BMO1 A 2020 4

Darn, it’s been such a while since I’ve posted here but here we go:

First, we notice that 2^{10} \approx 10^3. So if we take 2^{22}, there’s no way that we can get a sixteen digit number. Similarly, we can do this for A=4 and find out that 4^{44} is way too big. Thus, A=3.

Now, let’s compute the LHS mod 100.
To do this we separate 100 into 4 and 25 and use the Chinese Remainder Theorem:
3^{33}+33 \equiv (-1)^{33} + 1 \equiv 0 \pmod {4}

Looking at remainders mod 25 isn’t so easy but one can simply just multiply and find that the pattern for powers of 3 \pmod {25} goes as follows:
3, 9, 2, 6, 18, 4, 12, 11, 8, -1
-3, -9, -2, -6, -18, -4, -12, -11, -8, 1
. . .

So, we can derive that 3^{33} \equiv -2 \pmod{25} and adding 33, which is 8 \pmod {25}, gets us 6.

Now, by CRT, we can find that 56 is both 0 \pmod {4} and 6 \pmod {25}.
So B=5 and E=6.

From here, since the LHS is divisible by 3 but not 9, the sum of the digits in the RHS must also do the same. Adding, we get 78 + 2D. Now D has to be a multiple of 3 for 78 + 2D to be a multiple of 3. However, we know that D cannot be 3, 6, or 9 (as A=3, E=6, and C=9), so thus D=0.

Putting everything together:
A=3, B=5, D=0, E=6

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nice solution XD

Thank you. Honestly, if I had a lot of time when taking the BMO, I’d just multiply the powers of 3 out XD (if the solution was harder to find).

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lmao true, computing 27^{11}+33 in the exam ought to be fun