*Can you crack the logic?* Question of IIM-B This is a very interesting and intelligent question.. If 3 3 3= E 7 7 7= E 8 8 8 = R 5 4 2 4= N Then *5 8 0 6 9= ?*
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Given : 3 3 3= E 7 7 7= E 8 8 8 = R 5 4 2 4= N
To Find : *5 8 0 6 9= ?*
Solution:
3 3 3= E
3 + 3 + 3 = 9
9 - ninE
E - last letter of ninE
7 7 7= E
7 + 7 + 7 = 21
21 - twenty onE
E - is last letter of twenty onE
8 8 8 = R
8 + 8 + 8 = 24
24 - Twenty FouR
R is last letter of twenty fouR
5 4 2 4= N
5 + 4 + 2 + 4 = 15
15 - FifteeN
N is last letter of fifteen
5 8 0 6 9 = ?
5 + 8 + 0 + 6 + 9 = 28
28 - Twenty eighT
T is last letter of Twenty eighT
5 8 0 6 9 = T
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*Can you crack the logic?* If 1 1 1 1= R 2 2 2 2= T 3 3 3 3= E 4 4 4 4 ...
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