DevSquare Logical Reasoning Cryptarithmetic Questions

KANSAS+OHIO=OREGON

K A N S A S

    O H I O

-----------

O R E G O N

given the value of O=5
Clearly,

• K + (nothing) is not equal to K


• K + (nothing) = O


• Thus, 1 carry must be coming from previous step


• K + 1 = O (O is given as 5)


• K = O - 1 = 5 - 1 = 4

Similarly,

• A + (nothing) = R


• This means 1 carry is coming from previous step


• Also, this is generating carry to next step


• Thus, A = 9


• R = A + 1(carry) = 0

rewriting the whole problem again

4 9 N S 9 S

    5 H I 5

-----------

5 0 E G 5 N

If we look at the 10's digit, we find that 9 + I = 5. This is only possible when -

• Case 1 :I = 6 (no carry from previous step)
  
  
  
  • 9 + 6 = 5 (1 carry to next step)
  
  
  
  


• Case 2 : I = 5 (1 carry from previous step)
  
  
  
  • 9 + 5 + 1(carry) = 5 (1 carry to next step)
  
  
  • 5 value is already taken by O so not possible thus
  
  
  • I = 6
  
  
  
  

Replace I = 6 and rewriting -

4 9 N S 9 S

    5 H 6 5

-----------

5 0 E G 5 N

Step at units digit S + 5 = N

• This generates no carry to next step as proved above


• So S + 5 < 10


• To make this possible


• Values of S ={0, 1, 2, 3, 4}
  
  
  
  • S can't be 0 or 4 as these values are taken by R and K
  
  
  • S can't be 1 as S + 5 = N
  
  
  • If S is 1 then N = 6
  
  
  • I is already 6 so, S can't be 1 also
  
  
  • Similarly, S can't be 4
  
  
  • N will become 9 and A is already 9
  
  
  
  

Let us do hit and trial with 2 and 3. Put S = 2 obviously N = 7. Let's rewrite the problem -

4 9 7 2 9 2     

    5 H 6 5 

-----------

5 0 E G 5 7

Now, 7 + 5 = E = 2 (1 carry to next step)

• This, is not possible As S already 2


• Thus, there must be 1 carry from previous step

7 + 5 + 1(carry) = E => E = 3

• Till now, R = 0, S = 2, K = 4, O = 5, I = 6, N = 7, A = 9


• Values pending = {1, 3, 8}

From hundreds place -

• 2 + H  + 1 (carry) = G


• From the above pending values, only one case is possible when


• H = 8, G = 1

Final solution will look like -

4 9 7 2 9 2

    5 8 6 5

-----------

5 0 3 1 5 7

KANSAS=497292
OHIO=5865
OREGON=503157
G+R+O+S+S=1+0+5+2+2=10

T H I S

    I S

-------

H E R E

• T + (nothing) is not equal to H


• Thus, there must be 1 carry from previous step


• Similarly, H + (nothing) is not equal to H, thus 1 carry from previous step


• Also, H + 1 (carry) >= 10 (It is generating 1 carry to next step)

This is only possible when

• H = 9


• When, H = 9, E = H + 1 = 0

Rewriting the whole problem

T 9 I S

    I S

-------

9 0 R 0

• Clearly, T = 8


• As T + 1 (carry) = 9

Also, S + S = 0

• This is only possible when S = 5

Rewriting the whole problem -

8 9 I 5

    I 5

-------

9 0 R 0

Now, I + I + 1(carry) = R is generating a carry, thus

• I = {5, 6, 7, 8, 9}


• Values : 5, 8, 9 are already taken


• Let's try I = 7
  
  
  
  • 7 + 7 + 1 = 5
  
  
  • 5 value is already taken
  
  
  
  


• Thus, 6 would be the value
  
  
  
  • 6 + 6 + 1 = 3
  
  
  • No conflict problem is solved -
  
  
  
  

8 9 6 5

    6 5

-------

9 0 3 0

8\times 0+ 6\times 3\times 9 - 5 = 157

 

  T O

+ G O

-----

O U T

The above 0 is O

Step 1

• The value of O is clearly 1 , as it is a carry


• Since, T + G is generating O is carry so value of O is 1.

Now, after substituting the values -

  T 1

+ G 1

-----

1 U T

Step 2

• Value of T = 1 + 1


• Thus, value of T = 2

  2 1

+ G 1

-----

1 U 2

Step 3

• Now, 2 + G > 10 (as its resulting a carry 1 on next)


• Now, possible values of G to get 1 carry at next step is - {G = 8 or 9}

If G is 9 then -

• U = 2 + 9 = 11


• So value of U becomes 1 and 1 goes to carry


• Now, value of O is already 1 so U value can not be 1 also.

