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I'll try to make sure of that.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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I'll try to make sure of that.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.

okay please see my reasoning onceAssume 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.

If in 4th column,

U+O+1with carry = R+10 (double digit number)

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?

This is the problem only.What happened tothisproblem??

Why so confused

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Now another problem

This I am trying for the last 4 hrs

YOUR+ YOU = HEART

kay with O = 3 , U=4, R=8

Heart = 10282 coming

If HEART = 10282 then A = 2 = T

Is that correct?!

Not correctConfusionis that you areNOTanswering my question!!!

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How do you know that?Not correct

Only one no can represnt one letter.How do you know that?

Please read https://www.faceprep.in/logical-reasoning/cryptarithmetic-problems/

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The statement above is incomprehensible to meOnly one no can represnt one letter.

One letter can be represented Only by one single digitThe statement above is incomprehensible to me

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.

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

\(\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:

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

What does this step means

This step I don't understand.

How did 3+U came.

You shifted U from rhs to LHS.

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

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?

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

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

It seems that you have considered U+R=8 in this step.−U+U+R≤−2+8

But why?

We know that U+R

So you have assumed that U+R= 8

3≤O this was the inequality . Then you have added U on both sides but why?3+U≤O+U=R≤6

What was your thought process from the beginning?

After adding,

3+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

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

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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.

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

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

I would not know how to answer that question!

You have infinite patience!!

which question ?I would not know how to answer that question!

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

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

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?

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

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

Assuming no carry in 4th column so

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

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

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

BUT this correct assumption out of many possibilities :