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I know that $$ 15 \equiv 1\pmod{7}, N \equiv 1\pmod{7},$$ but cannot proceed further.

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closed as off-topic by C. Falcon, TheGeekGreek, Nosrati, TravisJ, user223391 Feb 21 '17 at 22:43

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  • $\begingroup$ I think this post could be closed as a duplicate of [this general version]. I have promised not to do it myself. In case you see a reason not to do so, I respect that, too. $\endgroup$ – Jyrki Lahtonen Feb 21 '17 at 20:27
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Note that $1309=7 \cdot 11 \cdot 17$. Let $N=15^{40}$.

Since $15 \equiv 1\pmod{7}$, $N \equiv 1\pmod{7}$.

Since $15^{10} \equiv 1\pmod{11}$, $N=\big(15^{10}\big)^4 \equiv 1\pmod{11}$.

Since $15 \equiv -2\pmod{17}$, $15^4 \equiv (-2)^4 \equiv -1\pmod{17}$ and $N=\big(15^4\big)^{10} \equiv (-1)^{10}=1\pmod{17}$.

Therefore $N-1$ is divisible by $7$, $11$, and $17$, and hence by $1309$.

The remainder is $1$.

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Note that $1309=7 \times 11 \times 17$. So $$15^{40} \equiv 1^{40} \equiv 1 \pmod {7}$$ Using $15 \equiv 1 \pmod {7}$. Then, $$15^{40} \equiv \left(15^{10}\right)^4\equiv 1 \pmod {11}$$ And $$15^{40} \equiv (15+17 \times 2)^{40} \equiv 7^{80} \equiv \left(7^{16}\right)^5 \equiv 1 \pmod {17}$$ From Fermat's Little Theorem. So $$15^{40} \equiv 1 \pmod {1309}$$ Using CRT. We are done.

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