Mathematical challenge
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This is a mathematical challenge, and it's been said
that:
If you're an engineer (or history major) , you should be able to solve
it in (under) three minutes, if you're an architect, in three hours;
if you're a doctor, in six hours; if you're an accountant, in three
months and if you're a lawyer, probably never.
If you're good at math or logical problems, solve this challenge and
the answer is the password to open the spreadsheet to which you can
add your name to the hall of fame.
What is the missing number in this logical series?
1, 2, 6, 42, 1806, ____???Anyone Knows?
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3263442
1806 x 1807 -
Its pretty easy.
What hall of fame are you talking about?
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When numbers get too big so fast, you use power or factoral.
When numbers get big relatively fast, you use multiply.
If not, use addition
For this question, you use k x k + k where k is the previous term
Some people know 45 x 45 = 2025 or even 42 x 42 = 1764 by heart.
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I am in the Science stream. And I came to my solution in less than 3min.
x * (x +1).
say starting with 1 and let x be the number before the unknown number that you're finding.
2nd in series will be:
1 * ( 1+1) = 1 *2 = 2
3rd in series will be
2 * (2 +1) = 2 * 3 = 64th in series will be
6 * (6 +1) = 6 * 7 = 42
5th in series will be
42 * (42 + 1) = 42 * 43 = 18066th in series wil be
1806 * (1806 + 1) = 1806 * 1807 = 3263442 -
Originally posted by Darkness_hacker99:
I am in the Science stream. And I came to my solution in less than 3min.
x * (x +1).
say starting with 1 and let x be the number before the unknown number that you're finding.
2nd in series will be:
1 * ( 1+1) = 1 *2 = 2
3rd in series will be
2 * (2 +1) = 2 * 3 = 64th in series will be
6 * (6 +1) = 6 * 7 = 42
5th in series will be
42 * (42 + 1) = 42 * 43 = 18066th in series wil be
1806 * (1806 + 1) = 1806 * 1807 = 3263442JC boy?
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Yep I'm with 3263442.
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Aiyah, why take the mathematical challenge so serious lor
What TS claims has no ground lah.
What hall of fame ? Play online games huh ?
The question is really so difficult meh ?
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Originally posted by HyuugaNeji:
JC boy?
I'm a Biomedical Sciences dude.
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more challenging puzzles, anyone?
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too easy
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Originally posted by Darkness_hacker99:
I am in the Science stream. And I came to my solution in less than 3min.
x * (x +1).
say starting with 1 and let x be the number before the unknown number that you're finding.
2nd in series will be:
1 * ( 1+1) = 1 *2 = 2
3rd in series will be
2 * (2 +1) = 2 * 3 = 64th in series will be
6 * (6 +1) = 6 * 7 = 42
5th in series will be
42 * (42 + 1) = 42 * 43 = 18066th in series wil be
1806 * (1806 + 1) = 1806 * 1807 = 3263442hihi!
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i am nobody, i took more time to look for the calculator in Windows than trying to understand the logic of the sequence, which really is quite easy.... because i tutor my sons when they were in primary school.
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Originally posted by mystiv:
more challenging puzzles, anyone?
Prove or disprove that 111...111 (n times) where n is a positive integer > 1 has an even number of positive divisors.
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i am stupid i cant solve it i want to die.. Waaaaahhhhhhhhhhh~
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oh for the record, i took less than 1 second. cos i just have look down at rednano's answer and i'll get the answer lawls.
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some other questions...pretty easy
1. What is the largest number you can made by 4 "1"s.
2. which is heavier, one pound of feathers or a pound of gold
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Originally posted by SBS n SMRT:
some other questions...pretty easy
1. What is the largest number you can made by 4 "1"s.
2. which is heavier, one pound of feathers or a pound of gold
11^11?
they weigh the same
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1111!?
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11 power 11.
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oh wait...11!^11!
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like that ah...
(11!^11! )!
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1) Conditions too vague so I came up with this 11/(1-1)
2) same
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Originally posted by Tancy2411:
1) Conditions too vague so I came up with this 11/(1-1)
2) same
1) is not infinity. It is undefined. -
Originally posted by greengoblin:
Prove or disprove that 111...111 (n times) where n is a positive integer > 1 has an even number of positive divisors.
it is clear that 111...111 (n times) can never be a perfect square. Hence it has an even number of divisors :)