Infosys Cryptarithmetic Quiz – 4

Solution:

Take SING to the LHS to make the equation NINA + SING = AGAIN

  N I N A

+ S I N G

-----------

A G A I N

• A = 1 or 2 as it is the carry generated from the previous step


• Given I + I = A =  1, thus I has to be either 5 or 0 with a carry


• By trail and error method, let I = 5


• Replace  the values in the equation

   1  

  N 5 N 1

+ S 5 N G

-----------

1 G 1 5 N

• N + N = 5 with a 1 (carry) = 15 which is not possible, thus I = 0

   1  

  N 0 N 1

+ S 0 N G

-----------

1 G 1 0 N

• N should be 5 as 5 + 5 = 10 = 0 with 1(carry)

   1  

  N 5 N 1

+ S 5 N G

-----------

1 G 1 5 N

• N + N = 5 with a 1 (carry) = 15 which is not possible, thus I = 0

   1

  5 0 5 1

+ S 0 5 G

-----------

1 G 1 0 5

• G has be 4  as 1 + G = 5 and it does not give a carry

   1

  5 0 5 1

+ S 0 5 4

-----------

1 4 1 0 5

• S has to be 9 as 5 + S = 4 with 1 as carry = 14

   1

  5 0 5 1

+ 9 0 5 4

-----------

1 4 1 0 5

Solution:

    d o w 

+   r e d

---------

  e r r e

• e = 1 which is the carry generated from previous step


• d + r = r , This can happen if d = 0 and r  = 9 or d = 9 and r = 0 along with 1(carry)


• r cannot be 9 as the value or r will change, if carry is generated from the previous step


• Thus d = 9 and r = 0


• Replace the values of e, d, r

   1    

 1   9 o w 

+   0 1 9

---------

  1 0 0 1

• Thus w has to be 2 as  w + 9 = 1 with 1 carry = 11 and 2 + 9 = 11


• Replace w with 2.

   1  1   

 1   9 o 2 

+   0 1 9

---------

  1 0 0 1

• o has to be 8 as 1(carry) + o + 1 = 0 with 1 (carry) = 10.

   1  1   

 1   9 8 2 

+   0 1 9

---------

  1 0 0 1

Solution:

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

• (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.

Solution:

• 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 the same values, so do it accordingly

  5 3 6 8 4

  7 6 0 4 2

-----------

1 2 9 7 2 6

Solution:

By symmetry, we take A is the maximum number of the three alphabets.

Let us say A can take a maximum value of 6.

Then 6! = 720 but we have taken maximum value is A is 6.

Therefore, the Maximum value of the digits is less than 6.

So A should take 5. Then 5! = 120

From the above we know that one the B and C should take 1 as their value as 120 consists of  1.

⇒ 5! + 1! = 121

If we take 4 as one of the number then 5!+ 1! + 4! = 145 or 1! + 4! + 5! = 145

Sum = 1 + 4 + 5 = 10

Solution:

  A B C D

+ E B C B

----------

A F G A G

• A = 1 carry generated from the previous stage


• A + E  = F, thus, 1 + E  = F


• E can be  8(with carry from previous stage) or 9(without carry from previous stage) and F has to be 0 as a 1(carry) is generated


• Let E = 8 and F = 0

 1 

   1 B C D

+1 8 B C B

----------

 1 0 G 1 G

• C + C = 1, thus C has to be 5 with 1 carry from its previous stage


• 5 + 5 + 1(carry) = 11 = 1 with 1(carry)

  1 1 1 

   1 B 5 D

+1 8 B 5 B

----------

 1 0 G 1 G

• B + B + 1(carry) = G, B can take the values {6, 7, 8 } as it has to generate a carry


• By trail and error method, take B = 6, then 6 + 6 + 1 = 13 = 3 with 1(carry)

  1 1 1 

   1 6 5 D

+1 8 6 5 6

----------

 1 0 3 1 3

• Thus D has to be 7 as 7 + 6 = 13 = 3 with 1 (carry)

  1 1 1 

   1 6 5 7

+1 8 6 5 6

----------

 1 0 3 1 3

Solution:

Let us say maximum value of A, B, C is equal to 4 then we dont get any satisfactory values for ABC

If we take maximum value is 5 then A^{3} = 5^{3} = 125 as this is a three digit number one of the number is equal to 1.

5^{3}+1^{3} = 126. Now for 3^{3} we get 153=1^{3}+5^{3}+3^{3}

From the above reasoning the other numbers satisfy the above relation are 370, 371 and 407. So total 4 numbers.

Solution:

• From the first row of multiplication, H =1 is clear, As HE x H = HE.


• Substitute H = 1 in all places.


• Now from the tenth's place, think about, the value of A. 1 + A = M.


• If M is a single digit number, then N = 1, which is impossible (Already we have given H = 1).


• So A = 9, Then M = 0, and N = 2.


• Now 1E x E = 119.


• So by trial and error E = 7.

     H H            1 7   

   x E H            7 1

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

     H E            1 7

   H H A        1 1 9 

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

 H N M E        1 2 0 7

Solution:

From the first two of multiplication, you can clearly say that B = 1, as JE x B = JE.

From the second row of multiplication, A = 0 as in the multiplication, second row should start from tenth's place, so A = 0.

Now in the hundred's place, J + Something = 10. When you add something to the single digit number that results in 10,  so J = 9.

So Now from the above table, E + 9 = D + 10 => E - D = 1.

Therefore, we cannot determine values of E and D, but we can say, that E and D are consecutive.

       

     J E          9 E

   X B B        X 1 1

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

     J E          9 E

   J E A        9 E 0

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

 B A D E      1 0 D E

Solution

F O R T Y

    T E N

    T E N

-----------

S I X T Y 

• Then N=5 or 0.


• If N=5 then T+1+2*E=T+10*p,


• Being p=0, 1 or 2. 1+2*E=10*p, impossible. Then N=0

F O R T Y

    T E 0

    T E 0

-----------

S I X T Y 

• Second column, E=5

F O R T Y

    T 5 0

    T 5 0

----------

S I X T Y

• S=F+1(carry)


• I=O+1 or I=O+2 and R+2*T+1=X+10*p (p=1 or 2)


• If p=1 then O=9 and I=0, impossible because N=0


• If p=2, O=9 and I=1.


• 1+R+2*T=X+20;


• R+2*T=X+19.


• If T=8,R=6,X=3,not possible.


• If T=8,R=7,X=4.


• Then F=2 and S=3

2 9 7 8 6

    8 5 0

    8 5 0

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

3 1 4 8 6