Are You Smarter Than A Sixth Grader (Taking Seventh Grade Math)?
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That's completely irrelevant. 761 => 1U, 2R, 3R, 4R, 5R, 6U, 7U. 752 => 1R, 2U, 3R, 4R, 5U, 6R, 7U. 743 => 1R, 2R, 3U, 4U, 5R, 6R, 7U. 7412 => 1U, 2U, 3R, 4U, 5R, 6U, 7U. .... Follow my previous post. Simply put the 1 first instead of the order I have. I have them in a more mathematical order that I used to split the numbers into the two groups. Order does not matter.
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That's completely irrelevant. 761 => 1U, 2R, 3R, 4R, 5R, 6U, 7U. 752 => 1R, 2U, 3R, 4R, 5U, 6R, 7U. 743 => 1R, 2R, 3U, 4U, 5R, 6R, 7U. 7412 => 1U, 2U, 3R, 4U, 5R, 6U, 7U. .... Follow my previous post. Simply put the 1 first instead of the order I have. I have them in a more mathematical order that I used to split the numbers into the two groups. Order does not matter.
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7 8 9
Mongo: Mongo only pawn... in game of life.
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Homework problem. Q: On the (x,y) coordinate system, you start at (0,0) and want to get to (14,14). At each step n, you move exactly n steps either right or up. So initially, you can only move 1 unit right or up, then 2, then 3, etc. How many paths are there to (14,14)? A: He got one on his own and he understood the symmetry and got 2. I told him my answer and why. Anyone wants to try?
I guess I missed something. I thought 'n' was a specific number to be repeated at each step so the answer would be only the factors of 14. Therefore only these 'n' steps would be allowed. 1,2,7, and 14 - 2R, 2U, etc... or 7R,7U ... So the number of paths would be 4
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Homework problem. Q: On the (x,y) coordinate system, you start at (0,0) and want to get to (14,14). At each step n, you move exactly n steps either right or up. So initially, you can only move 1 unit right or up, then 2, then 3, etc. How many paths are there to (14,14)? A: He got one on his own and he understood the symmetry and got 2. I told him my answer and why. Anyone wants to try?
At each point you can move either right or up. The setup makes the step sizes irrelevant to the problem - they will always add to 28 (14 horizontal + 14 vertical), (1 + 2 + 3 + 4 + 5 + 6 + 7). That means we will reach (14,14) from (0,0) in exactly 7 steps. At each of the 7 steps you can either move right or up, so you have two choices at every step node. The answer then becomes: 2^7 = 128 possible paths. If you were to visualize it, you would simply draw a complete grid between (0,0) and (14,14)
"There are only 10 types of people in the world - those who know binary and those who don't."
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Ha ha, I did too. I was with you all the way. I was all, "Yeah, you tell 'im, Bassam!" Until you said that and I rechecked my answers. :sigh:
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Ha ha, I did too. I was with you all the way. I was all, "Yeah, you tell 'im, Bassam!" Until you said that and I rechecked my answers. :sigh:
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Homework problem. Q: On the (x,y) coordinate system, you start at (0,0) and want to get to (14,14). At each step n, you move exactly n steps either right or up. So initially, you can only move 1 unit right or up, then 2, then 3, etc. How many paths are there to (14,14)? A: He got one on his own and he understood the symmetry and got 2. I told him my answer and why. Anyone wants to try?
Shouldn't it be 16? If there are 8 distinct groups of 3-4 numbers that sum to 14, one path will start with the first move going up and the next will start with the first move going right. Incidentally, here is my code:
function combine(a, min) {
var fn = function(n, src, got, all) {
if (n == 0) {
if (got.length > 0) {
all[all.length] = got;
}
return;
}
for (var j = 0; j < src.length; j++) {
fn(n - 1, src.slice(j + 1), got.concat([src[j]]), all);
}
return;
}
var all = [];
for (var i = min; i < a.length; i++) {
fn(i, a, [], all);
}
all.push(a);
return all;
}function arr_sum(arr) {
var n = 0;
for (var i = 0; i < arr.length; i++) {
n += arr[i];
}
return n;
}
k=[1,2,3,4,5,6,7];
combine(k, 1).filter(function(v, i) {
return 14 == arr_sum(v)
}) -
Shouldn't it be 16? If there are 8 distinct groups of 3-4 numbers that sum to 14, one path will start with the first move going up and the next will start with the first move going right. Incidentally, here is my code:
function combine(a, min) {
var fn = function(n, src, got, all) {
if (n == 0) {
if (got.length > 0) {
all[all.length] = got;
}
return;
}
for (var j = 0; j < src.length; j++) {
fn(n - 1, src.slice(j + 1), got.concat([src[j]]), all);
}
return;
}
var all = [];
for (var i = min; i < a.length; i++) {
fn(i, a, [], all);
}
all.push(a);
return all;
}function arr_sum(arr) {
var n = 0;
for (var i = 0; i < arr.length; i++) {
n += arr[i];
}
return n;
}
k=[1,2,3,4,5,6,7];
combine(k, 1).filter(function(v, i) {
return 14 == arr_sum(v)
}) -
I think there are 16 distinct paths, as below R1U2R3R4U5R6U7 U1R2U3U4R5U6R7 R1R2U3U4R5R6U7 U1U2R3R4U5U6R7 R1R2U3R4U5U6R7 U1U2R3U4R5R6U7 U1U2R3U4R5R6U7 R1R2U3R4U5U6R7 U1U2R3R4U5U6R7 R1R2U3U4R5R6U7 U1R2U3U4R5U6R7 R1U2R3R4U5R6U7 R1U2U3U4U5R6R7 U1R2R3R4R5U6U7
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I think there are 16 distinct paths, as below R1U2R3R4U5R6U7 U1R2U3U4R5U6R7 R1R2U3U4R5R6U7 U1U2R3R4U5U6R7 R1R2U3R4U5U6R7 U1U2R3U4R5R6U7 U1U2R3U4R5R6U7 R1R2U3R4U5U6R7 U1U2R3R4U5U6R7 R1R2U3U4R5R6U7 U1R2U3U4R5U6R7 R1U2R3R4U5R6U7 R1U2U3U4U5R6R7 U1R2R3R4R5U6U7