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ElitmusZone » Cryptarithmetic Multiplication 27

# Cryptarithmetic Multiplication 27

 1. Value of E ? (a) 5 (b) 6 (c) 7 (d) 8
 2. Value of O ? (a) 5 (b) 6 (c) 7 (d) 8
 3. Value of G + O + T + O ? (a) 16 (b) 17 (c) 18 (d) 19

in 3rd multiplication O*E = O, which means that either O is 5 and E is an odd no. or E is 6 nd O is an even number
now check option 1 for case: O= 5, E odd no. whch gives only one possible value of E= 7.
now check the third multiplication by using this vlues nd the question will be solved :
A G 7
* 5
_______________
G 5 T 5

2. vikas solanki

you can proceed like this

0 + 0 + 0 =0,,this can only be possible if o =5 ,think with no carry from previous if we will consider carry thn no same value is possbile to give 0 +0 =9 since max carry from a +g can be 1 only so 0=5 for sure
now O*E=E MEANS E IS ODD NUMBER AND IN ABOVE PROBLEM O+CARRY =E,THN CARRY MUST BE 2 AND E =7
O*A =O,A ALSO ODD NUMBER POSSIBLE VALUES ARE{3,9} BECAUSE IT CANT BE {1,7,9}
NOW SEE THE PROBLEM A*A G E = 3 DIGIT NUMBER MEANS A<=3 ,SINCE A*A IS LAST NUMBER AND GIVING NO CARRY
SO VALUE OF A=3
NOW PROCEED WITH THESE VALUES