Word prob: A factory worker uses machines to sort

[Bladesofhalo]

I have probelm I would like to solve, but I just can't seem to get the right answer. Here it is :?

"A factory worker uses machines to sort cards into pile. On one occasion a machine operator obtained the following curious result. When a box of cards was sorted into 7 equal groups, there were 6 cards left over; when a box of cards was sorted into 5 equal groups, there were 4 cards left over; and when it was sorted into 3 equal groups, there were 2 left over. If the machine cannot sort more than 200 cards at a time, how many cards were in the box?''

 

[Denis]

one oh four

 

[soroban]

Hello, Bladesofhalo!

A factory worker uses machines to sort cards into pile.

On one occasion a machine operator obtained the following curious results.
When a box of cards was sorted into 7 equal groups, there were 6 cards left over.
When a box of cards was sorted into 5 equal groups, there were 4 cards left over.
When it was sorted into 3 equal groups, there were 2 left over.
If the machine cannot sort more than 200 cards at a time, how many cards were in the box?

The approach depends on what techniques you've been taught.
If you're familiar with Modular Arithmetic, the solution is straightforward.
If you know only basic algebra, we can baby-talk our way through it.

Let \(\displaystyle N\) = number of cards in a box.

We are told that: \(\displaystyle \,\begin{array}{ccc}(1)\;N\:=\:7a\,+\,6 \\ (2)\;N\:=\:5b\,+\,4\\ (3)\;N\:=\:3c\,+\,2\end{array}\;\) where \(\displaystyle a,\,b,\,c\) are positive integers.

Equate (2) and (3): \(\displaystyle \:5b\,+\,4\:=\:3c\,+\,2\;\;\Rightarrow\;\;c\:=\:\frac{5b\,+\,2}{3}\;\;\) (4)

Since \(\displaystyle c\) is an integer, \(\displaystyle 5b\,+\,2\) must be a multiple of 3.
. . Testing a few numbers, we find that: \(\displaystyle \:b\:=\:2,\,5,\,8,\,\cdots\)
. . That is, \(\displaystyle b\) is one less than a multiple of 3: \(\displaystyle \:b\:=\:3m\,-\,1\;\;\) (5)

Substitute into (4): \(\displaystyle \:c\:=\:\frac{5(3m\,-\,1)\,+\,2}{3}\;\;\Rightarrow\;\;c\:=\:5m\,-\,1\;\;\) (6)


Equate (1) and (2): \(\displaystyle \:7a\,+\,6\:=\:5b\,+\,4\;\;\Rightarrow\;\;a\:=\:\frac{5b\,-\,2}{7}\)

Substitute (5): \(\displaystyle \:a\:=\:\frac{5(3m\,-\,1)\,-\,2}{7}\:=\:\frac{15m\,-\,7}{7}\)

Since \(\displaystyle a\) is an integer, \(\displaystyle m\) must be a multiply of 7: \(\displaystyle \:m\,=\,7k\;\;\) (7)
Then: \(\displaystyle \:a\:=\:\frac{15(7k)\,-\,7}{7}\;\;\Rightarrow\;\;\L a\:=\:15k\,-\,1\)

Substitute (7) into (5): \(\displaystyle \:b\:=\:3(7k)\,-\,1\;\;\Rightarrow\;\;\L b\:=\:21k\,-\,1\)

Substitute (7) into (6): \(\displaystyle \;c\:=\:5(7k)\,-\,1\;\;\Rightarrow\;\;\L c\:=\:35k\,-\,1\)


So we have: \(\displaystyle \L\:\left\{\begin{array}{ccc}a\:=\:15k\,-\,1 \\ b\:=\:21k\,-\,1 \\ c\:=\:35k\,-\,1\end{array}\)

\(\displaystyle \text{If }k\,=\,1\text{, we have: }\a,b,c)\:=\14,\,20,\,34)\;\;\Rightarrow\;\;\L N\,=\,104\)

\(\displaystyle \sout{\text{If }k\,=\,2\text{, we have: }\a,b,c)\:=\29,\,41,\,60)\;\;\Rightarrow\;\;N\,=\,209}\) . . . too big

 

[TchrWill]

I have probelm I would like to solve, but I just can't seem to get the right answer. Here it is :?

"A factory worker uses machines to sort cards into pile. On one occasion a machine operator obtained the following curious result. When a box of cards was sorted into 7 equal groups, there were 6 cards left over; when a box of cards was sorted into 5 equal groups, there were 4 cards left over; and when it was sorted into 3 equal groups, there were 2 left over. If the machine cannot sort more than 200 cards at a time, how many cards were in the box?''

Somewhat longer than soroban's but it gets you there.

Let N be the number we seek

1--N/7 = A + 6/7 or N = 7A + 6
2--N/5 = B + 4/5 or N = 5B + 4
3--N/3 = C + 2/3 or N = 3C + 2
4--Equating (1) and (2) yields 7A + 6 = 5B + 4 or 5B - 7A = 2
5--Dividing through by the lowest coefficient results in B - A - 2A/5 = 2/5
6--(2A + 2)/5 must be an integer
7--We want a unit coefficient for A so multiply by 3 yielding (6A + 6)/5
8--Divide by 5 again yielding A + A/5 + 1 + 1/5
9--(A + 1)/5 must be an integer k making A = 5k - 1
10--Substituting (9) back into (4) yields 5B - 35k + 7 = 2 or B = 7k - 1
11--k.....1.....2.....3.....4.....5.....6.....7.....8.....9.....10...
.....A.....4.....9...14....19...24...29...34....39...44.....49
.....B.....6...13...20....27...34...41...48....55...62.....69`
12---Equating (1) and (3) and (2) and 3 produce
....A = 3k - 1 and C = 7k - 1 for (1) = (3) and
....B = 3k - 1 and C = 5k - 1 for (2) = (3)
13--For (1) = (3)
....k.....1.....2.....3.....4.....5.....6.....7.....8.....9.....10...
....A....2.....5.....8....11....14...17...20...23...26.....29...
....C....6....13...20...27....34...41...48...55...62.....69...
14--For (2) - (3)
....k.....1.....2.....3.....4.....5.....6.....7.....8.....9.....10...
...B....2.....5.....8....11...14....17...20...23....26....29
...C....4.....9....14...19...25....29...34...39....44....49
15--Upon examination of the three tables, we find A = 14 & B = 20, A = 14 & C = 34, and B = 20 & C = 34 the matching values.
16--Checking; N = 7(14) + 6 = 5(20) + 4 = 3(34) + 2 = 104