Logical reasoning

Subhotosh Khan

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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.
I'll try to make sure of that.
 

Saumyojit

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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.
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?



What happened to this problem??
This is the problem only.
Why so confused
 

Subhotosh Khan

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JeffM

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Yes, you have but two possible cases. So try one.

\(\displaystyle \text {ASSUME } x = 0.\)

\(\displaystyle \therefore U + R = T + 10*0 = T\\

0 + O + U = O + U = R\\
0 + Y + O = Y + O = A + 10.\)
\(\displaystyle \text {But } Y = 9 \implies 9 + O = A + 10 \implies O = A + 1.\)

All the remaining letters, A, R, T, O, U fall between 2 and 8 because H = 1, E = 0, and Y = 9.

\(\displaystyle 2 \le A \implies 3 \le O.\)

\(\displaystyle U + R = T \le 8 \implies 2 \le U \le 6 \text { and } 2 \le R \le 6.\)

\(\displaystyle \therefore O + U = R \le 6 \implies 3 \le O = R - U \implies 3 + U \le R \le 6 \implies U \le 3 \implies U = 2 \text { or } 3.\)

\(\displaystyle U = 3 \implies O = R - 3 \le 6 - 3 \implies 3 \le O \le 3 \implies O = 3 = U, \text { which is impossible.}\)

\(\displaystyle \therefore U = 2 \implies A = 3, \ O = 4, \ R = 6, \text { and } T = 8.\)

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:

Saumyojit

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∴O+U=R≤6⟹3≤O=R−U⟹3+U≤R≤6⟹U≤3⟹U=2 or 3.
3+U≤R≤6

What does this step means
This step I don't understand.
How did 3+U came.
You shifted U from rhs to LHS.
 

JeffM

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Did you understand

\(\displaystyle 2 \le A \text { and } O = A + 1 \implies 3 \le O\)?

How about

\(\displaystyle x = 0, \ T \le 8, \text { and } U + R = T + 10x \implies U + R = T \le 8\)?

\(\displaystyle U \ge 2 \implies - U \le - 2 \implies -U + U + R \le -2 + 8 \implies R \le 6.\)

Still with me?

\(\displaystyle 3 \le O, \ O + U = R, \text { and } R \le 6 \implies\)

\(\displaystyle 3 + U \le O + U = R \le 6 \implies 3 + U \le 6 \implies 3 + U - 3 \le 6 - 3 \implies U \le 3.\)
 

Saumyojit

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−U+U+R≤−2+8
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+U O+U=R≤6
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.
 

Subhotosh Khan

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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



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

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

I would not know how to answer that question!

You have infinite patience!!
 

Saumyojit

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Did you understand

\(\displaystyle 2 \le A \text { and } O = A + 1 \implies 3 \le O\)?

How about

\(\displaystyle x = 0, \ T \le 8, \text { and } U + R = T + 10x \implies U + R = T \le 8\)?

\(\displaystyle U \ge 2 \implies - U \le - 2 \implies -U + U + R \le -2 + 8 \implies R \le 6.\)

Still with me?

\(\displaystyle 3 \le O, \ O + U = R, \text { and } R \le 6 \implies\)

\(\displaystyle 3 + U \le O + U = R \le 6 \implies 3 + U \le 6 \implies 3 + U - 3 \le 6 - 3 \implies U \le 3.\)
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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