Tuesday 4 September 2012

SIMPLE... BUT FUNNY... THAT IS MATHEMATICS

Problem: 
Can you form “100” using all digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 exactly once. Any mathematical symbols can be used. (There is no restriction in the number of symbols used)

Answer(s):
123 + 45 – 67 + 8 – 9 = 100
123 – 45 – 67 + 89 = 100
12 – 3 – 4 + 5 – 6 + 7 + 89 = 100
12 + 3 – 4 + 5 + 67 + 8 + 9 = 100
98 + 1 + (27/54) + (3/6) = 100
(1 x 2) + 34 + 56 + 7 – 8 + 9 = 100
1 + 2 + 3 + 4 + 5 + 6 + 7 + (8 x 9) = 100
1 + (2 x 3) + 4 + 5 + 67 + 8 + 9 = 100
(-1 x 2) – 3 – 4 – 5 + (6 x 7) + (8 x 9) = 100
123 + 4 – 5 + 67 – 89 = 100
123 – 4 – 5 – 6 – 7 + 8 – 9 = 100
(1 + 2 – 3 – 4) x (5 – 6 – 7 – 8 – 9) = 100
(9 + 8 + 7 + 6 – 5) x (4 + 3 – 2 – 1) = 100
19  + (3 x 2) + 4 + 5 + 6 +78 = 100
87 + 9 + 6 + 5 – (3 x 2) - 14  = 100
98 – 7 + 6 + 5 – 4 + 3 – 2 + 1 = 100
79 + (8 x 3) – 4 + 6 – 5 x (2 – 1) = 100
69 + 8 + 7 + (5 x 4) – 3 – 2 + 1 = 100
97 + 8 – 6 + 4 – 5 + 3 – 2 + 1 = 100
83 + 45 – 29 + 7 – (6 x 1) = 100
(78 x 2) – 65 + 4 – 3 + 9 – 1 = 100
147 – 85 + 23 + 6 + 9 = 100
783/9 + 6 + 5 + 4 – 3 + 2 – 1 = 100
852/6 – 39 – 7 + 41 = 100
85 + 9 + 6 + 7 – 3 – 4 x (2 – 1) = 100
196 – 87 + 25 – 34 = 100
12 + 34 + 56 – 7 + 8 - √9 = 100
1 + (23 x 4) + 5 – 6 + 7 – 8 + 9 = 100
9 – 8 + 7 + 65 + 4 + (23 x 1) = 100
(9 x 8 x 7) + 6 + 5 + 4 + (3 x 2) + 1 = 100
2– 9 – 8 – 6 – 5 + 4 – 3 – 1 = 100
87 + 9 + 3 + (4/5) + (1.2/6) = 100
-1 – 2 – 3 + 4 + 5 + 6 – 7 + 98 = 100
182 – 97 + 6 +5 + 4 – 3 = 100

246 - (8 x 7) – 59 – 31 = 100
(9 x 4) + (6 x 3) + (8 x 7) - (5 x 2 x 1) = 100
√9 + 24 + 53 + 8 + 7 + 6 - 1 = 100
265 – 79 – 84 – 3 + 1 = 100
(1² x 8 x 9) + 67 – 35 - 4 = 100
1 x (9.8 + .2) x (4 – 3 + 7 - 6) x 5 = 100
(6 x 9) x (5 - 4) + 18/2 + 37 = 100
3² + (8 x 7) + (6 x 9) – 14 - 5 = 100
(4 – 3) x 2 + (7 x 5) + √81 + (6 x 9) = 100
36/4 + (8 x 7) + 25 + 9 + 1 = 100


98 + 2 + 4 + 3 – 7 + 1 + 5 – 6 = 100
6C3 + 8C4 + 9 + 1 + 5 + 2 – 7 = 100
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Can you find any other solution …
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Courtesy : http://manooss123.blogspot.in/

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