# Logical reasoning

#### Saumyojit

##### Full Member
Last edited by a moderator:

#### Dr.Peterson

##### Elite Member
If,
MAC + MAAR = JOCKO, find the value of 3A + 2M + 2C.

a) 31b) 36c) 33d) 38

Also, does "3A" here mean 3 times the digit A, or 3 tens plus A? (The answer is obvious.) Mixing notations without comment is dangerous, and I've never seen it in a cryptarithm.

#### Saumyojit

##### Full Member

Also, does "3A" here mean 3 times the digit A, or 3 tens plus A? (The answer is obvious.) Mixing notations without comment is dangerous, and I've never seen it in a cryptarithm.
See I tried this way
A=1 M=13 R=18 ....

MAC= 13*100+1*10+3*1=1313
MAAR = 13*1000+1*100+1*10+18=13128

ON ADDITION I get 14441 on LHS

On rhs JOCKO gives 115425

So it is not matching.

Then , another method I did was
MAC =1313
MAAR= 131118
THEN I DID ADDITION BUT the sums on both the sides doesn't match.

What is given that's it

#### Dr.Peterson

##### Elite Member
Are you saying that in your culture, this type of puzzle doesn't require each letter to represent a single digit? Maybe you'd better show us a solved example, to show what it means. (One trouble with puzzles of the sort you've been asking about is that they don't come with instructions. The usual cryptarithm does.)

And are you saying that you tried to solve it by totally random guessing?

The traditional way to solve these is to determine any "obvious" facts (e.g. J = 1; why?), and possibilities for "carries", then write a system of equations if necessary. I've done that, and solved the problem without too much trouble.

#### Saumyojit

##### Full Member
Yes,
J=1
O = Zero
Remaining digits {2,3,4,5,6,7,8,9}
Equation: A+A+1= K
R+C= 10

A+M=C with carry 1

These are the information I could get after trying for 12 hrs .

U have solved it please let me know how u did it.
I am depressed

#### Dr.Peterson

##### Elite Member
Yes,
J=1
O = Zero
Remaining digits {2,3,4,5,6,7,8,9}
Equation: A+A+1= K
R+C= 10

A+M=C with carry 1

These are the information I could get after trying for 12 hrs .

U have solved it please let me know how u did it.
I am depressed
You are (I think correctly) assuming carries into every column except the M+A. This might not be true, but is a good hypothesis to start with.

So the equations you get are

C + R = 10
2A + 1 = K
M + A = C + 10
M + 1 = 10

There are 4 equations for 7 unknowns, so some guesses will be needed (in order to find solutions where the variables are distinct single digits.

We know the value of M. Replace M with that value in the other three equations.

Given a value of A, we can find C, K, and R. Write equations for these, and find what values of A yield valid values for the others, either using inequalities, or just trial and error. For example, if A=2, then you will find that C=1, which is not allowed. Keep going.

The work you show is good; it would be helpful if you had shown us what kinds of things you tried that failed, rather than just your resulting mental state. But it is true that puzzle solving requires patience; that's why I don't recommend it for everyone.

#### Saumyojit

##### Full Member
Are you saying that in your culture, this type of puzzle doesn't require each letter to represent a single digit? Maybe you'd better show us a solved example, to show what it means. (One trouble with puzzles of the sort you've been asking about is that they don't come with instructions. The usual cryptarithm does.)

And are you saying that you tried to solve it by totally random guessing?

The traditional way to solve these is to determine any "obvious" facts (e.g. J = 1; why?), and possibilities for "carries", then write a system of equations if necessary. I've done that, and solved the problem without too much trouble.
Okay it was easy.
I didn't got this equation M+1=10
That's was the problem.

After I got m=9
Then 2...8 was reaming.
Then by hit and trial Answer came

#### JeffM

##### Elite Member
Trial and error may be more expeditious, but pure logic can give a unique answer on the assumptions that (1) each distinct letter represents a distinct decimal digit and (2) MAC > 99, MAAR > 999, and JOCKO > 9999.

$$\displaystyle 100M +10A + C + 1000M + 100A + 10A + R = 10000J + 1000O + 100C + 10K + O \implies$$

$$\displaystyle (C + R) + 10(A + A) + 100(M + A) + 1000M = O + 10K + 100C + 1000O + 10000J.$$

It may seem that you have one equation to find 7 unknowns. But in fact, you have as well over 14 inequalities and a restriction to integer solutions.

