An Interesting Equation of Powers

[@wesleydeng71]

For the 1st part consider mod 6. Since 4^a==4(mod 6) if a>0, one gets a=c=0 right away.

[@mahmud9930]

Yes, good point.
4^aΞ4(mod 6) a>=1
If we consider that,
(4)+(-1)^b+(0)Ξ(1)^d (for mod 6) it is obvious that there is no solution. So a=0.
1+5^b+6^c=7^d and we can see that c is also 0. Then equation is too easy to solve:))

[@sadeekmuhammadryan4894]

The results you derive are most often useful. I want to master this trick.

[@youtubeuser6581]

If you see these types of problem its always modulo

[@sadeekmuhammadryan4894]

@@youtubeuser6581 Right, modular arithmetic is a very handy tool.

[@youtubeuser6581]

@@sadeekmuhammadryan4894 plus, questions like these usually have really trivial solutions. Your best bet before trying anything is to do a or b or c modulo something, and note that it only satisfies RHS if the solution is 1 or 2 or 0, maybe 3. Just a few simple tips to weasle your way through the math competition.

[@sadeekmuhammadryan4894]

@@youtubeuser6581 Thank You for the advice. 😁

[@mcwulf25]

The trivial solution is so often the only solution!

[@piman9280]

...... which is why this problem was so disappointing!

[@yoav613]

Nice.i do not think there is more simple solution than this.

[@ankit..901]

if x[x[x[x]]]=88 then x=a/b where gcd(a,b)=1 find a+b ([•] represents greatest integer function)...TRY THIS

[@eduardciuca217]

The solution is x=22/7. It's a well known value. So a+b=22+7=29

[@alexandercenteno7418]

Find the sum of the N terms that satisfy that 1^3 + 2^3 + ... + n^3 divided by n+5 gives as residue 17 help please

[@yakupbuyankara5903]

a=1/2,b=3,c=3,d=3.

[@bryanchaler8128]

What branch of math is this? I stopped math at de.

[@browhat6935]

Number theory and modular arithmetic specifically

[@AEROXYDHYA]

a=1 b=1 c=2 and d =3

[@AEROXYDHYA]

This can also be one of the solution

[@alikagazade5126]

do not do that. Only look mod 3

[@willbishop1355]

Another one where the only solution is the trivial one. Kind of boring.

[@nigeldbd]

wow bro you're so smart please have my children

[@piman9280]

@@nigeldbd It's Will, not Wlihemina.