1. ## Farmer's Story Problem

A cowboy was asked how many cows were on the ranch. He replied that he was unsure, but he knew that when he ocunted them by twos, threes, fours, fives, or sixes, he always had one left over. The only way he could acoid this was to count by sevens; he then had none left over. What is the smallest number of cows on the ranch?

2. Originally Posted by swimalot
A cowboy was asked how many cows were on the ranch. He replied that he was unsure, but he knew that when he ocunted them by twos, threes, fours, fives, or sixes, he always had one left over. The only way he could acoid this was to count by sevens; he then had none left over. What is the smallest number of cows on the ranch?

we know that it needs to be a multiple of 7 from all of the other clues we know it can't be even (because two can't divide it) and the last digit need to be a one because when divided by 5 it needs one left over.

here is our list,

56,63,70,77,84,91

If you check all the other conditions you will see that they hold.

I hope this helps.

edit: 4|91 12(3)+3
so 91 is out.

Sorry.

3. ## farmers story

posted by swimalot

count by sixes until you find a number divisible by each counting multiple.this is 60 so the farmer has 61 cows

4. Originally Posted by bjhopper
posted by swimalot

count by sixes until you find a number divisible by each counting multiple.this is 60 so the farmer has 61 cows
Originally Posted by swimalot
The only way he could acoid this was to count by sevens; he then had none left over.
I am sorry this is incorrect becuse of the above statement 7 must divide the number of cows

$7|61$ gives $61=7\cdot 8 +5$

He would have 5 cows left over. So 61 can't be an answer

5. Hello, swimalot!

A cowboy was asked how many cows were on the ranch.
He replied that he was unsure, but he knew that when he counted them
by twos, threes, fours, fives, or sixes, he always had one left over.
When counted by sevens, he had none left over.
What is the smallest number of cows on the ranch?
Let $N$ = number of cows on the ranch.

The LCM of 2, 3, 4, 5, 6 is $60$
. . Hence: . $N \:=\:60a + 1$ ... for some integer $a.$

Since $N$ is divisible by 7: . $60a + 1 \:=\:7b$ ... for some integer $b.$

Solve for $b\!:\;\;b \:=\:\frac{60a+1}{7} \;=\;8a + \frac{4a+1}{7}$

Since $b$ is an integer, $4a + 1$ must be divisible by 7.

The first time this happens is: $a = 5$

Therefore: . $N \;=\;60(5)+1 \;=\;\boxed{301}$

6. Originally Posted by TheEmptySet
we know that it needs to be a multiple of 7 from all of the other clues we know it can't be even (because two can't divide it) and the last digit need to be a one because when divided by 5 it needs one left over.

here is our list,

56,63,70,77,84,91

If you check all the other conditions you will see that they hold.

I hope this helps.
Hello Tessy

91 doesn't work for ... four

91=88+3

7. You could use the Chinese Remainder Theorem, Topsquark has a lovely example in post #5 here http://www.mathhelpforum.com/math-he...nt-modulo.html

8. I will do the Chinese Remainder Theorem for you:

we know that if we divide it by 1, 2, 3, 4, 5, 6 that we have a remainder of 1, and when divided by 7, there is a remainder of 0. So let x be the number of cows, then we have:

$x \equiv 1 (mod2)$
$x \equiv 1 (mod3)$
$x \equiv 1 (mod4)$
$x \equiv 1 (mod5)$
$x \equiv 1 (mod6)$
$x \equiv 0 (mod7)$

Now, 2, 4, 6 are not pairwise coprime, I am not sure how to handle this, so I will remove the 4 and the 6, and then test them afterwards.

This gives us:
$x \equiv 1 (mod2)$
$x \equiv 1 (mod3)$
$x \equiv 1 (mod5)$
$x \equiv 0 (mod7)$

So because 2*3*5*7 = 210, we need to find a pair of integers $(r_i, s_i)$ such that:

$2r_1 + \left ( \frac{210}{2} \right )s_1 = 1$

$3r_2 + \left ( \frac{210}{3} \right )s_2 = 1$

$5r_3 + \left ( \frac{210}{5} \right )s_3 = 1$

$7r_4 + \left ( \frac{210}{7} \right )s_4 = 1$

Using Topsquark's numbers (I got different ones the last time I did it, but it doesn't matter, because any solution will work)
$(r_1, s_1) = -51, 1)$
$(r_2, s_2) = -23, 1)$
$(r_3, s_3) = -25, 3)$
$(r_4, s_4) = -17, 4)$

Then
$x \equiv 1 \cdot \left ( \frac{210}{2} \right )s_1 + 1 \cdot \left ( \frac{210}{3} \right )s_2 + 1 \cdot \left ( \frac{210}{5} \right )s_3 + 0 \cdot \left ( \frac{210}{7} \right )s_4 ~~~~(mod 210)$

Simplify and substitute in the values
$x \equiv \left ( \frac{210}{2} \right ) \cdot 1 + \left ( \frac{210}{3} \right )\cdot 1+ \left ( \frac{210}{5} \right )\cdot 3 ~~~~(mod 210)$

Simplify
$x \equiv 301 (mod 210)$

And because $301 \equiv 91 (mod 210)$
$x \equiv 91 (mod 210)$
We choose this one because it is the smallest positive number which is congruent to 301 (mod210).

Now, I omitted the 4 and 6 because I didn't know how to deal with them as they were not pairwise coprime, but we should test each of them to make sure it works: $1 \equiv 91 (mod 4)$ this is false (1=0*4+1, and 91=22*4+3) so we know that $x\not{=}91$ So we try the next number that satisfies $x \equiv 91 (mod 210)$

x = 91 + 1*210 = 301

and we test this one
$1 \equiv 301 (mod 4)$ This is true (1=0*4+1, and 301=75*4+1)
$1 \equiv 301 (mod 6)$ This is true (1=0*6+1, and 301=50*6+1)

So the answer is 301 cows.