You are not logged in.
Pages: 1
Hi,
268 numbers are written around a circle. The 17th number is 3, the 83rd is 4 and the 144th is 9, The sum of every 20 consecutive numbers is 72. Find the 210th number.
If every 20 consecutive numbers equal 72, then every 260 consecutive numbers equal 936.
Now take a17 (the 17th number), then the sum of all the 17th-to-8th numbers equals 936. The sum of 18th-to-9th numbers also equals 936. We can see that the a17=a9. While repeating the process I realize the pattern, a17=a9=a1=a261=a253=a245, etc..., that is, the index is subtracted by 8 each time. How can I prove that this pattern continues without calculating each term?
And is there a shortcut for solving the problem?
Thanks for help.
Last edited by atran (2014-01-01 10:35:25)
Offline
Offline
Hm, I seem to be getting .
Is that the answer or am I doing something wrong?
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Offline
The right answer is anonimnystefy's.
Offline
Actually, I totally missed Nehushtan's hide box. If I had seen it I wouldn't have much doubt in my answer.
Have you been able to get the result?
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Offline
Not yet, I'm working on it... Soon will finish. I've programmed a program for the problem and seen a pattern.
Offline
I've written all the code, except the sort-algorithm block. This is written in C. I hope I haven't made mistakes.
#include <stdio.h>
// The number of unknowns is 67
int main()
{
float n[268];
int i, j, u[67], v[67];
for(i=0; i<268; i++) {
n[i]=0;
}
for(i=0; i<67; i++) {
u[i]=-1;
v[i]=-1;
}
i=16;
do {
n[i]=3;
i+=20;
if(i>267) i-=268;
} while(i!=16);
i=82;
do {
n[i]=4;
i+=20;
if(i>267) i-=268;
} while(i!=82);
i=143;
do {
n[i]=9;
i+=20;
if(i>267) i-=268;
} while(i!=143);
i=209;
j=0;
do {
u[j]=i;
i+=20;
if(i>267) i-=268;
j++;
} while(i!=209);
SelectionSort(u, 67);
for(i=0, j=0; i<268; i++) {
if(n[i]!=0) printf("n[%i]\t: %f\n", i, n[i]);
else {
printf("n[%i]\t: Unknown\n", i);
v[j]=i;
j++;
}
}
for(i=0; i<67; i++) {
if(u[i]!=v[i]) printf("All unknowns are not equal.\n");
}
return 0;
}
void SelectionSort(int a[], int array_size)
{
int i;
for (i = 0; i < array_size - 1; ++i)
{
int j, min, temp;
min = i;
for (j = i+1; j < array_size; ++j)
{
if (a[j] < a[min])
min = j;
}
temp = a[i];
a[i] = a[min];
a[min] = temp;
}
}
The output becomes:
n[0] : 3.000000
n[1] : Unknown
n[2] : 4.000000
n[3] : 9.000000
n[4] : 3.000000
n[5] : Unknown
n[6] : 4.000000
n[7] : 9.000000
n[8] : 3.000000
n[9] : Unknown
n[10] : 4.000000
n[11] : 9.000000
n[12] : 3.000000
n[13] : Unknown
n[14] : 4.000000
n[15] : 9.000000
n[16] : 3.000000
n[17] : Unknown
n[18] : 4.000000
n[19] : 9.000000
n[20] : 3.000000
n[21] : Unknown
n[22] : 4.000000
n[23] : 9.000000
n[24] : 3.000000
n[25] : Unknown
n[26] : 4.000000
n[27] : 9.000000
n[28] : 3.000000
n[29] : Unknown
n[30] : 4.000000
n[31] : 9.000000
n[32] : 3.000000
n[33] : Unknown
n[34] : 4.000000
n[35] : 9.000000
n[36] : 3.000000
n[37] : Unknown
n[38] : 4.000000
n[39] : 9.000000
n[40] : 3.000000
n[41] : Unknown
n[42] : 4.000000
n[43] : 9.000000
n[44] : 3.000000
n[45] : Unknown
n[46] : 4.000000
n[47] : 9.000000
n[48] : 3.000000
n[49] : Unknown
n[50] : 4.000000
n[51] : 9.000000
n[52] : 3.000000
n[53] : Unknown
n[54] : 4.000000
n[55] : 9.000000
n[56] : 3.000000
n[57] : Unknown
n[58] : 4.000000
n[59] : 9.000000
n[60] : 3.000000
n[61] : Unknown
n[62] : 4.000000
n[63] : 9.000000
n[64] : 3.000000
n[65] : Unknown
n[66] : 4.000000
n[67] : 9.000000
n[68] : 3.000000
n[69] : Unknown
n[70] : 4.000000
n[71] : 9.000000
n[72] : 3.000000
n[73] : Unknown
n[74] : 4.000000
n[75] : 9.000000
n[76] : 3.000000
n[77] : Unknown
