Latest Elitmus Aptitude Question SOLUTION: AGE * OAT --------- SOAR HOG- GOTO-- --------------- GECOIR i want to know the procedure to solve this type of queqtions.

1) AGE * T = SOAR
2) AGE * A = HOG ( i.e A is a small no.)
3) AGE * O = GOTO ( here E * O gives O, so O can be either 0 or 5, but O can not be 0 as AGE * 0 = 000 (OOO); this implies O is 5 & E is an odd no.)
so putting O as 5 we have AGE * 5 = G5T5 (from here we have G * 5 does not generates a carry as A * 5 gives G5 and that's possible only if G=1 (A1E*5=15T5) which implies A=3)
so now the crypt is like ->
31E
x 53T
--------
S53R
H51x
15T5xx
----------
1EC5IR ( from here we have I = (3+1) = 4 and C = (5+5+5) = )

Again from 31E * 3 = H51 ( i.e. E * 3 = 1 + carry =2 as next 1 * 3 =3 + carry =5 which left us with the fact that E * 3 = 21 so E = 7. and again we have 3 * 3 = 9 == H; one more thing 317 * 5 = 1585 so T=
now we have
317
x 538
----------
S53R
951x
1585xx
----------
17C54R
and now it looks easier
1) 317 * 8 = 2536 (so S = 2, R = 6)

2) So, the result will be 170546.(and not 170596 i have mistyped before but now I've Edited it) and Thus C= 0 (My Favorite Digit)

1.G
2.S
3.A
4.I
5.O
6.R
7.E
8.T
9.H

Hence The Problem Is Solved. QED

___317
___538
-------
__2536
__951x
1585xx
------
170546

AGE
* OAT
---------
SOAR
HOG-
GOTO--
--------
GECOIR

first of all split the question inn 3 parts
AGE
T
------
SOAR

AGE
A
----
HOG

AGE
O
----
GOTO

AGE * o giving o at unit place
it makes to possibilites
1) o=5 E=1,3,7,9
2) E=6 o=2,4,6

now see the sum
u can get from there that o + 1 or 2(carry) = E
so we can easily get o=5 and E=7

than put this values back
AG7
5
----
G5T5

u can get G=1 T=8 A=3 from here.

it is not so much tough.
u can refer elitmuszone.com site to learn this.

please explain the ans clearly i don't understand how 317 comes

G MAX 1 AND G NOT EQUAL TO 0 SO "G=1"
SINCE A*A=H HENCE A

can anyone explain the concept to solve these type of questions?

bhai koi to kaidey se explain kar do ho kaise rahe hai is type k ques
I mean to say from where 317 538 came & how G max 1 or min 0 is assumed????
please explain this in detail...

http://cryptarithms.awardspace.us/primer.html

try to get concept from here.

Thankyou ABHISHEK for the link.... :-)

170546 is the ans.

3*O+Carry=O,
we can have only 3 no (9,4,5).
(1)
If O=9,
then see in 2nd last column O+Carry=E and We can't take any carry from (O+Carry=E ) , otherwise existence of G won't be there. so we reject 9.
(2)
If O=5,
look at A G E * A= HOG
so A can have value 0,1,2,3.
we neglect 0 and 1 (if we will take then A G E * A should be equal to A G E * A ).
(a)
take A =2
2 G E * 2= H 5 G
from multiplication it is clear that each digit gives only 1 digit after multiplication. so, to
maintain O=5 , G can have only two values (2,7). G can't be 2 (because A has been already taken
as 2.) now if G =7 then H will also become 5. so we reject 7 too.
(b).
take A=3,
so, 3 G E *3= H 5 G so H =9 [ no carry because it will lead to 4 digit formation.]
now, O=5, A=3 , H=9 ,
look at 3 G E * 5= G 5 T 5 , E can have value (1,3,5,7,9).
u can't 3 , 5 and 9 . u also can't take 1. so E= 7
look at 3 G 7* 3= 9 5 G . so G =1
so we have O=5, A=3 , H=9 , E=7 , G=1 , I=4
by putting these values u will get answer 3 1 7
* 5 3 8
2 5 3 8
9 5 1
1 5 8 5
1 7 0 5 4 6