After expanding the above term binomially, I can well guess that the majority terms are irrational, but i'm unable to find any proper method of solving this sum, after repeated trials.

Please help

Thank You

  • $\begingroup$ Cute disguise for a Diophantine equation. $\endgroup$ – Brian Tung May 7 '15 at 21:17
  • $\begingroup$ Each of the $101$ terms in the expansion has the form ${100\choose i}{\sqrt[5]{3}}^i{\sqrt[3]{7}}^{100-i}$. Such a term is irrational unless $i$ is a multiple of $5$ and $100-i$ is a multiple of $3$. Can you go from there? $\endgroup$ – Steve Kass May 7 '15 at 21:19
  • $\begingroup$ @BrianTung: I'd say a cute disguise for a Chinese Remainder. $\endgroup$ – Pedro M. May 7 '15 at 21:21
  • $\begingroup$ @Pedro: to-MAY-to, to-MAH-to? :-) $\endgroup$ – Brian Tung May 7 '15 at 21:22
  • $\begingroup$ @BrianTung: Fair enough :-). Although a "Diophantine Equation" sounds scarier, somehow. $\endgroup$ – Pedro M. May 7 '15 at 21:25

Hint: The binomial sum is $$\sum_{k = 0}^{100} {100 \choose k} 3^{k/5}7^{(100-k)/3},$$ and the rational terms are exactly the integer terms, i.e., when both $k/5$ and $(100-k)/3$ are integers.

  • $\begingroup$ @InfinitelyUndefined Have you learned abound congruences? You need to solve the system $k \equiv 0 \pmod 5$ and $k \equiv 100 \pmod 3$ for $k$, by using the Chinese Remainder Theorem. $\endgroup$ – Pedro M. May 7 '15 at 22:13
  • $\begingroup$ No i haven't.Ok i'll try to learn it.There i wanted to mean (100-6-3+2).Sorry for that. $\endgroup$ – Infinitely Undefined May 7 '15 at 22:24
  • $\begingroup$ @InfinitelyUndefined: I assumed this was a problem in a class involving congruences and Chinese Remainder. However, this particular case is simple enough to allow for simpler solutions. For instance, one can write $k = 5m$ and ask when $100 - k = 5(20 - m)$ is a multiple of $3$. $\endgroup$ – Pedro M. May 7 '15 at 23:00
  • $\begingroup$ Thank you,& something tells me that my previous logic was also not entirely wrong.**3|(100-k)** & 5|k.So,the no (100-k) should be such that it is divisible by both 3 & 5 i.e 15. [otherwise 'k' will not be divisible by 5,the vise-versa is impossible as you said that we cannot always find such 'm' such that 3|(100-15m)] .There are 6 such (100-k)'s which are divisible by 15.(15,30,45,60,75,90).Also (100-k) can be 0.So,from here 'k' corresponds to be(85,70,55,40,25,10,100 respectively.) which are all divisible by 5.So,there are 7 rational terms & (100-7)=93 irrational terms. $\endgroup$ – Infinitely Undefined May 8 '15 at 7:22
  • $\begingroup$ @InfinitelyUndefined: You're right, well spotted! You can argue that $100 - k$ must be one of the seven multiples of $15$, hence there are $7$ rational terms. Note, however, that this works because $100$ is a multiple of $5$ (if you want to play around with the problem, you can try to see what happens if the exponent is $101$ instead). $\endgroup$ – Pedro M. May 8 '15 at 16:36

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