last one I posted.

Thanks in Advance

The letters A,B,C and D each stand for a different Single Digit. Use the clues below to find the digit. Use the clues to find the digits.

Clues

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

What number does D Stand for?

Guys above are right. D=6, but A=1, B=2, C=3

c-note

CA+CB=DC:

a = 1 b= 2 c=3 d=6

OPTIONS
a = 1,2,3,4,5,6,7,8,9
b = 1,2,3,4,5,6,7,8,9
c = 1,2,3,4,5,6,7,8,9
d = 1,2,3,4,5,6,7,8,9

TOLD C IS ODD > 1, B,D EVEN
a = 1,2,3,4,5,6,7,8,9
b = 2,4,6,8
c = 3,5,7,9
d = 2,4,6,8

dc is an odd number (ends in odd number)
cb is an even number (ends in even number)
therfore ca must be odd-> a must be odd

a = 1,3,5,7,9
b = 2,4,6,8
c = 3,5,7,9
d = 2,4,6,8

CA+CB=DC
so D will be roughly twice C, only options are c=3 d=6

a = 1,3,5,7,9
b = 2,4,6,8
c = 3
d = 6

now a+b = c (3)
only options are 1,2

a = 1
b = 2
c = 3
d = 6

Paddycrumlinman

c-note wrote: »

CA+CB=DC:

a = 1 b= 2 c=3 d=6

OPTIONS
a = 1,2,3,4,5,6,7,8,9
b = 1,2,3,4,5,6,7,8,9
c = 1,2,3,4,5,6,7,8,9
d = 1,2,3,4,5,6,7,8,9

TOLD C IS ODD > 1, B,D EVEN
a = 1,2,3,4,5,6,7,8,9
b = 2,4,6,8
c = 3,5,7,9
d = 2,4,6,8

dc is an odd number (ends in odd number)
cb is an even number (ends in even number)
therfore ca must be odd-> a must be odd

a = 1,3,5,7,9
b = 2,4,6,8
c = 3,5,7,9
d = 2,4,6,8

CA+CB=DC
so D will be roughly twice C, only options are c=3 d=6

a = 1,3,5,7,9
b = 2,4,6,8
c = 3
d = 6

now a+b = c (3)
only options are 1,2

a = 1
b = 2
c = 3
d = 6

Great explanation on this one. Thanks a million now I have to explain it to my 7 year old.... :eek:

FoxT

C >1
C is odd
C+C<10 since the answer is a 2-digit number.
Therefore C=3

---

D could be = C+C or C+C+1 if A+B >9

BUT D is even, so it cannot be = C+C+1

Therefore

D=6

---

A + B = C

B is Even

So A=1, B=2

---

Verify:

CA+CB=DC
31+32=63

Next!


- FoxT

KINGVictor

It is
A-1
b-2
c-3
d-6

irelandrover

i see i got the right answer in the wrong way, i took the CA to mean C multiplied by A

starbelgrade

D = D

ascanbe

A to the B to the C to the D, Paddycrumlinman's heading for a Maths degree
D to the C to the B to the A, the AH crew get solutions with a little delay

starbelgrade

ascanbe wrote: »

A to the B to the C to the D, Paddycrumlinman's heading for a Maths degree
D to the C to B to the A, the AH crew get solutions with a little delay

Word.

ascanbe

starbelgrade wrote: »

Word.

True dat.

Saila

Paddycrumlinman wrote: »

Great explanation on this one. Thanks a million now I have to explain it to my 7 year old.... :eek:

now you can explain why your '7 year old' is up at 00:30, waiting for you to do his homework hmmm...

Tiny Explosions

Paddycrumlinman wrote: »

Great explanation on this one. Thanks a million now I have to explain it to my 7 year old.... :eek:

This seems like a fairly difficult math problem for a 7 year old!

I'd hate to see his/her's maths homework questions when their 10 years old.:eek:

DOC09UNAM

Saila wrote: »

now you can explain why your '7 year old' is up at 00:30, waiting for you to do his homework hmmm...

The time is not the same for everyone.

alwaysadub

OP-Does your son think you figure these out by yourself, or does he know you ask a load of Irish people on the internet for help?

Thank fcuk i didn't go to school in America, cos they are hard questions for a 7year old!

AnonoBoy

"I beat the smart kids! I beat the smart kids!"

ascanbe

alwaysadub wrote: »

OP-Does your son think you figure these out by yourself, or does he know you ask a load of Irish people on the internet for help?

Thank fcuk i didn't go to school in America, cos they are hard questions for a 7year old!

If you'd asked me that when i was 7, i'd still be trying to figure it out now.

Sheeps

D = 7

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.