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RMO 2024 Problem 6 | Case work in Combinatorics with a Switch | Math Olympiad

  Рет қаралды 1,212

Cheenta Academy for Olympiad & Research

Cheenta Academy for Olympiad & Research

Күн бұрын

Пікірлер: 17
@Vabadrish
@Vabadrish Ай бұрын
In todays paper I couldn't solve any other problem(ofc becoz of skill issue😅) But i solved this one(which was supposed to be the toughest one) I never prepared for rmo separately....im satisfied with the verry little ive achieved but i deeply enjoyed solving this problem 🙃
@ArnavAnoopNarayanan
@ArnavAnoopNarayanan 15 күн бұрын
i have a doubt in 11:28 cant j equals m as well
@dakcom-mk6mp
@dakcom-mk6mp Ай бұрын
Nice
@dibyasatyamghosh1639
@dibyasatyamghosh1639 Ай бұрын
Sir in the exam I also thought along the same lines but there is a problem that 2^(j+1)
@Cheenta
@Cheenta Ай бұрын
Though 2^{j+1} < 3*2^j, it does not divide 3*2^j. Hence that won't be counted.
@dibyasatyamghosh1639
@dibyasatyamghosh1639 Ай бұрын
@Cheenta aah! I missed that could have gotten it fully correct
@dibyasatyamghosh1639
@dibyasatyamghosh1639 Ай бұрын
Sir how much partial can I expect in this problem?
@izukumidoriya7636
@izukumidoriya7636 Ай бұрын
how many did you solve except this problem? also which grade are you in may I know?
@dibyasatyamghosh1639
@dibyasatyamghosh1639 Ай бұрын
@@izukumidoriya7636 expect this I solved 1,3 and 5 fully .And I am in 12th
@radheshyamdubey1052
@radheshyamdubey1052 Ай бұрын
Sir please solve kv jnv paper there are 400 people from kv jnv who had a different paper sir
@manojmalv
@manojmalv Ай бұрын
f(p)=1 f(p^k)=2
@Illusioner_
@Illusioner_ Ай бұрын
f(p^k) is not necessarily 2. Think about f(27), {1,3,27}, {1,3,9,27}, {1,9,27} and {1,27} all work.
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