# Number Problems

Benjamin makes up present boxes for children, each box must contain two balloons, One whistle and one tube of sweets. Balloons are sold in packets of 40, whistles are sold in packets of 6 and tubes of sweets n packets of 25 What is the smallest number of whole packets of each Benjamin must use to fill up the boxes without any balloons or whistles or tubes of sweets being left over? How Many Present boxes will he make up?

-

Hint: You want to find $n$ such that $\frac n{20}$ (the number of packets of balloons bought), $\frac n6$, and $\frac n{25}$ are all integers. The least common multiple of the denominators is ?

-
I'm rubbish at maths and i need an answer quickly:/ –  Shannon Sep 12 '12 at 16:09
@Shannon: do you know what prime numbers are and how to factor 6, 20, and 25? You could look at en.wikipedia.org/wiki/Least_common_multiple –  Ross Millikan Sep 12 '12 at 16:40
Yeah i know what prime numbers are and I can do factors, but i'll still have a look haha:-) –  Shannon Sep 12 '12 at 16:45
Let's take it up where Ross left off. You need $n/20,n/6,n/25$ to be whole numbers. Let's write $$n/20=a,\quad n/6=b,\quad n/25=c\tag1$$ Then let's go $$n=20a,\quad n=6b,\quad n=25c\tag2$$ Still with me? OK, then $$20a=6b=25c\tag3$$ Let's take the first part of that: $20a=6b$. 20 and 6 are both even, so let's divide through by 2: $$10a=3b\tag4$$ The right side is a multiple of 3, so the left side is a multiple of 3, so $a$ is a multiple of 3: $$a=3d\tag5$$ Similarly, the left side is a multiple of 10, so the right side is a multiple of 10, so $b$ is a multiple of 10: $$b=10e\tag6$$ So now equation (4) becomes $(10)(3)d=(3)(10)e$, that is, $30d=30e$, and we can cancel to get $$d=e\tag7$$ Good!
Now go back to $6b=25c$, the second half of equation (3). Combining it with equation (6), we get $(6)(10)e=25c$. Dividing through by 5, $$12e=5c\tag8$$ Right side's a multiple of 5, so left side's a multiple of 5, so $e$ is a multiple of 5, and similarly $c$ is a multiple of 12: $$e=5f,\quad c=12g\tag9$$ Put these into (8) and you get $$f=g\tag{10}$$ We've come pretty far from $a,b,c$: time to get back to them! (5) and (7) tell me $a=3e$, then (9) tells me $a=15f$, then (10) says $$a=15g\tag{11}$$ (6) and (9) tell me $b=50f$, so from (10), $$b=50g\tag{12}$$ And (9) says $c=12g$. Going back up to (2), we get $$n=300g,\quad n=300g,\quad n=300g\tag{13}$$ Now $g$ can be any whole number we want, so let's make it as small as we can: $g=1$. Well, now you've got your $n$, so you've got the number of packets of each of the three supplies.