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# Cryptarithmetic Division Problem 3

First we will convert this question into multiple, Cryparithmetic Multiplication Problems –

Before that in 4th row you see ANML and the divisor is also ANML, so value of D = 1 clearly.

PT

As,

A N M L

x D

--------

A N M L

thus, D = 1.

C 1 N K B

_______________

A N M L |P K N L B C 1 E

P 1 B A

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

L C B B

A N M L

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

C M M M C

C C B K N

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

B K C L 1

B M N B M

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

1 L B N E

1 B L E 1

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

B M A P

Check Unit digit method here.

Now, Find the multiplication with most substituted values,

it is -

A N M L

x B

--------

1 B L E 1

Now, using Unit digit's method possible ways of getting 1 are

(check the link above for Unit digit method to find all rules)

3 x 7 = _1 (21 only last digit considered)

7 x 3 = _1 (21 only last digit considered)

Thus, values are

either L = 3, B = 7 and L = 7, B = 3

using case 1 first,

A N M 3

x 7

---------

1 7 3 E 1

Now, if you consider A x 7, this results in 17.

There is only one possibility, for A x 7 + carry = 17

when A = 2 and carry = 3( A = 1 not possible as

7 + 6(max carry ie 7 x 9 =63) results only in 13 which is <17.

So now eqn is

2 N M 3

x 7

---------

1 7 3 E 1

Now, we know what carry from 7 x N was 3, this is only possible when

7 x N = _3 when N = 4( 28 + some carry >=2)

or 5( 35 + some carry less equal to 4)

Taking N = 4,

2 4 M 3

x 7

---------

1 7 3 E 1

Now, at unit's digit 3 value and 3 as carry.

It is possible for 7 x 4 + 5(carry) = 33

Now to get 5 carry from last step

you have to multiply 7 with either 7 or 8

Now, take M = 8

2 4 8 3

x 7

---------

1 7 3 E 1

Now carry from unit's 7 x 3 will be 2

and 7 x 8 + 2 carry = give value of E = 8. and M also 8. So, not possible.

Sim. M = 7 is not possible(try yourself)

Now try L = 7 and B = 3

A N M 7

x 3

---------

1 3 7 E 1

Now A = 4 to get 13 with 1 carry from prev. multiplication,

1

4 N M 7

x 3

---------

1 3 7 E 1

Now, to give 1 carry and 7 at unit's place

value of N = 5 with 2 carry from previous operation

1 2

4 5 M 7

x 3

---------

1 3 7 E 1

Now, 3 x M can give 2 carry for M={6, 8, 9}

Let's do hit and trial. Take M = 6.

1 2 2

4 5 6 7

x 3

---------

1 3 7 E 1

Now, 3 x 7 gives 2 carry to next mult.

and

3 x 6 + 2 as carry = 20 so E = 0.

1 2 2

4 5 6 7

x 3

---------

1 3 7 0 1

E = 0, D = 1, A = 4, N = 5, M = 6 its easy to find other values too.

PT

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