This type of problems can be solved by back-tracking technique after finding one letter (number).

### Approach (Method):

The best method which I follow is from bottom to top multiplication (You can do in vice-verse), but I found it was easy.

If you see the question, start from bottom to top like D*(ABC) and E*(ABC)

It is given that E * C = C this happens only with numbers of 5 & 6 let’s see how?

### Explanation:

1 * 5 = 5

3 * 5 = _5 (Actually 15 but concentrate on last digit)

7 * 5 = _5 (35) I am not using 5 * 5 because it’s already assigned to C we should not repeat.

9 * 5 = _5 (45) so, from this **don’t** conclude C = 5 & E= (1, 3, 7, 9).

Because there is another number with these properties Ex: – number (6)

6 * 2 = _2 (12)

6 * 4 = _4 (24)

6 * 8 = _8 (48)

From this **don’t **conclude E = 6 & C = (2, 4, 8).

Let, note all the possible cases and try. For Example take C = 5 and E = odd number (1, 3, 7, 9)

A | B | C | |

D | E | ||

F | E | C | |

D | E | C | |

H | G | B | C |

Now Replace C with 5 and redraw the following table.

And also place E as odd number (3, 7, 9) I didn’t write 1 here because it’s not going to be 1 Let’s take first **E = 3** and fill the table.

A | B | 5 | |

D | 3 | ||

F | 3 | 5 | |

D | 3 | 5 | |

H | G | B | 5 |

If you see the colored portion then sum 3 + 5 = B (8) lets assign 8 to B

And redraw the table, if u put B = 8, let’s check from first

3 * 5 = 5 (carry 1)

3 * 8 = 24 + 1 = 5 (carry 2) but in the table it is showing 3 it is contradictory, then we can understand that **E! = 3.**

So Next I will take **E = 7 **and again fill the table.

A | B | 5 | |

D | 7 | ||

F | 7 | 5 | |

D | 7 | 5 | |

H | G | B | 5 |

A | 2 | 5 | |

D | 7 | ||

F | 7 | 5 | |

D | 7 | 5 | |

H | G | 2 | 5 |

If you see the red color portion

Sum of 7 + 5 = 2 (carry 1) so we got **B = 2.**

Assign **B = 2** and fill the table.

Now see the row which I highlighted and think what value we can keep for D after clear Observation we can get **D = 3 **satisfy the given condition and put D = 3.

A | 2 | 5 | |

3 | 7 | ||

F | 7 | 5 | |

3 | 7 | 5 | |

H | G | 2 | 5 |

1 | 2 | 5 | |

3 | 7 | ||

8 | 7 | 5 | |

3 | 7 | 5 | |

4 | 6 | 2 | 5 |

To get 375 in second row the value of A is

**A = 1 **so put value of A in table

If you keep **A =** **1** it’s very simple

**F = 8, G = 6, H = 4.**

Hope everyone can understand the solution…… Thanks for spending your valuable time to read this 🙂

## 46 responses to “Multiplication Problem 1”

sirisantosh

February 1st, 2013 at 15:43

Got it without seeing ur solution

A=1,B=2,C=5,D=3,E=7,F=8,G=6,H=4

yash.bhanu

February 2nd, 2013 at 06:12

Very good 🙂

Pranay Das

February 5th, 2013 at 10:06

Sm people r born with Intellignce….and some keeps on figuring it out through out their life..Gud one.:)

yash.bhanu

February 5th, 2013 at 10:08

Mr pranay Das

Thanku 🙂

rrrrrrrrrrrrrrr

March 18th, 2013 at 02:00

please tell me a simpler way…………

yash.bhanu

March 18th, 2013 at 03:28

Mr.rrrrrrrrrrrrrrrrr

This is the simplest way .. once u try to do the simplest question like https://cryptarithmetic.wordpress.com/2013/02/01/multiplication-problem-6/ (multiplication-problem-6) . Then u can easily understand the how to solve it

Ravindra

March 19th, 2013 at 06:10

I got a doubt regarding the explanation of the problem, at the beginning of the solution you considered E*C=C and you told that it would happen only with 5 &6 but what about 1*3=3,1*4=4,1*6=6,1*7=7,1*8=8,1*9=9

can u explain why you eliminated them..

Thank u in advance 🙂

Ravindra

March 19th, 2013 at 15:04

HI

In the starting of your explanation for E*C=C why you excluded the options 1*2=2,1*3=3,1*4=4,1*6=6,1*7=7,1*8=8,1*9=9..

yash.bhanu

March 23rd, 2013 at 07:39

we should not take value as 1 if we take it will be mistake

Ravindra

March 23rd, 2013 at 17:51

can u please explain it..the mistake..plzzz

i thought i could do it after seeing it here..but i faced it in the exam 😦

Ashfaque Ahmed

March 24th, 2013 at 06:21

A small addition i would like to make : take ABC X E = FEC and ABC X D = DEC

now we can directly see that C has to be = 5 ..why??? because D and E both can’t be = 6 at the same time , and from this we can also that D & E lies in { 1,3,7,9 } but D & E can’t be =1 & 0 so D & E lies in { 3,7,9}. now we can proceed further….

yash.bhanu

March 24th, 2013 at 07:32

Yes Mr.Ashfaque Ahmed,

what you told is exactly correct you can proceed in that way also, Every person has their own way. But we took it as simple and easily understandable by normal person who is not having any idea up-on like this problems. So we are elaborating the explanation.

