Latest Elitmus Aptitude Question SOLUTION: D-M-H * A-N-P __________________ H M D D H D Z S B M Z P __________________ H H D B Z D ___________________ 1)Find 2A+1 ? a)7 b)11 c)13 d)15 Find the numbers cor

kindly please tell me from where have to prepare question like this?? kindly please tell me please

Q. What is a crypt arithmetic problem?

ans]

: It is a mathematical problem in which each letter represents a digit (for example, A = 5).

Our aim is to find the value of each letter.

No two letters represent the same digit (If A = 5, B cannot be 5).

The problem could be quite challenging but could be made very easy if solved with certain rules, tricks and presence of mind.

Follow these rules:

1. 0 * n = 0

2. 1 * n = n

3. 5 * odd number = (x)5 where x can be 0/1/3/4,

like 5 * 1 = 05 (x = 0)

5 * 3 = 15 (x = 1)

5 * 7 = 35 (x = 3)

5 * 9 = 45 (x = 4)

4. even number * 6 = (x)(even number) where x can be 1/2/4

like 2 * 6 = 12 (x = 1)

4 * 6 = 24 (x = 2)

8 * 6 = 48 (x = 4)

Let's understand the way how we will solve these questions with an example:



F A C

* H E I

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

E E A G

C J F E

D I J D

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

E B F E F G



Step 1:

Determine the characters which can be 0. Although this may not be useful in all cases but keeping the track of it may be useful in many cases.

In our example the multiplicand is FAC and multiplier is HEI

Rule 1: Any leading character will not be 0, so neither of F/H/E/C/D is 0

Rule 2: If any digit among E/I in multiplier is 0 then the corresponding row in multiplication will be all the same digit,

but in our case no row is either EEE/III so neither of E/I is 0

Rule 3: In multiplicand FAC, if C is 0 then the first,second and third row of multiplication will contain the trailing character as C,

but we don't have trailing character in all row as C, so C is no 0

Result of Rule 1, Rule 2 and Rule 3: Characters which can be 0 are - A/B/G/J

Note: Although the rule says that A can be 0 but if you come across situation where you need to consider one among these suspected character as 0,

then consider A(middle character of multiplicand) in the last. Or you can assume it to be non zero for simplicity.

Concluding the step 1: None of the character among multiplier, multiplicand and all leading character in each row will be 0.

So Characters which can be 0 are - B/G/J

Step 2:

Find whether we have any of the following property(in general for all problems):

I * C = I

I * C = C



E * C = E

E * C = C



H * C = H

H * C = C

Remember, they will give problem which has one of these property and this will be starting point of our approach to solve it.

We see that we have the property E * C = E

Now from the rule 3 and 4 we notice that either E is 5 and C is an odd number or E is an even number and C is 6

We will now proceed with the first case when E is 5 and C is an odd number and will consider the second case if we are not able to solve by this assumption.

Also we will have to think differently from here for each problem but practising more and more problem will make it easier.



F A C

* H 5 I

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

5 5 A G

C J F 5

D I J D

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

5 B F 5 F G



Now look carefully, the last column from left says: D + carry = 5. Now think what could be the maximum carry from the last sum C + I = B ?

Remember the sum of two number will never give a carry more than 1 and sum of three numbers will never give a carry more than 2 except in one case:



U V W

* X Y Z

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

9 9 9 9

9 9 9 9

9 9 9 9

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

1 1 0 9 8 8 9



Here the 2nd column from left gives a carry equals to 2 when both numbers of the column are same and are equal to 9(not in our case)

So we see that in D + carry = 5, the carry must be 1 and hence D has to be 4



F A C

* H 5 I

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

5 5 A G

C J F 5

4 I J 4

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

5 B F 5 F G



Also C can be either 3/7/9 and can't be 1 because if C was 1 then trailing character of each row would have been I,5 and H respectively.

Try to minimize the number of values that a character can have. Let's do it for C.

Focus on all C in our problem and see how they are generated. Look at second row of multiplication where:



F A C

* 5

--------

C J F 5



This multiplication give a number which will always be smaller than 5000 so C can be 3 only.



F A 3

* H 5 I

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

5 5 A G

3 J F 5

4 I J 4

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

5 B F 5 F G



Focus on the second row of multiplication again:



F A 3

* 5

--------

3 J F 5



We can see that F can be either 6 or 7.

Let's proceed with F = 6 and will consider F = 7 if we are stuck and are not able to proceed with F = 6



6 A 3

* H 5 I

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

5 5 A G

3 J 6 5

4 I J 4

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

5 B 6 5 6 G



Now see the second column from right A+5=6, which implies that A must be 1 and since we haven't gotten any character yet with value 1,

accept the value and proceed further.



6 1 3

* H 5 I

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

5 5 1 G

3 J 6 5

4 I J 4

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

5 B 6 5 6 G



6 1 3

* 5

--------

3 J 6 5

=> J = 0



6 1 3

* H 5 I

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

5 5 1 G

3 0 6 5

4 I 0 4

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

5 B 6 5 6 G



6 1 3

* H

--------

4 I 0 4



=> H must be 8 so that when multiplying H with 3 gives unit place digit as 4.



6 1 3

* 8 5 I

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

5 5 1 G

3 0 6 5

4 I 0 4

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

5 B 6 5 6 G



6 1 3

* 8

--------

4 I 0 4



=> I = 9



6 1 3

A is 6.

4 9 3
* 6 7 8
------------
3 9 4 4
3 4 5 1
2 9 5 8
--------------
3 3 4 2 5 4
--------------

****493
***x678
-------------
***3944
**3451
*2958
-------------
*334254


notice the first line of the multiplication process..
(DMH)*P=HMDD
clearly, if you start considering P as 1,2,3.. the range of H becomes very shrinked from 1 to P-1.
therefore we shall start with P=9 and then come down decrementing by 1.
P=9 does not stand in anyway. so lets start with P=8 now, and then will get a match with H=3,M=9 and consequently D=4.
replace all the values obtained above in the equation and then compute the rest of the value by simple multiplication and addition process..

1) c ----- 13

493*678
-------
3944
3451*
2958**
--------
33425

493
*678
------------
3944
3851
2958
------------
334254 so 2A+1=13 ans is c

how A become 6? tell me murthy how understand this cryptography questios
explain step by step

M MUST BE 8 or 9.because M+H=H AS it cannot have carry more than 1 or 2.
B and H must be consecutive number.
H+Z+2=10(AS sum of two no. cannot be 20 it is only possibile when H & Z both should be 9 and it is not possible.)
by this we conclude M must be 9.
H+Z+2=10 so H+Z=8
so possible value can be 6,2 : 7,1 :5,3 :8,0
and B & H must be consecutive no.
so by hit and trial H=3 and Z=5 so B=2
493
678
_____________
3944
3451
2958
_____________
334254

493
678
..............
3944
3451
2958
...........
334254


2A+1=2*6+1=13

493
*678
----------------
3944
3451
2958
--------------
334254
------------------



2A+1 = 13

Ans = c) 13