$S = 2^2 + 4^2 + 8^2 + ... ... + 256^2 + 512^2$, then what is the second digit from the end of $S$ ?
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2Welcome to stackexchange. Please [edit] the question to show us what you tried and where you are stuck. Then perhaps we can help. Have you worked out the answer for the first few smaller problems like this? – Ethan Bolker Jan 12 '19 at 15:43
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1The sum is clearly less than $9\cdot 1000^2$ which has only $7$ digits, so what does "tenth place digit" even mean here? ... ooh, do you mean the "tens digit", i.e. the second-to-last digit? – hmakholm left over Monica Jan 12 '19 at 15:46
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Yes, I meant tens digit. Second digit from the last. – Adnan Toky Jan 12 '19 at 15:52
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Your sum is $$ (2^1)^2 + (2^2)^2 + \cdots + (2^8)^2 + (2^9)^2 = 4^1 + 4^2 + \cdots + 4^8 + 4^9 $$ which is a finite geometric series that we can throw the standard formula at to get $$ 4\cdot \frac{4^9-1}{3} $$ To evaluate this modulo $100$, use that $3^{-1} \equiv 67 \pmod{100}$ and $4^9 = (2^9)^2 = 512^2 \equiv 12^2 \pmod{100}$, so you're looking for the tens digit of $$ 4\cdot(12^2-1)\cdot 67 $$
hmakholm left over Monica
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It's probably worth explaining how you spotted $4^9\equiv 12^2$ (other than working them out), or the benefit of writing $12^2$ instead of $144$ or $44$. – J.G. Jan 12 '19 at 16:05
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@J.G.: The $4^9$ in the geometric sum was $5\underline{12}^2$ in the question's wording, so that's pretty immediate. (If not for this coincidence, I would have had to look it up; the powers of 2 between 65536 and 1048576 do not come up often enough for me to remember them). – hmakholm left over Monica Jan 12 '19 at 16:10
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Did you just happen to know the value of 1/3 mod 100? For hand or mental calculation you could also use $4^9-1=64^3-1=(64-1)(64^2+64+1)=63x4161$. – marty cohen Jan 12 '19 at 16:11
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1@martycohen: I noticed without much thought that $3\cdot 33=99\equiv -1\pmod{100}$, so the reciprocal of $3$ must be $-33\equiv 67$. – hmakholm left over Monica Jan 12 '19 at 16:13
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@Toky: A more detailed argument would be that multiplying an integer by $201$ does not change its last two digits, so $$ 4\cdot\frac{4^9-1}{3} \equiv 4\cdot(4^9-1)\cdot\frac{201}3 \pmod{100} $$ given that we know the left-hand-side is an integer. – hmakholm left over Monica Jan 12 '19 at 16:17
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Hint:
Let us find $$R=\dfrac{4^9-1}3\pmod{\dfrac{100}4}$$
$4^9=2^{18},2^9=512\equiv12\pmod{25},2^{18}\equiv12^2\equiv-6$
$3^{-1}\equiv-8\pmod{25}$
$R\equiv(-6-1)(-8)\equiv6\pmod{25}$
$4R\equiv6\cdot4\pmod{25\cdot4}$
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