Now, obv the value of G has to be 8, lets verify this also

• U = 2 + 8 = 10


• So value of U becomes 0, there is no conflict.

So final values are -

T = 2, O = 1, G = 8, U = 0

    2 1

+   8 1

-----------

  1 0 2

Now O + U + T = 1 + 0 + 2 = 3

After replacing the value E = 5

  T 5 5

+ L 5 T

------

  A L L

• In the above example


• 5 + 5 = L


• L can either be 0 or 1 (1 carry from previous step)

If L = 0

• Then, 5 + T must also be 0


• This means, T must also be 5 which is not possible

Then L = 1

• 5 + T = 1( 11 as 1 carry to next step)


• So T has to be 6


• T = 6

Rewriting the problem -

  6 5 5

+ 2 5 6

------

  A 1 1

Thus, A = 6 + 2 + 1(carry) = 9

A + L + L = 9 + 1 + 1  = 11

    U S A

+ U S S R

-----------

P E A C E

Step 1

• Clearly, P = 1, U = 9, E = 0 (1 carry from previous stage)


• Let's replace the values here

    0 S A



+ 9 S S R



-----------



1 0 A C 0

Step 2

• A + R = 0


• This is possible if, A = 5, R = 5, but, both can't take same values


• So its possible with (8,2), (7,3), (6,4), (4,6), (3,7), (2,8)


• Let's do hit and trial and take (2,8) and replace the new values.

      1

    0 S 2

+ 9 S S 8

-----------

1 0 2 C 0

Step 3

• Now, assuming S value to be 3


• We have S + S + 1 (carry) = C


• 3 + 3 + 1 => C = 7 (unoccupied)


• All the values are found out we just need to verify

      1

    0 3 2

+ 9 3 3 8

-----------

1 0 2 7 0

Step 4

• Values, are replaced and all the operations work just fine

Value of P + E + A + C + E = 1 + 0 + 2 + 7 + 0 = 10

        O N E

        O N E

        O N E

    +   O N E

      -----------

        T E N  

4 x O = T

4 x N = E

4 x E = N

STEP 1:


Assume E = 1, then N = 4

Then 4N must give 1 in the units place. 4N = 4(4) = 16 which is 6 in the units place

Therefore E is not equal to 1

Assume E = 2, Then N = 4 x 2 = 8

Then 4N must give 2 in the units place. 4N = 4(8) = 32 which is 2 in the units place

Thus E = 2 and N = 8, with a carry of 3

Step 2

Let O = 1, then T = 4 x 1 = 4 + 3(carry) = 7

We cannot take O other than 1 as it gives a carry

Thus,

    3

     1 8 2

     1 8 2

     1 8 2

 +   1 8 2

    --------

     7 2 8

Thus T = 7 , E = 2, N = 8

T + E + N = 7 + 2 + 8 = 17

Putting T = 0 as given

  E A 0

  E A 0

  E A 0

--------

B E E 0

• As T=0, no carry for A+A+A = (3A).

Since, the step above can generate following possible carries C = {0, 1, 2}

  C

  E A 0

  E A 0

  E A 0

--------

B E E 0

• There will be no carry generate from units place to tens place as all values are 0

Possible values of A and E can be calculated by

• 3A= E (No Carry to next step) - Eq1


• 3A = 10 +E (1 Carry to next step) - Eq2


• 3A = 20 + E (2 Carry to next step) - Eq3

Here,

• Largest carry generated by addition of three one digit number is 27(9+9+9).


• Hence value of E is less than 7 for Eq 3


• For Eq 3

Assume value of E is 7.

• Therefore value of A=9


• Now, C + E + E + E = BE.


• (2) + (7) + (7) + (7) = 23.


• This contradicts are resultant should also be 7, while its 3

We tried A = 9, for which E was 7 and resulted in contradiction.

Let us now try, A = 8

• If A = 8


• Then, 3A = E + (carry)


• Putting A = 8, E = 4 (carry = 2)

Trying the same again,

• Now, C + E + E + E = BE.


• (2) + (4) + (4) + (4) = 14.


• Hence value satisfied with our prediction.


