In 1556, NiccoloÁ Tartaglia of Brescia claimed that the sums 1 + 2 + 4,
1 + 2 + 4 + 8, 1 + 2 + 4 + 8 + 16, etc. are alternately prime and com-
posite. Show that his conjecture is false.
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2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
The sum k=0∑n2k=2n+1−1.
If we look at the remainders when those powers of 2 are divided by 7, we observe a pattern
n
2n
2nmod7
2
4
4
3
8
1
4
16
2
5
32
4
6
64
1
⋮
⋮
⋮
We see that this pattern repeats in a cycle of 3.
Tartaglia conjectured that for odd n, 2n−1 is a prime.
However, since 2n leaves a remainder of 1 every third time, every sixth time this remainder will occur at an odd number (for 26k+3,k∈N) and 2n−1 will be divisible by 7.
The first time this occurs is for 1+2+⋯+128+512 and infinitely many times after that.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
The sum k=0∑n2k=2n+1−1.
If we look at the remainders when those powers of 2 are divided by 7, we observe a pattern
We see that this pattern repeats in a cycle of 3.
Tartaglia conjectured that for odd n, 2n−1 is a prime.
However, since 2n leaves a remainder of 1 every third time, every sixth time this remainder will occur at an odd number (for 26k+3,k∈N) and 2n−1 will be divisible by 7.
The first time this occurs is for 1+2+⋯+128+512 and infinitely many times after that.
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Correct