Logical reasoning

JeffM said:
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.
Click to expand...
I'll try to make sure of that.
 
JeffM said:
Assume y = 1

∴x+U+O=R+10⟹1+O+Y=A+10z⟹O+10=A+10⟹O=A, which is impossible.∴x+U+O=R+10⟹1+O+Y=A+10z⟹O+10=A+10⟹O=A, which is impossible.\displaystyle \therefore x + U + O = R + 10 \implies 1 + O + Y = A + 10z \implies O + 10 = A + 10 \implies O = A, \text { which is impossible.}

Therefore y = 0, and thus x + U + O = R.
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okay please see my reasoning once


If in 4th column,
U+O+1with carry = R+10 (double digit number)
OR
U+O= R+10 (double digit number)

THEN,
there will be a Carry of 1 in 3rd column

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

THIS IS NOT POSSIBLE.
SO There will not be any carry in 3rd Column so result of 4th column will be a Single digit no.

SO,
O+Y=A+10 THIS EQUATION CAN BE FORMED of 3rd column.Its a fact now .

And from 4th column there are two possibilities
U+O+1 with carry = R

or

U+O = R


You did the same thing .

Now how to proceed futher .
HIT and trial?



Subhotosh Khan said:
What happened to this problem??
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This is the problem only.
Why so confused
 
Yes, you have but two possible cases. So try one.

[MATH]\text {ASSUME } x = 0.[/MATH]
[MATH]\therefore U + R = T + 10*0 = T\\ 0 + O + U = O + U = R\\ 0 + Y + O = Y + O = A + 10.[/MATH][MATH]\text {But } Y = 9 \implies 9 + O = A + 10 \implies O = A + 1.[/MATH]
All the remaining letters, A, R, T, O, U fall between 2 and 8 because H = 1, E = 0, and Y = 9.

[MATH]2 \le A \implies 3 \le O.[/MATH]
[MATH]U + R = T \le 8 \implies 2 \le U \le 6 \text { and } 2 \le R \le 6.[/MATH]
[MATH]\therefore O + U = R \le 6 \implies 3 \le O = R - U \implies 3 + U \le R \le 6 \implies U \le 3 \implies U = 2 \text { or } 3.[/MATH]
[MATH]U = 3 \implies O = R - 3 \le 6 - 3 \implies 3 \le O \le 3 \implies O = 3 = U, \text { which is impossible.}[/MATH]
[MATH]\therefore U = 2 \implies A = 3, \ O = 4, \ R = 6, \text { and } T = 8.[/MATH]
Do you see why?

YOU = 942. YOUR = 9426. HEART = 10368.

942 + 9426 = 10368.

As Dr. Peterson keeps saying, these take perseverance in addition to logic.
 
Last edited:
Did you understand

[MATH]2 \le A \text { and } O = A + 1 \implies 3 \le O[/MATH]?

How about

[MATH]x = 0, \ T \le 8, \text { and } U + R = T + 10x \implies U + R = T \le 8[/MATH]?

[MATH]U \ge 2 \implies - U \le - 2 \implies -U + U + R \le -2 + 8 \implies R \le 6.[/MATH]
Still with me?

[MATH]3 \le O, \ O + U = R, \text { and } R \le 6 \implies[/MATH]
[MATH]3 + U \le O + U = R \le 6 \implies 3 + U \le 6 \implies 3 + U - 3 \le 6 - 3 \implies U \le 3.[/MATH]
 
JeffM said:
−U+U+R≤−2+8
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It seems that you have considered U+R=8 in this step.
But why?

We know that U+R ≤ 8
So you have assumed that U+R= 8

JeffM said:
3+U ≤ O+U=R≤6
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3≤O this was the inequality . Then you have added U on both sides but why?
What was your thought process from the beginning?
After adding,
3+U ≤ O+U

Then , you have assumed that R= 6 but why? What were you trying to achieve ?

Then, you substituted R=6 in O+U => 3+U ≤ 6

Then, you subtracted 3 ( but why?) to get U≤ 3.
 
Saumyojit said:
It seems that you have considered U+R=8 in this step.
But why?

We know that U+R ≤ 8
So you have assumed that U+R= 8



3≤O this was the inequality . Then you have added U on both sides but why?
What was your thought process from the beginning?
After adding,
3+U ≤ O+U

Then , you have assumed that R= 6 but why? What were you trying to achieve ?

Then, you substituted R=6 in O+U => 3+U ≤ 6

Then, you subtracted 3 ( but why?) to get U≤ 3.
Click to expand...

Saumyojit said:
Then, you substituted R=6 in O+U => 3+U ≤ 6

Then, you subtracted 3 ( but why?) to get U≤ 3
Click to expand...
@JeffM

I would not know how to answer that question!

You have infinite patience!!
 
JeffM said:
Did you understand

[MATH]2 \le A \text { and } O = A + 1 \implies 3 \le O[/MATH]?

How about

[MATH]x = 0, \ T \le 8, \text { and } U + R = T + 10x \implies U + R = T \le 8[/MATH]?

[MATH]U \ge 2 \implies - U \le - 2 \implies -U + U + R \le -2 + 8 \implies R \le 6.[/MATH]
Still with me?

[MATH]3 \le O, \ O + U = R, \text { and } R \le 6 \implies[/MATH]
[MATH]3 + U \le O + U = R \le 6 \implies 3 + U \le 6 \implies 3 + U - 3 \le 6 - 3 \implies U \le 3.[/MATH]
Click to expand...

Okay I gave it a thought and came to another approach combining mostly from ur assumption and thinking.

Assuming no carry in 4th column so Summation of U+ R will be a single digit and obviously T will be less than equal to 8 and greater than equal to 2 .

NOW,
U + R=T implies that U+O= R

Some constrains on U and R is both are grater than equal to 2 and lesser than equal to 6 as We cannot get Sum i.e t=8
by (7,1) Combination(1 is already taken).

NOW, i have the set {2,3,4,5,6,7,8}
Now i have to find a possible combination of U and O keeping in mind that it adds up to R which is less than 7 and the value of O is one greater than A . SO from the set i can see that U=2 and O=4 fits nice and A=3 .

Then i check U +R=T . 2+6=8 . No conditions broke . DONE



BUT this correct assumption out of many possibilities :

1: NO carry in 4th column or assuming Summation of U+ R to be SINGLE DIGIT is the most vital thing which set the path to the Approach.
 
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