1. | Value of W ? | |||

(a) 6 | (b) 7 | (c) 8 | (d) 9 |

2. | Value of 2F + P ? | |||

(a) 12 | (b) 13 | (c) 14 | (d) 15 |

3. | Value of 3G + F ? | |||

(a) 20 | (b) 21 | (c) 22 | (d) 23 |

```
W P D
x G K I
K F P P
G G Z M
F G F I
G D W D F P
Here, you don't have a single clue to start solving the Cryptarithmetic problem. You have to use Unit Digit Method (Hit and Trial Approach) for solving this Cryptarithmetic problem.
Here you have one clue.
W P D
x G K I
K F P P
G G Z M
F G F I
G D W D F P
G = F + 1(carry)
Now, firstly go through Unit Digit Method and try to solve the problem by your own.
```

```
1. D
2. C
3. D
W=9, P=4, D=3, G=6, K=7, I=8, F=5, Z=0
9 4 3
x 6 7 8
7 5 4 4
6 6 0 1
5 6 5 8
6 3 9 3 5 4
```

(Solution have been given considering you as beginner in Cryptarithmetic) W P D x G K I K F P P G G Z M F G F I G D W D F P Now you have to divide the problem into three parts. As, (1) W P D x I K F P P (2) W P D x K G G Z M (3) W P D x G F G F I At this stage, you have to analyse all three parts and find which has maximum number of clues.(i.e. maximum number of variables in the repetition) In this problem you can choose (1). You can see P is repeated three times. Hence, take (1) (1) W P D x I K F P P Now you have to start hit and trial with the possible value of P={1, 2, 3, 4, 5, 6, 7, 8, 9} Firstly take P=1 Possible ways of getting 1 at Unit Digit Case 1 - D=1, I=1, P=1 Rejected ( D = I = P) Case 2 - D=7, I=3, P=1 Needs to be checked further Case 3- D=3, I=7, P=1 Needs to be checked further Case 4- D=9, I=9, P=1 Rejected ( D = I) [In Cryptarithmetic, each variable should have unique and distinct values.] Now check for Case 2 and Case 3 Put P=1 (1) W 1 D x I K F 1 1 Case 2 D=7, I=3, P=1 (1) W 1 7 x 3 K F 1 1 [K F 5 1] Rejected Case 3 D=3, I=7, P=1 (1) W 1 3 x 7 K F 1 1[ K F 9 1] Rejected [Relax it's going to take some time to understand the whole concept. Please read.. again!] Now take P=2 Possible ways of getting 2 at Unit Digit Case 1 D=1, I=2, P=2 Rejected (P = I) Case 2 D=2, I=1, P=2 Rejected (D = P) Case 3 D=6, I=2, P=2 Rejected (I = P) Case 4 D=2, I=6, P=2 Rejected (D = P) Case 5 D=3, I=4, P=2 Needs to be checked Case 6 D=4, I=3, P=2 Needs to be checked Case 7 D=4, I=8, P=2 Needs to be checked Case 8 D=8, I=4, P=2 Needs to be checked Case 9 D=7, I=6, P=2 Needs to be checked Case 10 D=6, I=7, P=2 Needs to be checked Case 11 D=8, I=9, P=2 Needs to be checked Case 12 D=9, I=8, P=2 Needs to be checked Put P=2 (1) W 2 D x I K F 2 2 Case 5 D=3, I=4, P=2 (1) W 2 3 x 4 K F 2 2 [ K F 9 2] Rejected Case 6 D=4, I=3, P=2 (1) W 2 4 x 3 K F 2 2 [ K F 7 2] Rejected Case 7 D=4, I=8, P=2 (1) W 2 4 x 8 K F 2 2 [ K F 7 2] Rejected Case 8 D=8, I=4, P=2 (1) W 2 8 x 4 K F 2 2 [ K F 9 2] Rejected Case 9 D=7, I=6, P=2 (1) W 2 7 x 6 K F 2 2 [ K F 6 2] Rejected Case 10 D=6, I=7, P=2 (1) W 2 6 x 7 K F 2 2 [ K F 8 2] Rejected Case 11 D=8, P=9, P=2 (1) W 2 8 x 9 K F 2 2 [ K F 5 2] Rejected Case 12 D=9, P=8, P=2 (1) W 2 9 x 8 K F 2 2 [ K F 3 2] Rejected Possible ways of getting 3 at Unit Digit Case 1 D=3, I=1, P=3 Rejected (D = P) Case 2 D=1, I=3, P=3 Rejected (I = P) Case 3 D=7, I=9, P=3 Needs to be checked further Case 4 D=9, I=7, P=3 Needs to be checked further Now check for Case 3 and Case 4 Put P=3 (1) W 3 D x I K F 3 3 Case 3 D=7 I=9, P=3 (1) W 3 7 x 9 K F 3 3 [ K F 3 3] for this case you have to check more. As you have taken I=9, P=3 and D=7 Put these values in main problem and write again. W 3 7 x G K 9 K F 3 3 G G Z M F G F 9 G 7 W 7 F 3 As, W 3 7 x G F G F 9 For getting 9 at unit digit, G should be equal to 7. As you have already taken D=7 so G cannot be equal to 7. Rejected [Relax it's going to take some time to understand the concept. Please read..... again !] Case 4 D=9, I=7, P=3 (1) W 3 9 x 7 K F 3 3 [ K F 4 3] Rejected Now take P=4 Possible ways of getting 4 at unit digit Case 1 D=1, I=4, P=4 Rejected (I = P) Case 2 D=4, I=1, P=4 Rejected (D = P) Case 3 D=2, I=2, P=4 Rejected (D = I) Case 4 D=7, I=2, P=4 Needs to be checked Case 5 D=2, I=7, P=4 Needs to be checked Case 6 D=3, I=8, P=4 Needs to be checked Case 7 D=8, I=3, P=4 Needs to be checked Case 8 D=4, I=6, P=4 Rejected (D = P) Case 9 D=6, I=4, P=4 Rejected (I = P) Case 10 D=9, I=6, P=4 Needs to be checked Case 11 D=6, I=9, P=4 Needs to be checked Case 12 D=8, I=8, P=4 Rejected (D = I) (1) W P D x I K F P P Case 4 D=7, I=2, P=4 (1) W 4 7 x 2 K F 4 4 [ K F 9 4] Rejected Case 5 D=2, I=7, P=4 (1) W 4 2 x 7 K F 4 4 [ K F 9 4] Rejected Case 6 D=3, I=8, P=4 (1) W 4 3 x 8 K F 4 4 For this case you have to check more. Put D=3, I=8, and P=4 and rewrite the problem W 4 3 x G K 8 K F 4 4 G G Z M F G F 8 G 3 W 3 F 4 At this stage you can easily predict the value of G=6 put G=6 in the main problem W 4 3 x 6 K 8 K F 4 4 6 6 Z M F 6 6 8 6 3 W 3 F 4 Here you can see value of F=5 put F=5 W 4 3 x 6 K 8 K 5 4 4 6 6 Z M 5 6 6 8 6 3 W 3 5 4 now you can easily solve the problem. W=9, M=1, K=6 9 4 3 x 6 7 8 7 5 4 4 6 6 0 1 5 6 5 8 6 3 9 3 5 4 [Relax - it's going to take some time to understand the whole concept. If you are facing any difficulty. Please go through the Cryptarithmetic Tutorial.]