• Thus, E=4 A=8 and B=1

Putting in equation - 2

  

  4 8 0

  4 8 0

  4 8 0

--------

1 4 4 0

Finally, T E E + T E E = 044 + 044 = 088

  C R O S S

  R O A D S

----------

D A N G E R

• Clearly, D = 1


• C + R can at max generate 1 as carry

Now, Let us look at the step S + S = R

• Clearly, R would be even, as sum of S + S will always be even


• So, possible values for R = {0, 2, 4, 6, 8}


• For R = 0
  
  
  
  • S needs to be 0
  
  
  • Both S and R can't be 0 thus, not possible
  
  
  
  


• For R = 2
  
  
  
  • S needs to be 1
  
  
  • D is already 1 so not possible
  
  
  
  


• R = 4
  
  
  
  • S = 2
  
  
  • Possible (Case 1)
  
  
  
  


• R = 6
  
  
  
  • S = 3
  
  
  • Possible (Case 2)
  
  
  
  


• R = 8
  
  
  
  • S = 4
  
  
  • Possible
  
  
  
  

Let us start hit and trial with S = 2 and R = 4

Putting all these values and rewriting -

Assuming C1, C2, C3 are carries, units, tens and hundreds will not have carry obv.

C1 C2 C3 

    C  4  O  2  2

    4  O  A  1  2

------------------

 1  A  N  G  E  4

• Clearly, E = 3 and C1 = 1


• Now, C2 + C + 4 = A (1 carry to next step)


• C2 + C + 4 >= 10


• Thus, C2 + C >= 6


• Possible Values for C2 = {0,1}


• C = {9, 8, 7, 6}

Trying Hit and Trial with

• C = 9, C2 = 1
  
  
  
  • C2 + C + 4 = A
  
  
  • 1 + 9 + 4 => A = 4 (1 carry)
  
  
  • Conflict as R already 4
  
  
  
  


• C = 9, C2 = 0
  
  
  
  • After, substituting A = 3
  
  
  • Conflict E = 3
  
  
  
  


• Let us try C = 8, C2 = 1
  
  
  
  • After, substituting, A = 3
  
  
  • Conflict E = 3
  
  
  
  


• C = 8, C2 = 0
  
  
  
  • A = 2
  
  
  • Conflict, S = 2
  
  
  
  

Similarly we can try all other values of C = 7, 6 with C1 = 1,0 all will have conflicts

Going back and trying Case 2, S = 3 and R = 6

C1 C2 C3 

    C  6  O  3  3

    6  O  A  1  3

------------------

 1  A  N  G  4  6

• Clearly, E = 4 and C1 = 1


• Now, C2 + C + 6 = A (1 carry to next step)


• C2 + C + 6 >= 10


• Thus, C2 + C >= 4


• Possible Values for C2 = {0,1}


• C = {9, 8, 7, 5} (4, 6 values already taken)

Taking C = 9 and C2 = 0

• C2 + C + 6 = A


• 0 + 9 + 6 => A = 5

Rewriting all values -

 1  0 C3 

    9  6  O  3  3

    6  O  A  1  3

------------------

 1  5  N  G  4  6

• Clearly, Step 6 + O = N is not generating any carry.


• Thus, 6 + O <=9


• O <=3, Possible values are O = {3, 2, 1, 0}

Let us try with O = 3

• C3 + 6 + 3 = N


• If C3 = 1
  
  
  
  • 1 + 6 + 3 = N
  
  
  • N = 0 (1 carry, not possible as C2 was found to be 0)
  
  
  
  


• If C2 = 0
  
  
  
  • N = 9
  
  
  • Not Possible as C is already 9
  
  
  
  

Let us try with O = 2

• C3 + 6 + 2 = N


•  If C3 = 1
  
  
  
  • N = 9, not possible as C = 9
  
  
  
  


• If C3 = 0
  
  
  
  • N = 8
  
  
  • Possible
  
  
  
  

Rewriting all values -

 1  0  0 

    9  6  2  3  3

    6  2  A  1  3

------------------

 1  5  8  G  4  6

• Values taken D = 1, O = 2, S = 3, E = 4, R = 6, N = 8,  C = 9


• Left values = {0, 5, 7}


• A = 5, G = 7, Clearly satisfies the conditions.

CROSS= 96233
ROADS=62513
DANGER=158746

Ans 31

  F I V E

  N I N E

---------

W I V E S

• Clearly, W = 1, as F + N = WI (2 digit number)

Now,

• N + N = E => 2N = E


• E + E = S => (2N) + (2N) = S


• 4N = S

Possible digits to satisfy this between  {1 to 9} would be -

• N = 1, E = 2, S = 4


• or N = 2, E = 4, S = 8

But, W already is 1. So, N can not be 1

Thus, N = 2, E = 4, S = 8 are only options....

Rewriting the problem -

  F I 2 4

  2 I 2 4

---------

1 I V 4 8

• F + 2 + carry(0/1) >=10 (as 1 carry to next step)


• To do this possible values of F are = {7, 8, 9}


• 8 is already taken by S, So, F = {7,9}

If we take F = 7

• F + 2 + Carry(0/1) = I


• 9 + Carry(0/1) = I
  
  
  
  • 9 + 0 (carry) = I (Case 1)
    
  
  
  • This is not possible as no carry to next step
  
  
  
  


• 9 + 1 (carry) = I (case 2)
  


• I = 0 not possible


• As step I + I = V should generate carry to next step i.e. Case 2

We must have done something wrong what?

