Cryptarithmetic Questions and Answers

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

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 = 0Also, 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 0Thus, W + I + V + E + S = 15

GUN+SUN= HUNT here H=1 is a carry over, U=6 as U+U=N where N = 2, also N+N=T where T= 4 this gives us 862 + 762 = 1624 i.e. 1+6+2+4= 13.

Substituting values we get and using hit and trial we get, S = 9, E = 5, N = 6, D = 7, M = 1, O = 0, R = 8, Y = 2

SEND + MORE = MONEY

9567 + 1085 = 10652

O + N + E = 0 + 6 + 5 = 11

WATC + MATC = WINNE by substituting the values with get 1278 + 9278 = 10556 where W=1 is a carryover.

Hence the value of WINNE is 10556 is the appropriate answer.

WORLD+TRADE=CENTER

  W O R L D
  T R A D E
-----------
C E N T E R
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

C+E+N+T+E+R = 27

From the given data, the value of A will be 1 because it is the only carry-over possible from the sum of 2 single-digit number. T maximum it can take only 9 and there should a carryover for T to give sum as 2 digit number. So T =9, P = 0, A = 1. T + T = 18, the value of E is 8 and 1 will be a carry over to the next column. That is 1 + A + A= L = 3. And finally H = 2. Hence, 819 + 9219 = 10038. A+P+P+L=E = 1+0+0+3+8 = 12.

M A C + M A A R ----------- J O C K O

Step 1

Here obviously J = 1 as max carry allowed it

(nothing) + M = O, i.e. 0 + M = O this should have had been M only

It means that there is 1 carry coming from previous step

Also this M + 1 (carry) = O, is generating 1 carry to next step

This means M + 1 > 10, this is only possible when M = 9

=> O = M + 1 = 9 + 0 = 0 (1 carry)

9 A C + 9 A A R ----------- 1 0 C K 0

Step 2

C + R = 0

Since, this is at units place there can be no carry from previous step

Possible values of (C, R) can be (2, 8), (3,7), (4, 6), (6,4), (7,3), (2,8)

Let's assume, first case C = 2 and R = 8 in that case C + R, will generate 1 carry to next step A + A = K. The new problem looks like this -

9 A 2 + 9 A A 8 ----------- 1 0 2 K 0

Step 3

Thus, A + A + 1 (Carry) = K

Also, 9 + A = 2, this is possible in two ways

If, A = 3 (no carry from previous step) 9 + 3 = 2 (1 carry to next step)

If, A = 2 (1 carry from the previous step) but C value is already assumed to be 2, so this is not possible thus, A has to be 3

Thus K = 3 + 3 + 1 = 7

All values are fetched and there is no conflict.

9 3 2 + 9 3 3 8 ----------- 1 0 2 7 0

Step 4

Thus, 3 A + 2 M + 2 C = 3(3) + 2(9) + 2(2) = 9 + 18 + 4 = 31

    H O W 
+ M U C H 
-------------
P O W E R

Step 1

P = 1, as it is the max carry generated

M + (nothing) = O

M has to be 9 to generate a carry, and O has to be o and 1(carry) is generated from the previous stage

    H 0 W 
+ 9 U C H 
-------------
1 0 W E R

Step 2

We need to generate a carry , thus take H + U = 10 + W

U and H can take the values between {2, 3, 4, 5, 6, 7, 8}

By trail and error method

Take U = 8 and H = 7, thus W = 15 = 5 and 1(carry)

On substituting the values in W + H = R, we get R = 12 = 2 and 1(carry)

     1    
 1  7 0 5
+ 9 8 C 7 
-------------
1 0 5 E 2

Step 3

Now to find C and E

0 + C + 1(carry) = E

C and E can take the values {3, 4, 6}

By trail and error method

Take C = 3, then E = 4

Thus all the conditions are satisfied.

     1    
 1  7 0 5
+ 9 8 3 7 
-------------
1 0 5 4 2

P + O + W + E + R = 1 + 0 + 5 + 4 + 2 = 12

     N O
+  G U N
     N O
----------
 H U N T

Step 1

Here H = 1,

from the NUNN column we must have "carry 1,"

so G = 9, U = zero.

Since we have "carry" zero or 1 or 2 from the ONOT column,

Correspondingly we have N + U = 10 or 9 or 8. But duplication is not allowed, so N = 8 with "carry 2" from ONOT.

Hence, O + O = T + 20 - 8 = T + 12.

Testing for T = 2, 4 or 6, we find only T = 2 acceptable, O = 7.

So we have 87 + 908 + 87 = 1082.

Y O U R
+ Y O U
-----------
H E A R T

Step 1
Obv H = 1
Y + nothing = E (not Y)
Thus there is 1 carry from previous step
Y + (1 carry) is generating 1 carry resulting H
Thus, Y + 1 > 10 => Y = 9
So, E = 0

H = 1, Y = 9, E = 0

9 4 U R
+ 9 4 U
-----------
1 0 A R T
Step 2

Possibilities for A

A can be A = 4 + 9 = 3
or A = 4 + 9 + 1 (carry) = 4
O is already 4 thus, A = 3

Step, 4 + 9 = A also, doesn't get any carry from previous step U + 4 = R

Step 3

U + 4 = R results in no carry.

Thus, U + 4 < 10 U < 6 4, 5, 1, 0 are already taken by O, A, H and E respectively So, U is either 5 or 2 Let's take U = 5 ( no carry) U + 4 = R 5 + 4 = R = 9 9 is already taken by Y Let's take U = 5 ( 1 carry) U + 4 + 1 = R 5 + 4 + 1 = R = 0 0 is already taken by E Thus, U = 2 Step 4 For tens position U + O = R i.e. 2 + 4 is 6 Guessing that there is no carry, T = R + U = 6 + 2 = 8 9 4 2 6 + 9 4 2 ----------- 1 0 3 6 8 Final Step If we put all these values they generate no conflicts, thus we have reached the solution Y + U + H + E =  9 + 1 + 0 + 2 = 12