## Multiplication Problem 1

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

Got it without seeing ur solution

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

Very good

excellent explanation … keep it up..:-)

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

Mr pranay Das

Thanku

please tell me a simpler way…………

Mr.rrrrrrrrrrrrrrrrr

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

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

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..

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

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

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….

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….

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…

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…

exactly !

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

Good buddy

aap to james bond ho

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

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

How it come.?

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

good

plz solve this problm:

DSP

LIE

—–

SPSS

PETA

PODE

—————–

PLEADS

5 4 3

6 7 8

———————-

4 3 4 4

3 8 0 1

3 2 5 8

————————

3 6 8 1 5 4

————————

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

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

it is too easy

ya your r8

thanks for the great site it helps me alot

u welcome

nice work get

Thnku

Please mail the detail solution

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

good wish .. will try to keep

Why you assumed b=(5+3)

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

good explanation…

Thnku

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)

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”)

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 ???

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

Good Explanation my friend

Thanku

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

WPD

*GKI

——

KFPP

GGZM-

FGFI–

——

GDWDFP

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

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

WPD

*GKI

——

KFPP

GGZM-

FGFI–

——

GDWDFP

its 943 *678