You can do it in another way in one(1) minute also by practice…

Good Luck For the exam….

ravindra

March 24th, 2013 at 07:45

HI Ahmed,

I have a doubt in the problem…why can’t we take these possibilities

1*2=2,1*3=3,1*4=4,1*6=6,1*7=7,1*8=8,1*9=9…

Kartheek.Bhanu

March 24th, 2013 at 07:59

We should not take the value 1 because (A B C) * E if we take E = 1 then

(A B C) * 1 = A B C but there is D E C .. That’s why we didn’t took…

Ashfaque Ahmed

March 24th, 2013 at 08:32

exactly !

hiiii

April 23rd, 2013 at 09:33

wow… nice.I understood such problems how to solve for the first time..Thank you!!!

Kartheek.Bhanu

April 25th, 2013 at 04:23

Good buddy 🙂

vamsi

May 11th, 2013 at 16:11

in choice With Choosing E=7,c=5 ..

u said that by observing marked position,value of d=3.

How it come.?

Rashmi T

May 12th, 2013 at 07:26

now understood how to solve this problem very well….. thank u… 🙂

Kartheek.Bhanu

May 13th, 2013 at 05:33

good 🙂

happy

June 8th, 2013 at 15:13

plz solve this problm:

DSP

LIE

—–

SPSS

PETA

PODE

—————–

PLEADS

Sanup Kumar

June 17th, 2013 at 21:03

5 4 3

6 7 8

———————-

4 3 4 4

3 8 0 1

3 2 5 8

————————

3 6 8 1 5 4

————————

Ateiv Jain

July 18th, 2013 at 17:09

How did u get the value of D=3 ??

Arul

October 18th, 2013 at 13:24

can u pls elaborate back tracking technique and pls provide more information about the this cryptarithmetic problems??? and i m the beginner..

chandra shekhar narayan

November 1st, 2013 at 13:45

it is too easy

Kartheek.Bhanu

November 1st, 2013 at 14:01

ya your r8 🙂

pratik

November 7th, 2013 at 10:32

thanks for the great site it helps me alot

Kartheek.Bhanu

November 7th, 2013 at 18:07

u welcome 🙂

njnj

November 14th, 2013 at 17:21

nice work get

Kartheek.Bhanu

November 15th, 2013 at 06:19

Thnku 🙂

shreyas

November 19th, 2013 at 15:45

Please mail the detail solution

$$vishal$$

November 26th, 2013 at 18:04

I would like know to know the solution of 4*4 multiply problem

Kartheek.Bhanu

November 26th, 2013 at 18:43

🙂 good wish .. will try to keep

Rahul

December 12th, 2013 at 21:46

Why you assumed b=(5+3)

or b=(7+5) in above cases???????????//

yo_crypto

December 15th, 2013 at 04:42

good explanation…

Kartheek.Bhanu

December 16th, 2013 at 14:09

Thnku 🙂

rishi

December 15th, 2013 at 10:49

is it compulsory that all digits have value among 1 to 9 which are multiplied like abc*def(a,b,c,d,e,f all are nonzero)

brahmaji

December 18th, 2013 at 00:33

How did you add 5+3=8(B)

(“If you see the colored portion then sum 3 + 5 = B (8) lets assign 8 to B”)

Manjunath

December 18th, 2013 at 19:19

When you multiplay (C*E) with values 5 & 7 i.e 5*7=35 and carry is 3… but you are not consodering this 3 carry for next addition.. you are telling the next addition will make 7*2 = 14 with D value 3 then its going to be 17 and hence E value equals 7.. but where does the carry from 5*7 = 35.. 3 goes ???

manoj

December 27th, 2013 at 12:09

can any one tell me how will be the verbal questions?

bhanu

January 5th, 2014 at 12:31

Good Explanation my friend

Kartheek.Bhanu

January 5th, 2014 at 12:43

Thanku

amrita

January 6th, 2014 at 15:42

can u solve in easier way …. wid full illustration plz…

WPD

*GKI

——

KFPP

GGZM-

FGFI–

——

GDWDFP

harsh saini

January 11th, 2014 at 14:15

ya it is understandable but some practice is required to get pretty handy at this

bharathi

January 24th, 2014 at 12:23

please suggest me some addition problems on this crypo…….i wany some websites also on this type of questions ……….send asap…….

vivek

July 1st, 2014 at 23:58

WPD

*GKI

——

KFPP

GGZM-

FGFI–

——

GDWDFP

its 943 *678