Question

abcde*a=6e
what are the values of the digit?

Answer 1

Answer:

Now , that's how a cookie crumbles!! Really awesome. I just was waiting for something like that , and it has come . If there is no Morgan Freeman out side then it's just a coincidence. Now this beautiful problem has a name “ cryptarithmatic problem” . We can simply assume the reason of naming: cryptogram+ arithmetic. It was first discovered in China and then spread all over the world . And we have experienced some mind blowing problems like , send+more=money ; point+zero=energy… and the list goes on . The second one is my personal favorite, probably because of the infectious influence of physics in my life . Whatever now come to this problem. There probably are so many ways to solve a problem of the mentioned group. But I personally like my own designed methods . So in this segment I won't just give the answer but will also discuss the methods. So let's roll it.

Method :1 ( manipulation and intuition)

Okay, In this case we will expand our imagination and manipulate numbers to get the answer. So, take a look of the problem,

abcdef×6=defabc.

So , here first we will try to go with a simple logic . If a>1 , then 6a≥12, so we will get a 7 digit number instead of 6 . So , a=1. Now 6a+c'=d (c'=carry) and 6d+c”=a. Now we will choose d such that the value of carry is smaller as to produce a bigger carry we have to produce a big number (ex.6×3=18: carry=1; 6×4=24, carry =2). So d=8 will satisfy our all condition ( you may try with d≥6 and u will get the same ) . Now 6f=c and 6c+c'=f . Here c=2 and f=7 fulfill these conditions having a carry 5 in the second condition. So' a’ must the last digit of a 5 digit number. Now as d=8 , a =1 so the number is 51, having a carry 3. So b is the last digit of a 3 digit number. Now c=2,f=7. So 4 is carry here (6×7=42). So e is such a number that implies, 6×e+4=b , where b is a three digit one . So e=5 and b =4 ( u can solve) . Now I m not wrong we have collected all the numbers . Let's check them ,

a=1 , b=4, c=2, d=8, e=5,f=7.

And our number is abcdef : 142857.

And 6×142857=857142. So, we are correct!!!

Method:2 ( building a structure of logic where Sherlock Holmes can breathe)

So , I have designed and developed it last night . The first method was a general approach but this is not . As u don't need to apply it if the problem is not too rigorous. So, let's roll it,

If we take any two arbitrary number like e and b then we can express the relation,

6e+c'=10×p+b…(1)

6b+c”=10×p’+a …..(2)

Now adding them we get,

5(e+b)=10(p+p')-(c'+c”)

Or, (e+b)=2(p+p')-{(c'+c”)/5}….(3)

Now this third one is our generalize form and we will now go ahead using it as a sword.

Now as e+b is an intiger so (c'+c”) is divisible by 5 . But c'+c” ≥ 10 ( 6×9=54 : highest carry 5). So c'+c”=5 or 10 . But as carry can be 0 and we are going to get a generalize form , so c'+c”≠10 as if c'=0 c”=10 which is impossible. So ,

c'+c”=5…..(4)

So the probable solution sets are := {1,4};{2,3};{0,5} ( without loosing generality) .

Now if the set is , {1,4} ( without loosing generality) .

e+ b=2(p+p')-1.

So, solution set of p, p' is , {1,4} ( as 6×7=42, 6×8=48 : Carry=4 and p=4 ; 6×2=12, 6×3=18, :carry =1 and p=1).

So , e+b=9.

So the solution sets are , {1,8}, {2,7},{4,5} ( {3,6} is excluded as 36+7=43. And we can't get '7′ as carry. )

And now we have chosen (e,b) arbitrarily so we have got our numbers. Have u noticed it !!

So , our number is 142857 .

U can check further for other sets of carries.

Ohuffff!! Lastly it is easy to calculate this in mind but it's bit difficult to express the thought process, and I have done that . So hope , it will help.

THANKS!!!

Answer 2

Answer:

4A=EorE−1<10⟹E∈{4,5,6,8,9},A∈{1,2}

4E≡A(mod10)⟹E=8,A=2

4D+3≡B(mod10)

4B<10,B∈{0,1,2}

B=1,4D+3=1(mod10)⟹D=2or7

21CD8×4=8DC12

If D=7,21C78×4=84(4C+3)12

Then 4C+3=30+C⟹C=9

21978×4=87912