$$\displaystyle C + R = O + 10x, \ x = 0 \text { or } x = 1.$$

$$\displaystyle 10(x + A + A) = 10(K + 10y), \ y = 0 \text { or } y = 1.$$

$$\displaystyle \therefore x + 2A = K + y.$$

$$\displaystyle 100(y + M + A) = 100(C + 10z), \ z = 0 \text { or } z = 1.$$

$$\displaystyle y + M + A = C + 10z.$$

$$\displaystyle 1000(z + M) = 1000O + 10000J.$$

$$\displaystyle z + M = O + 10J.$$

$$\displaystyle \text {But } z \le 1, \ J \ge 1, \ O \ge 0, \text { and } M \le 9 \implies z + M \le 10 \text { and } 10 \le O + 10J.$$

$$\displaystyle \therefore z = 1,\ J = 1, O = 0, \ \text { and } M = 9.$$

$$\displaystyle O = 0 \implies C + R = O + 10x = 10x \implies R = 10 - C C \text { or } R = - C, \text { the latter being impossible.}$$

$$\displaystyle \therefore x = 1 \text { and } R = 10 - C.$$

Moreover, A, C, K, and R are integers > 1 and < 9 because O is zero, J is one, and M is 9.

$$\displaystyle \text {Furthermore, } y + M + A = C + 10z \implies y + 9 + A = C + 10 \implies y + A = C + 1.$$

$$\displaystyle y = 1 \implies 1 + A = C + 1 \implies A = C, \text { which is impossible.}$$

$$\displaystyle \therefore y = 0 \text { and } C < A - 1 \implies 3 \le A \le 8 \text { and } 2 \le C \le 7.$$

$$\displaystyle x + 2A = K + 10y \implies 1 + 2A = K \implies A = \dfrac{K - 1}{2}.$$

$$\displaystyle K \text { is even} \implies A \text { is not an integer, which is impossible.}$$

$$\displaystyle \therefore K = 3, 5, \text { or } 7.$$

$$\displaystyle K \le 5 \implies A \le 2, \text { which is impossible.}$$

$$\displaystyle \therefore K = 7 \implies A = 3 \implies C = 2 \implies R = 8.$$

$$\displaystyle MAC= 932,\ MAAR = 9338, \text { and } JOCKO = 10270.$$

Let’s check

$$\displaystyle 932 + 9338 = 10270. \ \checkmark$$

#### Saumyojit

##### Full Member
Another problem

W W W + D O W N =
E R R O R.
Find the value of D + O + W + N = ?

a)12b) 21c) 22d) 27

I have tried a bit .

Facts from the question I can derive

D+1(Carry)=10
D=9
E=1
R=0

W+N =10
W+W+1=O

If, W+n=10

Then W + O+ 1= 10

AS n and O are different

2 Assumption :

W+O +1=10
Or

W+O =10

Now how to proceed further?

Last edited:

#### Saumyojit

##### Full Member
The above problem is solved but help with below

There is another problem

PLAYS + WELL= BETTER

P+B+W=?

B=1, P=9,E=0

Y+L=10
OR
Y+L+1=10

A+E+1= T
OR
A+E+1=T+10

L+W+1=T+10
OR

L+W=T+10

#### JeffM

##### Elite Member
The above problem is solved but help with below

There is another problem

PLAYS + WELL= BETTER

P+B+W=?

B=1, P=9,E=0

Y+L=10
OR
Y+L+1=10

A+E+1= T
OR
A+E+1=T+10

L+W+1=T+10
OR

L+W=T+10
This general type of puzzle has two assumptions:

(1) each distinct letter represents a distinct decimal digit, and
(2) the initial digit of each number is not zero.

But all the puzzles that you have shown have an additional common feature that makes them much easier to solve than puzzles of this general type, namely that these all add a number with n digits to a number with (n + 1) digits to get a number with (n + 2) digits. Given our assumptions, the first digit of the (n + 1) digit number is nine, and the first and second digit of the (n + 2) digit number are one and zero respectively.

I shall demonstrate this with respect to this problem, but you should develop a general proof using induction.

$$\displaystyle p = 1000W + 100E + 10L + L, \ q = 10000P + 1000L + 100A + 10Y + S, \\ \text {and } r = 100000B + 10000E + 1000T + 100T + 10E + R.[$$
We assume that all letters are digits and W, P, and B > 0.

$$\displaystyle \therefore 100000 \le r \le 999999, \ 10000 \le q \le 99999, \text { and } 1000 \le p \le 9999.$$

$$\displaystyle \text {ASSUME } P < 9 \implies q \le 89999 \implies r = q + p \le 89999 + 9999 = 99998 .$$

$$\displaystyle \text {But that contradicts } 100000 \le r \implies P \not < 9 \implies P = 9 \implies 90000 \le q \le 99999.$$

$$\displaystyle \therefore q + p = r \le 99999 + 9999 = 109998 \implies 100000 \le q \le 109998 \implies B = 1 \text { and } E = 0.$$

So the problem easily reduces to

$$\displaystyle 90000 + 1000L + 100A + 10Y + S + 1000W + 100E + 10L + L = 100000 + 1000T + 100T + R.$$

#### Saumyojit

##### Full Member
Another problem

W W W + D O W N =
E R R O R.
Find the value of D + O + W + N = ?

a)12b) 21c) 22d) 27

I have tried a bit .

