LCM hcf question A group of Red Cross members was practising for their National Day parade march past . If they marched in 2s , one pupil is without a partner . If they march in 3s , 5s , or 7s, there will be one pupil still without a partner . Calculate the least number of pupils in the contingent .
I'm a little confuse by how to find LCM for this . If they march in 4s, there will still be one pupil without a partner ? 
 A: Let us assume your desired answer is $n$.
Now, we know that $n-1$ is the multiple of $2, 3, 5$ and $7$. As $n$ is the smallest possible number holding this property, $n-1$ should be smallest possible number holding "it's" property. Thus $n-1$ must be the $lcm(2,3,5,7)=2.3.5.7$
Thus, $n=(2.3.5.7)+1=210+1=211$
As, $211=208 +3$, there will be three students who do not have a fourth partner
A: You've got two questions going on here. The second is straightforward: if they march in $4$s, you know immediately from the condition on $2$s that either one or three pupils will be left at the end. This doesn't give you an answer to the first immediately, though.
This should show up with the Euclidean Algorithm and Chinese Remainder Theorem if you're using a textbook. If not, you want to follow any example you find in a search for "Chinese Remainder Theorem". In case this is assigned homework, I won't write a full solution, but rather the first few steps to get you going. The first step is really what gets the ball rolling (and the fact that $2$, $3$, $5$, and $7$ are pairwise coprime makes it nice).
Let $n$ be the number of pupils marching. Set up a system of congruences:
$$n\equiv 1 \pmod{2}\\
n\equiv 1 \pmod{3}\\
n\equiv 1 \pmod{5}\\
n\equiv 1 \pmod{7}$$
Then you know there are integers $s,t,u,v$ such that
$$n=1+2s\\
n=1+3t\\
n=1+5u\\
n=1+7v$$
and you can substitute the first into the second:
$$1+2s\equiv1 \pmod{3}.$$
This implies $2s\equiv 0 \pmod{3}$, so since $2$ and $3$ are coprime, $s\equiv 0 \pmod{3}$. You can rewrite $s$ as a multiple of $3$, then, so $n=1+2s$ becomes $n=1+2(3t)$. With $n=1+6t$, rinse and repeat through $5$ and $7$.
