Logical reasoning

Saumyojit said:
This problem

AA + BB + CC = ABC

A+B+C =?

A =1
AA=11

10*A+A + 10*B +B +10*C +C


11A+11B +11C = 11(A+B+C)

ABC= 100*A +10*B +1*C (SUM)
How to proceed further
Click to expand...
Are you sure you can conclude that A=1? This time there are three numbers being added, not just two.

As for how to proceed, you've seen that there are many ways to work on one of these; and the main benefit of doing them is to learn perseverance. So keep trying. If necessary, just try every possible number.

In fact, since the LHS is 11(A+B+C), you might just try each possible value of A+B+C (there aren't many, and you can eliminate some) and see what you get. Each problem is different, and this one is particularly so!
 
JeffM said:
It is relatively easy to show that A = 2.
Click to expand...
That's odd, because my solution has A=1!

But I'll add that the digital root is one way to drastically reduce the number of possibilities; in fact, that almost directly tells you what A+B+C is, without even solving.
 
Dr.Peterson said:
That's odd, because my solution has A=1!

But I'll add that the digital root is one way to drastically reduce the number of possibilities; in fact, that almost directly tells you what A+B+C is, without even solving.
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Hmm I’ll check my work.

I am assuming that each distinct letter represents a distinct decimal digit. I am further assuming that A, B, and C all exceed zero.

A + B + C = C + 10x
x + A + B + C = B + 10y
y + 0 + 0 + 0 = A

And x and y are non-negative integers less than 3.

If x = 0, A + B + C = C which is impossible on the assumption that neither A nor B is zero.

If y = 0, A = 0, which is impossible on the assumption that A is not zero.

If A = 1, then y = 1 and A + B + C = 1 + B + C < 19 so x = 1.
1 + B + C = C + 10x and x + 1 + B + C = B + 10.
So 1 + B = 10x, which means B = 9 and x = 1.
Therefore 1 + 1 + 9 + C = 9 + 10 or C = 19 - 11 = 8.

11 + 99 + 88 = 11(1 + 9 + 8) = 11 * 18 = 190 + 19 = 198. Looks good.

Not sure where I blundered.
 
Here's my quick way:

Thinking in terms of casting out nines (digital root), the sum of digits in the sum AA+BB+CC is 2(A+B+C); this must be congruent (mod 9) to the sum of the digits of the sum, ABC, which is A+B+C. Therefore their difference, A+B+C, must be a multiple of 9 between 0 and 27. This can be only 9 or 18.

But also, the sum is 11A+11B+11C = 11(A+B+C). This is therefore either 11*9 = 99 or 11*18 = 198. Only the latter has three digits, so A=1, B=9, C=8.
 
Dr.Peterson said:
Here's my quick way:

Thinking in terms of casting out nines (digital root), the sum of digits in the sum AA+BB+CC is 2(A+B+C); this must be congruent (mod 9) to the sum of the digits of the sum, ABC, which is A+B+C. Therefore their difference, A+B+C, must be a multiple of 9 between 0 and 27. This can be only 9 or 18.

But also, the sum is 11A+11B+11C = 11(A+B+C). This is therefore either 11*9 = 99 or 11*18 = 198. Only the latter has three digits, so A=1, B=9, C=8.
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I have cleared the last two doubts.

Now ,

GUN + NO + NO = HUNT

value of every letter?


H=1

U+N or U+N+1=0 OR U + N+2=0

O+O+N=T + 10 OR T
O+O+N=T+20


G+1=U+10*H (ASSUMPTION)
 
Oh for goodness sake. Yes, H obviously is 1 because

99 + 99 = 198 and 198 + 999 = 1197.

So U is either 0 or 1. But H is 1 so U is 0.

So G, N, O, and T are distinct digits greater than 1.

N + O + O = T + 10x

x + U + N + N = N + 10y

y + G = 10.

So G is either 8 or 9 because y cannot exceed 2.

If G is 8, y must be 2. And

[MATH]x + U + N + N = N + 10y \implies x + N = 20, \text { which is impossible because } x + N \le 2 + 9 = 11 < 20.[/MATH]
So G = 9 and y = 1.

[MATH]x + U + N + N = N + 10 \implies x + N = 10 \implies N = 8 \text { and } x = 2 \text { because } G = 9.[/MATH]
So N + O + O = T + 10x means 2O = T + 20 - 8 = T + 12.

And that means O > 5 but < 8. If O is 6, T is zero, which is impossible because U is zero. So we end up

G = 9,
H = 1.
N = 8.
O = 7
T = 2.
U = 0.

Let’s check. 908 + 87 + 87 = 1082.

It is all about realizing that some letters are not zero, that each letter represents a distinct digit, and that certain equations are true. You cannot solve these only by getting equations. You have to test the equations against the other constraints.
 
JeffM said:
Oh for goodness sake. Yes, H obviously is 1 because

99 + 99 = 198 and 198 + 999 = 1197.

