Brain Teasers
No New Digits
What 6 digit number when multiplied by either 2, 3, 4, 5 or 6 has no new digits which appear?
Answer
142857.142857 * 2 = 287514
142857 * 3 = 428571
142857 * 4 = 571428
142857 * 5 = 714285
142857 * 6 = 857142
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I wouldn't call this hard because it's something that you either know off the top of your head or you're just going to spend a lot of time finding it. The answer is what's known as a cyclic number... it's actually a very common lesson in Programming 101 classes to have students find a cyclic number when they're learning control statements.
Personally, I think this teaser is very good and highly underrated. I gave it the highest "Fun" rating.
Thank you!
I am the dumb one as I do not even know what it is, or what we are supposed to do. Never was any good at math, but this one is way over my head. I like the first comment as it makes me feel better.
I agree with BAbe. Went straight to the answer.
This puzzle was terrific! I think it would go well in the Logic section.
If I multiply a six digit number by 6 and get a six digit number, the first digit must be 1.
If every number must include a 1, then the last digit should produce a 1 when multipled by 2,3,4,5, or 6. So, the last digit must be 7 (3 x 7 = 21).
Now I can find all the digits by looking at multiples of 7 (7, 14, 21, 28, 35, 42).
My digits are 741852. 1 is first, seven is last. Now you need to arrange 2458 in the middle.
When multiplying by 2, you'll be "carrying the one" on most digits. You can not carry a 1 before the 4, because 2*4 + 1 = 9 and 9 is not one of my digits. After 2*2+1 = 5, nothing is carried. So, 2*4 = 8, and 2*2+1 must come just before it.
142??7 x 2
285??4
Test 5 and 8 to find the final solution. Then confirm it works for x3, x4, x5 and x6.
If I multiply a six digit number by 6 and get a six digit number, the first digit must be 1.
If every number must include a 1, then the last digit should produce a 1 when multipled by 2,3,4,5, or 6. So, the last digit must be 7 (3 x 7 = 21).
Now I can find all the digits by looking at multiples of 7 (7, 14, 21, 28, 35, 42).
My digits are 741852. 1 is first, seven is last. Now you need to arrange 2458 in the middle.
When multiplying by 2, you'll be "carrying the one" on most digits. You can not carry a 1 before the 4, because 2*4 + 1 = 9 and 9 is not one of my digits. After 2*2+1 = 5, nothing is carried. So, 2*4 = 8, and 2*2+1 must come just before it.
142??7 x 2
285??4
Test 5 and 8 to find the final solution. Then confirm it works for x3, x4, x5 and x6.
Snowdog, that was unmitigated brilliance!
Excellent analysis Snowdog!
The digits are from the fraction one seventh.
1/7 is 0.142857 with all six digits repeating.
The digits are from the fraction one seventh.
1/7 is 0.142857 with all six digits repeating.
A scip model for solving this teaser can be found at
https://gist.github.com/saska-gist/
35a0351d088d8699a8f8c34fdf1a5422
https://gist.github.com/saska-gist/
35a0351d088d8699a8f8c34fdf1a5422
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