What functions can I use to find how many integer terms the following expression has?

$$\left (\sqrt[19]{19}+\sqrt[95]{95}\right )^{1995}$$

Edit 1

In the development of that binomial, at some point there are terms that are whole numbers, I need to know if in those 1996 terms of development, how many are integers. Which Mathematica functions should I use to find this out?

Edit 2

"integer terms" = an example with other numbers $\sqrt{8}\sqrt{2}=\sqrt{16}=4$

put on hold as off-topic by Szabolcs, David G. Stork, QuantumDot, LCarvalho, Öskå Nov 15 at 6:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question cannot be answered without additional information. Questions on problems in code must describe the specific problem and include valid code to reproduce it. Any data used for programming examples should be embedded in the question or code to generate the (fake) data must be included." – Szabolcs, David G. Stork, QuantumDot, LCarvalho, Öskå
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    Please clarify what you mean by "integer terms". – Henrik Schumacher Nov 9 at 20:01
  • 2
    I voted to close this because people are clearly guessing at what you want. I will retract the vote if the question is clarified. – Szabolcs Nov 9 at 20:06
  • Indeed.... a VERY useless title. – David G. Stork Nov 9 at 23:47
  • @David G. Stork So, what would be the useful title for this case? – zeros Nov 10 at 1:52
  • What is "terms of development"? – QuantumDot Nov 11 at 1:49
 IntegerLength@IntegerPart[(19^(1/19) + 95^(1/95))^1995]    


  • After OP clarified, I am pretty sure that this is not what they wanted. – Henrik Schumacher Nov 9 at 23:00

Not sure what you mean. I interpret it as if you were looking for the number of integer summands after binomial expansion of the sum.

a = 19^(1/19);
b = 95^(1/95);
Count[Table[Binomial[1995, k] a^k b^(1995 - k), {k, 0, 1995}], _Integer]


If my assumption is right, then this will be a somewhat less expensive way to compute this number:

k = PadRight[IntegerPartitions[1995, 2]];
{i, j} = Transpose[Join[k, Reverse /@ k]]; 
Count[Mod[i, 19] + Mod[j, 95], 0]


And it is actually

Count[Mod[j, 95], 0]


since 19 is a divisor of 95.

This might also give you a hint on how to solve this exercise on paper.

  • That is what I understood too, but others interpret it differently. – Szabolcs Nov 9 at 20:07
  • Yeah, it sounds like a Math Olympics exercise from 1995... – Henrik Schumacher Nov 9 at 20:07
  • Henrik Schumacher You've hit the nail on the head, that's exactly what I needed, thank you very much, but that 690 that you got before What does it represent? – zeros Nov 10 at 2:04
  • 690 is the number of digits of the integer part of the number, so the number of digits in front of the decimal dot of the number's decimal representation. – Henrik Schumacher Nov 10 at 6:54
Ceiling[Log[10, (19^(1/19) + 95^(1/95))^1995]]
(* 690 *)
  • After OP clarified, I am pretty sure that this is not what they wanted. – Henrik Schumacher Nov 9 at 23:00

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