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lbatch · Feb 4, 2009, 10:03 AM

If a collection of pencils is placed in rows of 4, there are 2 pencils left; if placed in rows of 5, there are 3 left; and if placed in rows of 7, there are 5 left. What is the smallest possible number of pencils in the collection?

Capuchin · Feb 4, 2009, 10:26 AM

x mod 4 = 2, x = 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50...
x mod 5 = 3, x = 8, 13, 18, 23, 28, 33, 38, 43, 48...
x mod 7 = 5, x = 12, 19, 26, 33, 40, 47...

You can continue until you find the same number in all 3 columns. You can ignore any odd ones for the last 2 formulae, because there are no odd numbers in the first sequence.

galactus · Feb 4, 2009, 11:36 AM

The LCD of the 'mods' 4,5,7 is 140.

So, we have:

$x\equiv -2(mod 140)$

$\frac{x+2}{140}=1$

Can you see what x is?

eawoodall · Feb 4, 2009, 02:14 PM

notice that the smallest that you can divide by four and get 2 is 4k+2.

similarly with 5g+3, and 7 h +5.

so some number makes all these equal.

4k+2=5g+3=7h+5=x.

if we let k = 1. 4k+2 = 4+2=6. 6=5g+3, 6/3 = 5g, 2 /5 = g. no.
k=2. 8+2=10. 10=5g+3, 7=5g, 7/5 = g. no.
k=3. 12+2=14. 14=5g+3, 11=5g. 11/5=g. not a whole number yet.
k=4. 16+2=18. 18=5g+3. 18-3=5g. 15=5g. 15/5 =5/5 (g). 3=g.
now to test for seven also. 7h+5=18. 18-5=7h, 13 =7h. 13/7 = h. no.

since every 20 numbers is the same pattern of four and fives. We skip ahead and add 20 to 18. 38=7h+5. 33=7h. 33/7=h. no.
continue to add 20 per time, 58=7h+5. 53/7=h. no
again, 78=7h+5. 73/7=h. no. 98=7h+5. 93/7=h. no.
so forth, 118=7h +5. 113/7=h. no. 138=7h+5. 133/7=h. 19=h.
138 works for all considtion.