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What is the largest power of 2 that divdies 17^17 + 15^15?

Apr 30, 2020

#1
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Let's break the sum of two numbers down to see if we can find a pattern:

\(17^1+15^1=32\) The highest power of 2 that can divide this is 32 or \(2^5\)
\(17^2+15^2=514\) The highest power of 2 that can divide this is 2 or \(2^1 \)

\(17^3+15^3=8288\) The highest power of 2 that can divide this is 32 or \(2^5 \)

\(17^4+15^4=134146\) The highest power of 2 that can divide this is 2 or \(2^1\)

Here you can see a pattern, the largest power of 2 that can divide the sum is either \(2^5\) or \(2^1\). You can also see that when the exponents in the sum are odd, the sum is divisible by \(2^5\) but when the exponents of the sum are even, the sum is divisible by \(2^1\)

In the sum of the numbers above, their exponents are 2 apart.

\(17^3+15^1=4928\) The highest power of 2 that can divide this is 64 or \(2^6\)

\(17^4+15^2=83746\) The highest power of 2 that can divide this is 2 or \(2^1\)

\(17^5+15^3=1423232\) The highest power of 2 that can divide this is 128 or \(2^7\)

\(17^6+15^4=24188194\) The highest power 2 that can divide this is 2 or \(2^1\)

\(17^7+15^5=411098048\) The highest power of 2 that can divide this is 64 or \(2^6\)

This pattern is different, but we can see that every other odd exponent in the sum has \(2^6\) as the largest power of 2 divisor.

Based on this pattern, \({17}^{17}+{15}^{15}\), the largest divisor that is a power of 2 is \(2^7\).

Apr 30, 2020
#2
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Sorry for the late response and long answer...I made an error in my explanation and had to correct it!

Apr 30, 2020
#3
+1

[17^17  +  15^15]  mod  2^8 =0

Apr 30, 2020
#4
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I guess my solution is wrong...can you elaborate a little more on yours please?

HELPMEEEEEEEEEEEEE  Apr 30, 2020
#5
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Sorry, don't really have any fancy explanation to put forward. Just wrote this short computer code and it came up with  1 , 2 , 3 , 4 , 5 , 6 , 7 , 8  as powers of 2 that divide the expression:

n=1; a=if((15^15 + 17^17) % 2^n==0, goto2, goto3);printn,", ",;n++;if(n<100, goto1, 0)

Guest Apr 30, 2020
#6
0

Ok...Thanks!

HELPMEEEEEEEEEEEEE  May 1, 2020