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thanks,it is very good site for cryptarithmetic problems

Take part 3 I.e W P D *G answer will b easier

take ques 3. and we know that according to ques F+1=G.

i.e. F and G are consecutive numbers.

now check optns in ques 3 by taking any two consecutive no. only option (d) satisfy the condition which means G=6 and F=5

now the question is just a piece of cake

its easier and wise to take help from the options when there is no hint in the question itself

FOR PRACT. PURPOSE AND IMPROVE YOUR CALCULATION TRY NOT TO LOOK FOR OPTION JUST USE HIT AND TRIAL

HERE THERE IS NO CLUE BUT WE CAN GUESS VALUE OF G >=6

WHY??

ANS::

G +G =D AND LEAVING A REMAINDER FOR F… SO G+G SHOLUD BE 8 WITH CARRY FROM PREVIOUS AS 2..

G+G SHOULD BE 10…MEANS G =5 BUT THIS IS NOT POSSIBLE AS IN MULTIPLICATION G*D=I THEN I WILL BE ZERO 0R 5 ..BUT I CANT BE ZERO(CRYPT RULE )AND CANT BE 5 SO G SHOULD BE GRATER THAN OR EQUAL TO 6

POSSIBLE VALUES ARE {6,7,8,9}

CHECK W P D

*K

_________

G G Z M 4TH PLACE WILL BE >=6 I.E. G AND IT CAN ONLY POSSIBLE IF K>=7

————- SO POSSIBLE VALUES OF K{7,8,9}

NOW YOU CAN DO HIT AND TRIAL

Nobody noticed I*W=I

W P D * I = K F P P

Now possibility of getting P=4, we can have two cases

Case 1: P=4, D=8 and I=3

Case 2: P=4, D=3 and I=8

So W 4 8 * 3 = K F 4 4 & W 4 3 * 8 = K F 4 4

Case 1:

Now, W 4 8 * 3 , the possible values of K F are 10, 13, 16, 19, 22, 25 & 28.

Now K can’t be 1 because anything multiplied by 1 will result same. Multiplication of W P D by K don’t give same result. So 10 to 19 are rejected.

22 cant be as K not equal to F

Now for 25 value of W ill be 8 which is not possible.

28 cant be because F and D can’t be same.

So case 1 is rejected.

Case 2:

Now W 4 3 * 8 , the possible values of K F are 11, 19, 27, 35, 43, 51, 59, 67 & 75

K can’t be 1. So 11 & 19 are rejected.

For 27 & 35, value of W will be 3 & 4 respectively both of which are assigned to D and P respectively. So 27 & 35 are rejected

43 cant be as 4 and 3 are already assigned to P and D.

ANything multiplied by 5 would result in 0 or 5. D * K = M. So if K=5 then M can’t be 5. Now P + M = F and there is no carry from previous side. So M is not 0. Since M is not equal to 0 or 5. So K can’t be 5. So 51 and 59 are rejected.

For 51, value of W=8. But I=8. So 51 is rejected.

Now for K F = 7 5, W =9. So it is accepted

someone please explain why F+1=G. Carry may be 2 ?