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ElitmusZone » Cryptarithmetic Problem 3

# Cryptarithmetic Problem 3

 1. Value of A + S + T + A + R + T ? (a) 21 (b) 26 (c) 24 (d) 25
 2. Find the value of 4R + T ? (a) 19 (b) 20 (c) 21 (d) 22
 3. Find the value of 2A + R ? (a) 5 (b) 6 (c) 7 (d) 8

1. nitish

In hint why A=1 is the only possible value of A.
A can be 6 as well, then {T,G,O}= {2 ,4 ,8} can take any values in these three,

2. Utsav Jain

Don’t GIve Up!!!!!

571
* 326
————————–
3426
1142
1713
————————–
186146

3. charan kumar

in this probelm your solution directly assumed as a = 1but why you can also take as a = 6 and then so you can assume as t, o,g, can belong to 2,4,8 as a*t = t, a*o = o, a*g= g implies that a can be 6 and so if t may be one among {2,4,8} so that 6 *2= _2 (last digit), 6*4 = _4 last digit, 6*8=_8 last digit then why cant we assume as a = 6 and solve the problem, you mentioned directly that a= 1 is the only alternative.

4. Saurav

why to stop for value of G=3 ?? it can be a possible case to take G=7 then I would be 3… isnt that possible?

5. Harsh Mishra

Guys, I also got the same thought. I guessed that this could help:
Note that in the second column from right the addition goes O+O=R i.e., 2O=R. Also if you’ve got A=6 then G, O, T must be in {2,4,8}.
Now,
2O = R
Then possible values of R can be
2*2=4, or
2*4=8, or
2*8=16 (ending with 6)
which contradicts all the values of G, O, T and A as well.

6. silambarasan

you are all correct A={6} T,O,G={2,4,8}

But if take sum at G+A+A=T

CASE 1:
A=6 , G=8
8+6+6=T (G+A+A=T)
T=20 (CASE IS WRONG BECAUSE T IS NOT BECOME 0)

CASE 2:
A=6, G=4
4+6+6=T (G+A+A=T)
T=16 (T IS NOT BECOME 6 BECAUSE WE ALSO TAKE A =6)

CASE 3:
A=6 G=2
2+6+6=T (G+A+A=T)
T=14 (TRUE T IS BECOME 4)

WE DECLARE T=4 & G=2 & O=8(BECAUSE 8 ONLY REMAINING IN THAT POSSIBILITIES SHOWN ABOVE)
THEN,
WE GO ANOTHER SUM O+O=R
8+8=16 (R IS NOT BECOMES 6 BECAUSE A=6 WE ALREADY TAKEN)

SO THEY CANNOT TAKE A=6 IN THIS PROBLEM