n[78] : 4.000000
n[79] : 9.000000
n[80] : 3.000000
n[81] : Unknown
n[82] : 4.000000
n[83] : 9.000000
n[84] : 3.000000
n[85] : Unknown
n[86] : 4.000000
n[87] : 9.000000
n[88] : 3.000000
n[89] : Unknown
n[90] : 4.000000
n[91] : 9.000000
n[92] : 3.000000
n[93] : Unknown
n[94] : 4.000000
n[95] : 9.000000
n[96] : 3.000000
n[97] : Unknown
n[98] : 4.000000
n[99] : 9.000000
n[100] : 3.000000
n[101] : Unknown
n[102] : 4.000000
n[103] : 9.000000
n[104] : 3.000000
n[105] : Unknown
n[106] : 4.000000
n[107] : 9.000000
n[108] : 3.000000
n[109] : Unknown
n[110] : 4.000000
n[111] : 9.000000
n[112] : 3.000000
n[113] : Unknown
n[114] : 4.000000
n[115] : 9.000000
n[116] : 3.000000
n[117] : Unknown
n[118] : 4.000000
n[119] : 9.000000
n[120] : 3.000000
n[121] : Unknown
n[122] : 4.000000
n[123] : 9.000000
n[124] : 3.000000
n[125] : Unknown
n[126] : 4.000000
n[127] : 9.000000
n[128] : 3.000000
n[129] : Unknown
n[130] : 4.000000
n[131] : 9.000000
n[132] : 3.000000
n[133] : Unknown
n[134] : 4.000000
n[135] : 9.000000
n[136] : 3.000000
n[137] : Unknown
n[138] : 4.000000
n[139] : 9.000000
n[140] : 3.000000
n[141] : Unknown
n[142] : 4.000000
n[143] : 9.000000
n[144] : 3.000000
n[145] : Unknown
n[146] : 4.000000
n[147] : 9.000000
n[148] : 3.000000
n[149] : Unknown
n[150] : 4.000000
n[151] : 9.000000
n[152] : 3.000000
n[153] : Unknown
n[154] : 4.000000
n[155] : 9.000000
n[156] : 3.000000
n[157] : Unknown
n[158] : 4.000000
n[159] : 9.000000
n[160] : 3.000000
n[161] : Unknown
n[162] : 4.000000
n[163] : 9.000000
n[164] : 3.000000
n[165] : Unknown
n[166] : 4.000000
n[167] : 9.000000
n[168] : 3.000000
n[169] : Unknown
n[170] : 4.000000
n[171] : 9.000000
n[172] : 3.000000
n[173] : Unknown
n[174] : 4.000000
n[175] : 9.000000
n[176] : 3.000000
n[177] : Unknown
n[178] : 4.000000
n[179] : 9.000000
n[180] : 3.000000
n[181] : Unknown
n[182] : 4.000000
n[183] : 9.000000
n[184] : 3.000000
n[185] : Unknown
n[186] : 4.000000
n[187] : 9.000000
n[188] : 3.000000
n[189] : Unknown
n[190] : 4.000000
n[191] : 9.000000
n[192] : 3.000000
n[193] : Unknown
n[194] : 4.000000
n[195] : 9.000000
n[196] : 3.000000
n[197] : Unknown
n[198] : 4.000000
n[199] : 9.000000
n[200] : 3.000000
n[201] : Unknown
n[202] : 4.000000
n[203] : 9.000000
n[204] : 3.000000
n[205] : Unknown
n[206] : 4.000000
n[207] : 9.000000
n[208] : 3.000000
n[209] : Unknown
n[210] : 4.000000
n[211] : 9.000000
n[212] : 3.000000
n[213] : Unknown
n[214] : 4.000000
n[215] : 9.000000
n[216] : 3.000000
n[217] : Unknown
n[218] : 4.000000
n[219] : 9.000000
n[220] : 3.000000
n[221] : Unknown
n[222] : 4.000000
n[223] : 9.000000
n[224] : 3.000000
n[225] : Unknown
n[226] : 4.000000
n[227] : 9.000000
n[228] : 3.000000
n[229] : Unknown
n[230] : 4.000000
n[231] : 9.000000
n[232] : 3.000000
n[233] : Unknown
n[234] : 4.000000
n[235] : 9.000000
n[236] : 3.000000
n[237] : Unknown
n[238] : 4.000000
n[239] : 9.000000
n[240] : 3.000000
n[241] : Unknown
n[242] : 4.000000
n[243] : 9.000000
n[244] : 3.000000
n[245] : Unknown
n[246] : 4.000000
n[247] : 9.000000
n[248] : 3.000000
n[249] : Unknown
n[250] : 4.000000
n[251] : 9.000000
n[252] : 3.000000
n[253] : Unknown
n[254] : 4.000000
n[255] : 9.000000
n[256] : 3.000000
n[257] : Unknown
n[258] : 4.000000
n[259] : 9.000000
n[260] : 3.000000
n[261] : Unknown
n[262] : 4.000000
n[263] : 9.000000
n[264] : 3.000000
n[265] : Unknown
n[266] : 4.000000
n[267] : 9.000000
Process returned 0 (0x0) execution time : 0.110 s
Press any key to continue.
According the my program, all the unknowns should be equal to the 210th number.
Now pick any set of 20 consecutive numbers, then it's plain algebra:
Last edited by atran (2014-01-01 13:34:51)
Offline
Pages: 1