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Major Grumblegutts' Challenge

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Major Grumblegutts stood in his command tent. “I need an exact figure of the number of men this division has available to go against the enemy, not counting those assembled here, of course”, he bellowed.

“Well sir, during the march, as the regular columns passed, I counted that there were five columns more then the number of men in the front line.”, replied one lieutenant.

“Yes, and after they had arrived at camp, I was told that an additional 845 men were added to the front line in order to dig trenches and that four other groups of the same amount were dispatched to other tasks”, said another lieutenant.

“Quite right, and I was in one of those of those groups when I took charge of it, and was given an addition of ten percent from another group to conduct a covert sortie”, offered a sergeant.

“Tell me about this sortie”, commanded Major Grumblegutts, “I understand it did not go well”.

“Well no sir”, the sergeant replied with his eyes downcast. “It did not. I stayed behind in camp to coordinate with the lieutenants. The entire sortie was ambushed and a number of men killed. Of the survivors a sixth managed to make it back to camp. Another eighth of the remaining perished on the enemy’s march back. After arriving at the enemy’s camp another man died of his wounds and a fourth of those remaining attempted to escape the following night, but only two out of three managed to do so and those that didn't were killed as an example. Those that remained were split up into four equal work gangs and sent to separate locations. Two of those gangs were liberated before reaching their destination and have been returned to us, but one in twelve is still not fit for duty”, the sergeant concluded.

Major Grumblegutts furiously scribbled some figures on the back of a discarded slip of paper. “Hhrrumph”, he grunted when finished. “If our intelligence is correct then the enemy has three men to our two, but we have the high ground. This will be a tough fight."

What is the size of the enemy force?

Obviously, it is fifty percent larger.

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Posted · Report post

Have the sergeant summoned in to report on the sortie after the exclusion comment? I believe that should work.

3549 in the other groups

1 sergeant stayed behind

88 returned in the first group from survivors

64 escaped

132 were rescued and are fit for duty

Soliders = 3834

Enemy = 5751

Was this what you were looking to get? I assume we are still not counting the Major and Lieutenants for the counts.

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Posted · Report post

No one wants to accept the challenge of determining the size of the Major's or his enemy's available forces?

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"four other groups of the same amount were dispatched to other tasks" Same amount as the 845 or just the same amount for the 4 groups? I'm guessing different than 845 since 10% has to come from one of them.

Is 845 + 4x = regular columns? Even if we find out x how do we determine the size of a column get the number of soliders on the front line?

The first part says 1/6 of the survivors returned to camp and implies that 5/6 of the survivors were sent to the enemy camp. Let s = survivors.

1/8 of 5/6 * s died on the way to the enemy camp. So 35/48 * s left alive in enemy camp.

35s/48 - 1 left alive

(35s/48 - 1)/4 tried to escape with 2/3 succeeding... So (35s/48-1)/6 made it back.

(35s/48 - 1) * 3/4 remain in the enemy camp

Half of that are freed... (35s/48 - 1) * 3/8

I'm assuming that 11/12 of that last set are fit for duty... (35s/48 - 1) * 11/32

Assume I find a value for s that will work... how will that tell me x * 1.1 since an unknown number of soliders died before the survivors were even considered?

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Posted · Report post

The soldiers marched in a rectangle that was "x" men wide by "x + 5" men deep. Upon arriving the men were split into 5 groups that were "x + 845"

The number of men in one group plus some additional men were sent on the sortie.

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Posted · Report post

Oh...so not a military column but using an actual column count with row count. That clears up a bit

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I found some numbers that would work for the sortie if I assume that none died before the 1/6 returned to camp and the 5/6 were captured.

A group of 480 would become 528 when 10% added.

88 returned to camp and 440 were captured

55 died on the way to the enemy camp leaving 385

384 were left after the one man died

96 tried to escape but only 64 succeeded leaving 288 in the enemy camp

they were split into four groups of 72

two groups totalling 144 were rescued but 12 of those 144 are still not fit for duty

But I can't get 845+4*480 to equal rank * file where rank + 5 = file.

845 + 4 * x will always be an odd number but rank^2 + 5 * rank will always be even. (odd * odd) + (odd * odd) = even or (even * even) * (odd * even)

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Posted · Report post

Just re-read your earlier post.

The rank = 65 and file = 70 for a total of 4550. rank^2 + 5 * rank = 5 * (rank + 845) => rank^2 = 4225

The sortie was 1001 men. The only number that will work for the survivors is 528. The next number that would work is 2064 but that is more than 1001.

3549 remained at camp.

88 returned in the first group from survivors

64 escaped

132 were rescued and are fit for duty

Which totals 3833... which can't be divided in half for the enemy to have 50% more...

What did I miss now?

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Posted · Report post

What did I miss now?

I stayed behind in camp to coordinate with the lieutenants.

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"I need an exact figure of the number of men this division has available to go against the enemy, not counting those assembled here, of course"

If the sergeant was part of the 4550 then by being assembled with the group talking he can't be added back to the 3833.

"I was in one of those of those groups when I took charge of it, and was given an addition of ten percent from another group to conduct a covert sortie"

I read this as though the sergeant was part of the 1001. Staying behind made the sortie 1000 which doesn't change the survivor count. This still leaves an odd number in camp. Either the sergeant would have to be added to the count or have come from a different group to make the camp count even.

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Posted · Report post

Seeing as there are only 2 lieutenants at the command that doesn't change the fact that we end up with an odd number, but if we factor in the major there would be 3830 men fit for battle and 3842 in camp total. I got the same numbers that curr3nt did. When I have a little more time to type things out I will show what I did. So I would say either the enemy has 5745 or 5763 soldiers.

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Ahh. I see the mistake I made. What if we decide that the sergeant was called into the meeting after it started. Does that make the last part solvable?

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Have the sergeant summoned in to report on the sortie after the exclusion comment? I believe that should work.

3549 in the other groups

1 sergeant stayed behind

88 returned in the first group from survivors

64 escaped

132 were rescued and are fit for duty

Soliders = 3834

Enemy = 5751

Was this what you were looking to get? I assume we are still not counting the Major and Lieutenants for the counts.

I believe that does it. Thanks for hanging in there. :thumbsup:

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