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| Can Anyone Attempt This Question by Dipocian(m): 5:21pm On Mar 29, 2015 |
can anyone attempt this question 2 Shares
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| Re: Can Anyone Attempt This Question by darksuit(m): 5:24pm On Mar 29, 2015 |
Booked Thinking |
| Re: Can Anyone Attempt This Question by Viktoh(m): 5:25pm On Mar 29, 2015 |
Using the above numbers we can make an equation (15-9) + (13-7) + (7-1) + (9-1) + (13-9) =30 from this solution we can get 6+6+6+8+4=30 ----------------------------------------- using permutation= 3!+3!+3!+3!+3!=30  |
| Re: Can Anyone Attempt This Question by PenSoftCorp(m): 5:36pm On Mar 29, 2015 |
those are odd nos in odd places(5) which is meant to give an odd no, but 30 is not an odd no. SO IT IS NOT POSSIBLE 1 Like |
| Re: Can Anyone Attempt This Question by ginawest(f): 5:41pm On Mar 29, 2015 |
its too hard jare 1 Like 1 Share |
| Re: Can Anyone Attempt This Question by Viktoh(m): 5:44pm On Mar 29, 2015 |
ginawest:i swear d tin noh easy |
| Re: Can Anyone Attempt This Question by darksuit(m): 5:47pm On Mar 29, 2015 |
PenSoftCorp:Shine ur eyes u can repeat d numbers |
| Re: Can Anyone Attempt This Question by phaulzoe(m): 6:54pm On Mar 29, 2015 |
Lolz....very simple 11+7+5+5+1+1=30 I rep #engineerz..we think faster than mathematician |
| Re: Can Anyone Attempt This Question by Viktoh(m): 7:01pm On Mar 29, 2015 |
phaulzoe:bro, tz 5 boxes not 6  |
| Re: Can Anyone Attempt This Question by PenSoftCorp(m): 7:19pm On Mar 29, 2015 |
darksuit:even if u repeat, it will still result to an odd no, but let me check again |
| Re: Can Anyone Attempt This Question by PatEinstEin(m): 9:33pm On Mar 29, 2015 |
Even if the guy is a magician, he can't solve that cos it's impossible Waiting for the superman that will claim to have the solution though  |
| Re: Can Anyone Attempt This Question by Dipocian(m): 9:53pm On Mar 29, 2015 |
Answer
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| Re: Can Anyone Attempt This Question by hatux: 12:30am On Mar 30, 2015 |
Will try tomorrow by GOD's grace... |
| Re: Can Anyone Attempt This Question by Viktoh(m): 12:59am On Mar 30, 2015 |
PatEinstEin:i did it up dr bruv |
| Re: Can Anyone Attempt This Question by ERCROSS(m): 1:30am On Mar 30, 2015 |
phaulzoe:OLODO.. Soro niyen...  |
| Re: Can Anyone Attempt This Question by PatEinstEin(m): 1:33am On Mar 30, 2015 |
| Re: Can Anyone Attempt This Question by phaulzoe(m): 1:26am On Mar 31, 2015 |
Wat du u mean guy |
| Re: Can Anyone Attempt This Question by phaulzoe(m): 1:28am On Mar 31, 2015 |
ERCROSS: Me olodo! Am correct after all |
| Re: Can Anyone Attempt This Question by lixom(m): 5:59am On Mar 31, 2015 |
And the answer is:... The simple addition of any of those five digits would not give thirty as the final answer because they are all odd numbers (2n+1 or 2n-1). No matter how you add odd numbers in an odd number times, it would not give you an even number as the answer. Except for the: 1. Introduction of other operators in the eqn, 2. One of the boxes would be left unfilled . 3. Recombination of the various digits and operators. If you agree to my solution, please like and share.... |
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