Math Olympiad | A Nice Exponential Problem | 95% Failed to solve!

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[RK-tfpq]

There is another way, a short cut. 2^a +4^b+16^c. (where ^ represents exponent) can be written as 1x2^a + 4x[4^(b-1)] + 16x[16^(c-1)] which is equal to 1x2^a + 4x[2^(2b-2)] + 16x[2^(4c-4)], because 4 = 2^2 and 16 = 2^4 . Here the sum of the constants in front of the exponents (that is, 1, 4 and 16) is 21. And it is given that this equation is equal is to 21x2^8. Thus each of the three exponent terms in the equation are all equal to 2^8, meaning that a = 8, 2b-2 = 8 and 4c-4 = 8, which gives b=5 and c=3.

[richardleveson]

Thank you - very enjoyable and well explained!

[billygraham]

Why the “condition” that a>b>c? Was that just stated in the statement of the problem or is that part of the method of solution? If it is part of the method at what point in the video was that explained?

[허공]

you made a big mistake assuming 2b>=a, 4c>=a and 4c>=2b.
there are 6 cases with considering size of a, 2b, 4c.
you consider only one case. there may be other solutions.

I found 3 more solutions as followed:
since 16^3 = 4096 < 5376 < 65536 = 16^4, 0 <= c <= 3
(case c = 3)
2^a + 2^(2b) + 4096 = 5376 => 2^a + 2^(2b) = 1280 = 2^8*5

if a>=2b, 2^a + 2^(2b) = 2^(2b)*(2^(a-2b)+1) = 2^8*5
2b=8 & 2^(a-2b)+1 = 5 => b = 4, a = 10
(10, 4, 3) is a solution.

if a<2b, 2^a + 2^(2b) = 2^a*(1 + 2^(2b-a)) = 2^8*5
a = 8 & 1 + 2^(2b-a) = 5 => a = 8, b = 5
(8, 5, 3) is a solution.

(case c = 2)
2^a + 2^(2b) + 256 = 5376 => 2^a + 2^(2b) = 5120 = 2^10*5

if a>=2b, 2^a + 2^(2b) = 2^(2b)*(2^(a-2b)+1) = 2^10*5
2b = 10 & 2^(a-2b)+1 = 5 => b =5, a = 12
(12, 5, 2) is a solution.

if a<2b, 2^a + 2^(2b) = 2^a*(1 + 2^(2b-a)) = 2^10*5
a = 10 & 1 + 2^(2b-a) = 5 => a = 10, b= 6
(10, 6, 2) is a solution.

(case c = 1)
2^a + 2^(2b) + 16 = 5376 => 2^a + 2^(2b) =5360 =2^4*235

if a>=2b, 2^a + 2^(2b) = 2^(2b)*(2^(a-2b)+1) = 2^4*235
2b =4 & 2^(a-2b)+1 = 235 => b = 2, a has no integer solution.

if a<2b, 2^a + 2^(2b) = 2^a*(1 + 2^(2b-a)) = 2^4*235
a = 4 & 1 + 2^(2b-a) = 235 => a = 4, b has no integer solution.

(case c = 0)
2^a + 2^(2b) + 1 = 5376 => 2^a + 2^(2b) = 5375
since RHS is odd, one term of LHS must be 1.
in any case, (a, b) has no integer solution.

Finally answer (a, b, c) = { (10, 4, 3), (8, 5, 3), (12, 5, 2), (10, 6, 2) }

[emopenna]

Aren't [10 4 3]; [10 6 2] and [12 5 2] also solutions?
And if it's not required that a, b and c are integers, can't we also add [8 6 2, 5] and [12 4 2, 5] as solutions either?

[archilk.khoperia]

10, 6, 2 ))) Просто за 2 минуты в векселе подобрал )))

[JPTaquari]

I started from 2^a, which is the closest you CAN get to 5376 is 4096 or 2^12

There was 1280 left, so, by the same reasoning, I go to the exponent 5 which is the maximum that gives: 4^5 = 1024

Now the problem is solved, as there are 256 or 16² left

Cool

[krishnanadityan]

I think of a general comment in that many people go to very many details which is unnecessary because the audience is class. Moreover, the presentation is very tidy contentwise as well by appearance so that alternate solutions fails to be noticed