There is another case that we never noticed -

• What if the below equations were never valid as they were generating carries
  
  
  
  • N + N = E => 2N = E
  
  
  • E + E = S => (2N) + (2N) = 6
  
  
  • 4N = S
  
  
  
  

Let us check

• What if E + E at units digit was generating a carry to next step


• Possible values to do this for E are = {5, 6, 7, 8, 9}

Let us try with E = 5

• E + E = S


• => S = 0


• Also, N + N + 1(carry) = E


• N + N + 1 = 5
  
  
  
  • Possible values of N to do this are N = {7, 2}
  
  
  
  

Let us try N = 7 and other values and rewrite the equation -

  F I 7 5

  7 I 7 5

---------

1 I V 5 0

• F + 7 = 1I


• => F + 7 + carry(0/1) >= 10

Let us assume the case where carry 1 is coming from previous step (Note 1)

• => F + 7 + 1 >= 10


• Equation is F + 7 + 1 = I

• Possible values for F are ={2, 3, 4, 6, 8, 9}


• F = 2 not possible as it will result I = 0, S is already 0


• F = 3 not possible as it will result I = 1, W already 1


• F = 4 is possible as I will be 2
  
  
  
  • But, step I + I + 1(Carry) = V will not generate carry as
  
  
  • 2 + 2 = 4 and carry is needed for (Note 1)
  
  
  
  


• F = 6, is possible and I = 4
  
  
  
  • But, again I + I + 1(Carry) = V will not generate carry
  
  
  
  


• F = 8, is possible and I = 6
  
  
  
  • I + I + 1(Carry) = V will generate carry
  
  
  • 6 + 6 + 1 = V = > V = 3
  
  
  
  

Putting all of these values in equation

  8 6 7 5

  7 6 7 5

---------

1 6 3 5 0

Thus, W + I + V + E + S = 15

Many websites give this solution which is incorrect
w=1
n=2
e=4
s=8
i=3
v=6
summation of wives =22.

    E V E R

  9 I N C E

-----------

1 0 R W I N

• Clearly, D = 1


• Clearly S = 9
  
  
  
  • As one carry must have been from previous step
  
  
  
  


• A = 0

2. Then,  we looks at E and I which results in R, as
9 is already occupied by S so E could be 8 and I
could be 7 and R will be 5 with carry 1.

3. Then we place the values of R=5 and E=8
resulting in N=3 with carry 1.

4. In ten’s digits – E=8 and and value of C could
be 3,4,6 (As all others are occupied) and to make I=7 (previously assigned in step 2) E must
be added by 9 which is already occupied by S.

So this hit ends here.

Now start again with reduced value of E or I,
repeat above steps until you get a correct
answer or a dead end.

5. Repeating above steps one time comes with
E=5 and I=7 which results in R=3 carry=1.

6. Placing R=3 and E=5 in unit digits and their
sum gives 8.

7. in ten’s digits place E=5 now you can add 2 to
make I=7 hence C=2.

8. in 100’s digits you have N=8 (from step 6)
and only digits that are left now are 4 and 6 hence. So placing V=4 you will get W=2 which
is not true because it is already assigned to C.
So we are left with only V=6 as the ultimate
choice and this leads the result in to W=4.

Hence this way we got the answer

    5 6 5 3

  9 7 8 2 5

-----------

1 0 3 4 7 8

Hence D+A+R+W+I+N = 1+0+3+4+7+8 =23

Since, SHE is a 3 digit number.

• Range of SHE is 100 - 999.


• In fact. true range would be 102 - 987


• Numbers can repeated in SHE

H can be anything between = {2, 3, 4, 5, .......}

Let us try hit and trial with 2

The equation will change as :

• (2E)^2 = S2E


• (HE)^H must be in the range 21 to 29


• so that result must be less then 999


• Do hit and trial and you will find answer is :


• (25)^2=625.

so S+H+E=6+2+5=13 ANS

WORLD+TRADE=CENTER

  W O R L D

  T R A D E

-----------

C E N T E R

ANSWER CENTER
so,start from last c the value is 1 always.
Next W+T=E right.
By adding numbers we should get 1 as a carry (which is C value).
so,take W as 5 and T as 7 and add now which is 12.

NO TWO ALPHABETS SHOULD HAVE SAME VALUES ALWAYS.
Do it accordingly.

  5 3 6 8 4

  7 6 0 4 2

-----------

1 2 9 7 2 6

53684+ 76042=129726