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all odd squares are $1 \pmod 8.$ This includes $n^4.$ Meanwhile, $$x_1^2 + \cdots x_n^2 \equiv n \pmod 8.$$ So it is necessary to have $n \equiv 1 \pmod 8.$ In the other direction, all numbers $k \equiv 3 \pmod 8$ are the sum of three odd squares. This is a result of Gauss, and equivalent to the fact that all positive integers ae the sum of three ...

5

With three digits $a,b,c$, You should be able to get at most six different $3$-digits numbers, and they are: $abc,acb,bac,bca,cab,cba$So when you add them up, the equation should be$$200(a+b+c)+20(a+b+c)+2(a+b+c)=222(a+b+c)=3231+n$$where $n$ is the unknown $6$th number. By quick estimation you can find that, When $a+b+c=15, n=99$, does not qualify; When $a+... 5 The sum of all six numbers is$222(a+b+c)$. Now you can check the multiples of$222$wich exceed$3231$, to find that$222\cdot18$does the job. 4 First you write the nonnegative integers in base$30,1,2,10,11,12,20,21,22,100,101,102,110,111,112,120,\dots$Then you add up the (base 3) digits in each number (presumably in base$10$)$0,1,2,1,2,3,2,3,4,1,2,3,2,3,4,3,\dots$Since we start from zero, we express$2011$in base$3$, getting$2202111_3$, then add the digits to get$9. 4 This may be over your head if you're in high school, but you want the maximum order of an element in the symmetric group on 12 letters. (In other words, you want Landau's function evaluated at 12.) This is 60, generated by the product of a 5-cycle, 4-cycle, and 3-cycle. (This is surely the way the people who wrote the test thought about this problem!) If ... 2 Let the numbers be arranged as follows: $$\begin{array}{ccccc} & a & & b & \\ 0 & & c & & 1000 \\ & d & & e & \end{array}$$ Then we have the relationships \begin{align*} 3a &= b + c \\ 3d &= c + e \\ 3b &= a + c + 1000 \\ 3e &= c + d + 1000 \\ 6c &= a + b + d + e + 1000. \end{align*} ... 2 The maximum sum of three square digits is9^2+9^2+9^2=243$, so first digit of any solution must be 0,1,2. But that means that the maximum sum of three square digits in any solution is$2^2+9^2+9^2=166$, so first digit must be 0 or 1. If the number is in the range 100-166, then the maximum sum of its squared digits is either$1^2+6^2+6^2=73$if the 2nd ... 1 Firstly, we let two girls be located as far a possible. Therefore, G1 stands at A and G2 stands at corner C. Next, we put G3 at corner D, which is 30m from A. Of course, this setting is not optimal. We then ask G3 walks x m towards G2 and stop at point E such that AE = EC. We then have$30^2 + x^2 = (30 + 20 – x)^2$. After solving for x, we get E is 34m ... 1 Starting off with @heropup's approach, we can speed up the calculations by appealing to symmetry: \begin{array}{ccccc} & a & & b & \\ 0 & & c & & 1000 \\ & d & & e & \end{array} Observe that we must have$a=d, b=e$, and$a = 1000-b$. This eliminates all variables but$b$and$c$(we might as well work with$b$... 1 If$DH$is the trapezium height, notice that $$EF=DH=\sqrt{DA^2-AH^2}=\sqrt{15^2-2.5^2}.$$ On the other hand, if$r$is the circle radius, we have: $$EF=OF+OE=\sqrt{r^2-10^2}+\sqrt{r^2-7.5^2}$$ By comparing and solving for$r$one gets$r^2=135$. 1 From the data, we see that this cyclic quadrilateral is an isosceles trapezium. Splitting the trapezium into two triangles, sharing a common side$x$which subtends angles$\theta$and$180-\theta$in each triangle respectively, we can apply the cosine rule and get $$x^2=625-600\cos\theta=450-450\cos(180-\theta)$$ from which we can deduce that $$\cos\theta=\... 1 Since all p, q , r cant be positive or negative .WLOG, we assume p,q\geq0 and r = -(p+q). So we have to prove$$a^2pq\leq (p+q)(b^2q + c^2p)$$. Dividing by pq,$$a^2\leq (p+q)(b^2/p + c^2/q)$$.(Work it out if one of them is 0). Expand right hand side to$$b^2 + c^2 + b^2q/p + c^2p/q$$. Applying AM-GM to the 2 right most terms,$$b^2q/p + c^2p/q\geq ... 1 This problem is in IMO 2009 shortlist. A solution can be found here: https://www.imo-official.org/problems/IMO2009SL.pdf 1 Here is yet another way to look at it: $$8000+400q+40r+4s=1000s+100r+10q+2 \iff 7998+390q=60r+996s.$$ Since the number on the LHS must end in an 8, the number on the RHS must also end in 8. This means that either$s=3$or$s=8$. If$s=3$then $$7998+390q=60r+2888 \iff 5010+390q=60r \iff 501+39q=6r.$$ Since the smallest the LHS can be is 501 and the ... 1 With$4\times pqrs=srqp$and$pqrs\geq 2000$, we infer that$s=8$or$s=9$. Together with the fact that$\times pqrs$ends in$p=2$, it must be that$s=8$. So we have $$4\times(2000+100q+10r+8)=8000+100r+10q+2\implies 13q=2r-1.$$ Possible values for$2r-1$are$\{-1,1,3,5,7,9,11,13,15,17\}$. It's obvious then that we must have$q=1$and$r=7$. So, overall, ... 1 Firstly, we notice that it must hold that$4s$must end at the digit$2$. Thus, there are two options for$s$. Either$s = 3$or$s = 8$. 1st case:$ s = 3$. In that case we have: $$8000 + 400q + 40r + 12 = 3000 + 100 r +10 q + 2.$$ But in that case the RHS is always less than$8000$, thus$s = 3$is rejected. Hence,$\boxed{s=8}.$In that case, we have:$...

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Neither of $x, y, z$ can be larger than $10$, so you have only $1000$ combinations to check by brute force, which is eminently feasible (and much quicker than trying to be clever). For the edited title: An algorithm that is not by brute force is not a good algorithm for this problem, because the additional time it will take to write it is much longer than ...

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