Quiz-1

KANSAS+OHIO=OREGON given the value of O=5
KANSAS=497292
OHIO=5865
OREGON=503157
G+R+O+S+S=1+0+5+2+2=10

t h i s
(8 9 6 5)
+ i s
(6 5)
------------------
h e r e
(9 0 3 0)

8*0+6*3*9-5 = 157

Basically you start with a few assumptions:
1. Each letter is a number from 0 to 9.
2. Each number can only be assigned to one of the letters.
We know that O+O=T and T has to be less than 10,
therefore O can be a number from 0 to 4.

G+T cannot be over 20
(as the max. number possible in an addition is 9+9=18
plus any carry over from a previous sum (which at the most can be 1)
and therefore O can only be 0 and 1.

If O is 0 then T is also 0 and one of our
assumptions is broken so O equals 1 and T
equals double that (2).

So GO+TO=G1+21=OUT=1U2, G+T=OU
so G has to be 8 or 9, if G=9
then U would be 1 and that breaks our second assumptions
so G=8 and U=0.
At the O+U+T=1+0+2=3

USA+
USSR=
PEACE
Here P is carry , P=1
when p=1, E=0 with carry 1 AND U=9
A+R=E=0 with carry 1.so, A=2 and R=8
U+S=A=2 with carry 1, S=3
S+S+1=C, 3+3+1=c=7
932+9338=10270
so,P+E+A+C+E=1+0+2+7+0=10
ANS 10

T E E
+ L E T
= A L L
SO, suppose E=5 and L =6, then L=1(carry 1),
again E +E i.e 5+5+(carry 1)=1, again (carry 1);
earlier we have taken T=6 and after summation L= 1
so, T+L=6+1+(carry 1)=8
so, A=8, L=1
A+L+L=8+1+1=10

Option (A)
Explanation
As it’s a sum of 3 numbers hence the maximum
value of D could be 2 or 1, but then we look at S
which could have maximum value 9 and if it
gets a carry of 1 then the value of A will be 0
and the D=1.

2. Then we looks at E and I which results in R, as
9 is already occupied by S so E could be 8 and I
could be 7 and R will be 5 with carry 1.

3. Then we place the values of R=5 and E=8
resulting in N=3 with carry 1.

4. In ten’s digits – E=8 and and value of C could
be 3,4,6 (As all others are occupied) and To
make I=7 (previously assigned in step 2) E must
be added by 9 which is already occupied by S.

So this hit ends here.

Now start again with reduced value of E or I,
repeat above steps until you get a correct
answer or a dead end.

5. Repeating above steps one time comes with
E=5 and I=7 which results in R=3 carry=1.

6. Placing R=3 and E=5 in unit digits and their
sum gives 8.

7. in ten’s digits place E=5 now you can add 2 to
make I=7 hence C=2.

8. in 100’s digits you have N=8 (from step 6)
and only digits that are left now are 4 and
6 hence. So placing V=4 you will get W=2 which
is not true because it is already assigned to C.
So we are left with only V=6 as the ultimate
choice and this leads the result in to W=4.

Hence this way we got the answer

….5 6 5 3
….E V E R
..9 7 8 2 5
+ S I N C E
———————-
D A R W I N
1 0 3 4 7 8

Hence D+A+R+W+I+N = 1+0+3+4+7+8 =23

arrange A-J(0-9) likewise K-L(0-9)
and U-Z(0-5)
the code is S=8,H=7,E=4 so 8+7+4=19

or

(HE)^H must be in the range 21 to 29
so that result must be less then 999
(3 digit ,as SHE) .so as per rule sol is
(25)^2=625.

so S+H+E=6+2+5=13 ANS

EAT+EAT+EAT = BEET.
As T=0, no carry for A+A+A(3A).
Possible values of A and E can be calculated by

1) 3A= E

2) 3A = 10 +E

3) 3A = 20 + E
Here Largest carry generated by addition of
three one digit number is 27(9+9+9).

Hence value of E is less than 7 for equation 3.

For Equation 3)

Assume value of E is 7. Therefore value of A=9
now carry + E + E + E = BE.
(2) + (7) + (7) + (7) = 23.
but 7 is not equal to 3. Contradict to our assumption.
Try another value of E as 4 for equation 3
E=4 therefore, A = 8.
now carry + E + E + E = BE.
(2) + (4) + (4) + (4) = 14.
hence value satisfied with our prediction.
hence E=4 A=8 and B=1
now TEE + TEE = TAA
044 + 044 = 088

In Above Equations A is integer. So take only
those values which are divisible by 3.

CROSS= 96233
ROADS=62513
DANGER=158746

Ans 31

https://www.sureshimr.in/2016/03/solution-of-problem-cross-roads-in.html

Solution here

WORLD+TRADE=CENTER
53684+ 76042=129726

ANSWER CENTER
so,start from last \'c\' the value is 1 always.
Next W+T=E right.
By adding numbers we should get 1 as a carry (which is \'C\' value).
so,take W as 5 and T as 7 and add now which is 12.

NO TWO ALPHABETS SHOULD HAVE SAME VALUES ALWAYS.
Do it accordingly.

4 digit + 4digit = 5 digit. fine
nine
=
wives
so clearly w=1 as f+n=i+10(it generates a carry over)
now,n+n = e,therefore 2n = e.
e+e=s, so we can write 2n+2n=s 0r 4n=s.
we can see that s is 4 times of n.
possible digit between 0-9 that satisfy this condition are
1,4 and 2,8 for n and e respectively.
but n cannot be one as the value of w is alreay 1 so the only choice we have is 2,8.
therefore n=2 and s=8.
from n=2 we can get e=4
now we can put i as 3,5,6,7,9
but if we put i as 5 or above then while
finding the value of \" f \" from eqn f+n=i+10 in
double digits so i has to be 3.
therefore i=3.
now i+i=v, so v=6
so finally we get
w=1
n=2
e=4
s=8
i=3
v=6
summation of wives =22.

L=1 E=5 T=6
So,
1 5 6
1 5 5 (+)
-------
3 1 1
-------
A=3 So, 3+1+1=5 ==> E