find the smallest value of N such that N mod 10=9, N mod 9 = 8, N mod 8 = 7, N mod 7 = 6 and so on till N mod 2 = 1 .
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1$-1$ would work, except that it isn't positive. – Ross Millikan Apr 30 '14 at 16:26
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Hint $\,\ 2,3,\cdots, 10\mid N\!+\!1 \iff {\rm lcm}(2,3,\cdots,10)\mid N\!+\!1$
Bill Dubuque
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Add to both sides of congruents $+1$. We know that $\text{lcm}(2,3,4,5,6,7,8,9,10)=2520$ and we get $N+1\equiv 0 \pmod {2520}$. Minimal value of $N$ will be $2519$
RFZ
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