Latest Elitmus Aptitude Question SOLUTION: Todays cryp question .. PDA *HAD ----------- NHHZ DNZE DIED --------- ANDAAZ ------------ What is 5(H-A) ?

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(#M40036054) ELITMUS Logical Reasoning question crypt Keep an EYE

Todays cryp question ..

PDA
*HAD
-----------
NHHZ
DNZE
DIED
---------
ANDAAZ
------------


What is 5(H-A) ?

Asked In Elitmus amit kumar singh (10 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

Chandra bhan Singh  (9 years ago)

1) As A*A=E so value of A can't be 5 & 6.
2) As well A=D+1. So value of D can't be 4 and 5.
3) Assume D=2, then A=3. We get H=4.
4) Value of P must be 7, bcz we got H=4. and 2*7=14.
this gives N=1.
Now with these value we can calculate rest of the part.

Solution is

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


Answer of 5(H-A)= 5(4-3)
= 5 * 1 =5.

Is this solution Helpfull? Yes (10) | No   

DHANANJAY Kumar  (10 years ago)

i got the hint from the last question asked: i.e H+A+N+D = ?
Options: 7, 8, 9, 10.
Therefore, its true that the sum of H+A+N+D will be less than or equal to 10.
soln:
-the first letter of ANDAAZ : A=D+1 ----------------------------(eq-1)
- then A*D = Z nd D*D=H or D*D +1 = H
- then in PDA * A : A*A = E nd D*A = Z or D*A +1 = Z
so A can be only 2 or 3 then E= 4 or 9
- from eq-1 : D+1 = A i.e D can be either 1 or 2 if A = 2 or 3.
- on solving for the two possible values we can the solution:

723
*432
---------
1446
2169
2892
--------
312336
---------
so, H+A+N+D=10
and 5(H-A)= 5(4-3)=5.

Is this solution Helpfull? Yes (7) | No (3)  

Kaustav Mitra  (10 years ago)

P D A
* H A D
-------------
N H H Z
D N Z E _
D I E D _ _
-----------------
A N D A A Z

P D A ---(i)---
* D
-------
N H H Z

P D A ---(ii)---
* A
-------
D N Z E

P D A ---(iii)---
* H
-------
D I E D

From final product ANDAAZ it is clear that D+1=A.
From (i) & (ii) ---> A x D = Z and Z less than 10.
From (ii) ---> Carry over D is 0, hence A x A = E and E less than 10.
Now A can be 1, 2 or 3.
A cannot be 1 as A x D = Z.
Taking A=2, then E=4, D=1 (since, D+1=A), Z=2. But Z cannot be equal to 2. Hence A cannot be equal to 2.
Taking A=3, then E=9, D=2, Z=6.
From (i) ---> D x D = H or H=4 And D x P = NH or P=7 and N=1.
Hence the result follows :::

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

Is this solution Helpfull? Yes (3) | No   

DHANANJAY Kumar  (10 years ago)

i got the hint from the last question asked: i.e H+A+N+D = ?
Options: 7, 8, 9, 10.
Therefore, its true that the sum of H+A+N+D will be less than or equal to 10.
soln:
-the first letter of ANDAAZ : A=D+1 (eq-1)
- then A*D = Z nd D*D=H or D*D +1 = H
- then in PDA * A : A*A = E nd D*A = Z or D*A +1 = Z
so A can be only 2 or 3 then E= 4 or 9
- from eq-1 : D+1 = A i.e D can be either 2 or 8 but * not possible since H+A+N+D

Is this solution Helpfull? Yes (1) | No   

Kaustav Mitra  (10 years ago)

P D A
* H A D
-------------
N H H Z
D N Z E _
D I E D _ _
-----------------
A N D A A Z

P D A ---(i)---
* D
-------
N H H Z

P D A ---(ii)---
* A
-------
D N Z E

P D A ---(iii)---
* H
-------
D I E D

From final product ANDAAZ it is clear that D+1=A.
From (i) & (ii) ---> A x D = Z & Z Carry over D is 0, hence A x A = E D x D = H => H=4 And D x P = NH => P=7 and N=1.
Hence the result follows :::

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

Is this solution Helpfull? Yes (1) | No   

DHANANJAY KUMAR  (9 years ago)

723*
432
1446
2169
2892
312336
Firstly, d+1=a,a=3and d=2 suits best in the equation PDA*D =NHHZ. after that would be solved.as soon.as u get the value of a and d.

Is this solution Helpfull? Yes (1) | No   

vishal kumar  (9 years ago)

easiest cryptarithmetic problem..........solved in first attempt

723
*432
------
1446
2169-
2892--
------
312336

Is this solution Helpfull? Yes (1) | No   

shakti singh   (10 years ago)

some improvement to dhanjay lumar's solution
as D= 1 or 2, now D=1 is not possible because if D=1 than 1*A= A ,so unit digit would be A instead of Z. hence D=2 onlyyy.

Is this solution Helpfull? Yes | No   

Raj Kumar Ray  (10 years ago)

PDA*A=DNZE, so possible E values = 1,4,9,6
Corresponding poss A values= 9,(2,8),(3,7),4
Now D=A-1;
aand H+E=A+10/A
considering H+E=A+10 (which usually happens)
H=(A+10)-E
= 8,4,8 (slight calculation is required eg- 9+10=19;19-1=18, not possible for value of H so ruled out)
so H= 8,4,8
possible CORRESPONDING A values 2,3,4
possible CORRESPONDING E values 4,9,6
now if A = 2, D=1 NOT POSSIBLE
if A=3,D=2, E=9 and H=4
putting the values in PDA*D=NHHZ, Z=6 & N=1 & P=7
thereby other values can be easily sorted out.

Is this solution Helpfull? Yes | No (1)  

sunil kumar  (9 years ago)

pda*had= 723*432
1446
2169
2892
------
312336

Is this solution Helpfull? Yes | No   

omsharan sharma  (9 years ago)

P=7,D=2,A=3,H=4,N=1,I=8,E=9,Z=6 These value get by solving 723*432
---------
1446
2169*
2892*
-------
312336
--------
ANDAAZ

Is this solution Helpfull? Yes | No (1)  

swati priya  (9 years ago)

723
432
.............
1446
2169
2892
...............
312336

5(H-A)=5

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Shushmit Yadav  (9 years ago)

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



5( H- A ) = 5 ( 4 - 3 ) = 5 * 1 = 5

Is this solution Helpfull? Yes | No   

Ehtesham  (9 years ago)

in this question see the last column from end : D + carry=A;
it means D and A will be consecutive nos. and A will b greater than D. D and A can not be 1( we can see it from its product value).now try with A=3 and D=2; here in the first attempt you will reach to the solution.

Ans:- 5

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