Facts from the question I can derive

D+1(Carry)=10
D=9
E=1
R=0

W+N =10
W+W+1=O

If, W+n=10

Then W + O+ 1= 10

AS n and O are different

2 Assumption :

W+O +1=10
Or

W+O =10

Now how to proceed further?
This is the type of approach that is needed.so please follow it

#### JeffM

##### Elite Member
First, you said that you had solved that problem.

Second, what I told you in my previous post would have made that problem much easier.

E = 1, R = 0, and D = 9 are instantaneous.

Third, if you have a teacher that has told you that you must solve these problems through some particular method, then follow that method. Don’t waste my time asking how to do such problems.

#### Saumyojit

##### Full Member
First, you said that you had solved that problem.

Second, what I told you in my previous post would have made that problem much easier.

E = 1, R = 0, and D = 9 are instantaneous.

Third, if you have a teacher that has told you that you must solve these problems through some particular method, then follow that method. Don’t waste my time asking how to do such problems.
Plays + well....
We need to solve this by carry and equation way

#### JeffM

##### Elite Member
Plays + well....
We need to solve this by carry and equation way
Well do so then. Why bother us?

#### Saumyojit

##### Full Member
Well do so then. Why bother us?
Can't you see that I tried in post #10 but couldn't solve it.

#### JeffM

##### Elite Member
First, what I previously told you is from the “carry and equation way.” It is just much faster.

$$\displaystyle S + L = R + 10w, \ w + Y + L = E + 10x, \ x + A + E = T + 10y,\\ y + L + W = E + 10z, \ z + P + 0 = E + 10u, \text { and } u + 0 + 0 = B.$$
Plus we know that u, w, x, y, and z are each either zero or one, that each distinct letter represents a distinct decimal digit, and that

$$\displaystyle B > 0, \ P > 0, \text { and } P > 0.$$

Now as I explained before

$$\displaystyle u + 0 + 0 = B > 0 \implies u = 1 \implies B = 1.$$

$$\displaystyle z + P + 0 = E + 10u \implies P = E + 10 \text { if } z = 0, \text { which is impossible because } P \text { and } E \text { are digits.}$$

$$\displaystyle \therefore z = 1 \implies 1 + P = E + 10 \implies P - 9 = E \implies P = 9 \text { and } E = 0 \text { because both are digits.}$$

You do not have to go through all this work if you simply recognize that it will obviously be true when you add a n digit number and a (n +1) digit number and get a sum that is a (n + 2) digit number.

Given that, you now have

$$\displaystyle S + L = R + 10w,\ w + Y + L = 10x, \ x + A = T + 10y, \text { and } y + L + W = 10.$$

Now solve it.

#### Saumyojit

##### Full Member
First, what I previously told you is from the “carry and equation way.” It is just much faster.

$$\displaystyle S + L = R + 10w, \ w + Y + L = E + 10x, \ x + A + E = T + 10y,\\ y + L + W = E + 10z, \ z + P + 0 = E + 10u, \text { and } u + 0 + 0 = B.$$
Plus we know that u, w, x, y, and z are each either zero or one, that each distinct letter represents a distinct decimal digit, and that

$$\displaystyle B > 0, \ P > 0, \text { and } P > 0.$$

Now as I explained before

$$\displaystyle u + 0 + 0 = B > 0 \implies u = 1 \implies B = 1.$$

$$\displaystyle z + P + 0 = E + 10u \implies P = E + 10 \text { if } z = 0, \text { which is impossible because } P \text { and } E \text { are digits.}$$

$$\displaystyle \therefore z = 1 \implies 1 + P = E + 10 \implies P - 9 = E \implies P = 9 \text { and } E = 0 \text { because both are digits.}$$

You do not have to go through all this work if you simply recognize that it will obviously be true when you add a n digit number and a (n +1) digit number and get a sum that is a (n + 2) digit number.

Given that, you now have

$$\displaystyle S + L = R + 10w,\ w + Y + L = 10x, \ x + A = T + 10y, \text { and } y + L + W = 10.$$

Now solve it.
Thanks.
Please watch this video from 28 minutes

And I have some doubts regarding this explanation

at 32:00 why he is keeping W as constant value '7' but not the other way round when u are assuming the values of L,W .
Why not L=7 , W= 6 or 8

#### Dr.Peterson

##### Elite Member
I watched parts of it; it isn't worth the time. He makes an interesting attempt to use Excel to keep track of things, but the orderliness falls apart at the end.

In particular, at 32:00 he does NOT say W is 7; he says that the PAIR "L,W" can have certain values, of which 7 is the largest and therefore is written second, under the W. And he ends up saying L is 7. So you're misunderstanding what he writes.

At any rate, this is certainly not a model of efficient thinking.

#### Saumyojit

##### Full Member
particular, at 32:00 h
Okay I will come back to this once I solve this another new problem

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