So U is either 0 or 1. But H is 1 so U is 0.

So G, N, O, and T are distinct digits greater than 1.

N + O + O = T + 10x

x + U + N + N = N + 10y

y + G = 10.

So G is either 8 or 9 because y cannot exceed 2.

If G is 8, y must be 2. And

[MATH]x + U + N + N = N + 10y \implies x + N = 20, \text { which is impossible because } x + N \le 2 + 9 = 11 < 20.[/MATH]
So G = 9 and y = 1.

[MATH]x + U + N + N = N + 10 \implies x + N = 10 \implies N = 8 \text { and } x = 2 \text { because } G = 9.[/MATH]
So N + O + O = T + 10x means 2O = T + 20 - 8 = T + 12.

And that means O > 5 but < 8. If O is 6, T is zero, which is impossible because U is zero. So we end up

G = 9,
H = 1.
N = 8.
O = 7
T = 2.
U = 0.

Let’s check. 908 + 87 + 87 = 1082.

It is all about realizing that some letters are not zero, that each letter represents a distinct digit, and that certain equations are true. You cannot solve these only by getting equations. You have to test the equations against the other constraints.
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Yes I did it without seeing your post.
Thanks btw.

Now another problem
This I am trying for the last 4 hrs

YOUR+ YOU = HEART

VALUE OF EACH LETTER

Y=9
E=0
H=1

R+U= T OR T+10
U+O=R OR
U+O+1=R

O= A+1
 
Saumyojit said:
Yes I did it without seeing your post.
Thanks btw.

Now another problem
This I am trying for the last 4 hrs

YOUR+ YOU = HEART

VALUE OF EACH LETTER

Y=9
E=0
H=1
Click to expand...
Good you seem to understand that piece of the puzzle.

Now I advise making the carry a variable.

So R + U = T + 10x, with x = 0 or 1.
x + U + O = R + 10y, with y = 0 or 1.
y + O + Y = A + 10z

But z = 1 so y + O + Y = A + 10.

You have four possible cases: x = 0 = y, x = 1 but y = 0, x = 0 but y = 1, or x = 1 = y.

Now try those cases looking for equality among different letters. Then you can reject that possibility.
 
Jeff said this,
R + U = T + 10x, with x = 0 or 1.
x + U + O = R + 10y, with y = 0 or 1.
y + O + Y = A + 10z



But,
SUMMATION OF U and O WITH CARRY 1 OR NO CARRY CAN ONLY BE A SINGLE DIGIT NUMBER.

IF DOUBLE DIGIT,
U+O+1 OR U+O= R+10
R+10 SHOULD NOT BE HAPPENING
THEN,
O+9+1= A+10
[We know Y=9]
O=A

THIS IS NOT POSSIBLE

SO,
O+Y=A+10.........ONLY THIS EQUATION CAN BE FORMED

.
 
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Saumyojit said:
Jeff said this,
R + U = T + 10x, with x = 0 or 1.
x + U + O = R + 10y, with y = 0 or 1.
y + O + Y = A + 10z
But,
SUMMATION OF U and O WITH CARRY 1 OR NO CARRY CAN ONLY BE A SINGLE DIGIT NUMBER.
Click to expand...
Why must summation of three digits be a single digit number?

1 + 5 + 6 = 12 = 2 + 10

And by the way, you have been told one problem per thread. This is post 37 in a thread that has many problems. DO NOT ADD ANOTHER PROBLEM TO THIS THREAD.
 
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Saumyojit said:
R + U = T + 10x, with x = 0 or 1.
x + U + O = R + 10y, with y = 0 or 1.
y + O + Y = A + 10z
Click to expand...


If,
U+O+1 OR U+O= R+10(double digit number)

THEN there will be a Carry of 1 in O+9...

O+9+1= A+10
[We know Y=9]
O=A

THIS IS NOT POSSIBLE

SO,
O+Y=A+10.........ONLY THIS EQUATION CAN BE FORMED


What's so difficult to understand??
 
What’s difficult to understand is your stating conclusions before you give any reasoning and your writing total nonsense. No one talked about multiplying ten by a two-digit number. The number to be multiplied by 10 is 0 or 1.

Possibly your reasoning was

Assume y = 1

[MATH]\therefore x + U + O = R + 10 \implies 1 + O + Y = A + 10z \implies O + 10 = A + 10 \implies O = A, \text { which is impossible.}[/MATH]
Therefore y = 0, and thus x + U + O = R.

Reasoning before conclusion.

Fine, now you have two cases x = 0 or x = 1.
 
Saumyojit said:
If,
MAC + MAAR = JOCKO, find the value of 3A + 2M + 2C.
Click to expand...

First, you must define what it is you are doing. does addition work on these letters the same way it works on single digits?
Are we to assume a Base 10 positional system? Tennis Racket + Cat Whisker doesn't mean anything. {Frog}{DogTongue} also doesn't mean anything

*** Unless you DEFINE it! ***
 
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