[math]\text{Prove that if}\,\,a,b_1,b_2,...,b_k\in\Z,\,\,b_i\,|\,a\,\,\text{for}\,\,i=1,...,k\\\\\text{and}\,\,\gcd(b_i,b_j)=1\,\,\text{for all}\,\,1\leq i< j\leq k\,\,\text{then}\,\,\red{b_1\cdot ...\cdot b_k\,\,|\,\,a}.\\\\\text{can someone give me some hint how to even touch it?}[/math]
number theory problem: Given integes a, b_1, b_2,..., b_k such that b_i|a and gcd(b_i,b_j)=1. Prove...
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Dr.Peterson
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You might start by considering the case k=2. What does it state, and how might you prove it?
In answering that, you might also tell us something about what facts you have learned that might be useful in proving it, so we can have a sense of your context.
Dr.Peterson
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Please correct what you wrote by stating exactly what you are proving (with the conditions clear), and then using those correct conditions.sorry there isn't information that gcd(a,b_i)=1
You may find @Steven G's suggestion useful, or you might have another approach in mind.
[math](1)\,\,b_1\,|\,a \Leftrightarrow \red{a=lb_1},\,\,l\in\Z\\\\(2)\,\,b_2\,|\,a \Leftrightarrow \blue{a=kb_2},\,\,k\in\Z\\\\\gcd(b_1,b_2)=1 \Leftrightarrow b_1x+b_2y=1\\\\\text{multiplying both sides by}\,\,a\,\,\text{we get:}\\\\ab_1x+ab_2y=a\\\\\text{now we can use (1) and (2):}\\\\\blue{kb_2}b_1x+\red{lb_1}b_2y=a\\\\b_1b_2(kx+ly)=a\\\\\text{so}\,\,b_1b_2\,|\,a\,\,\square\\\\\text{I got it!}[/math]
Dr.Peterson
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Now try extending it to k=3, and see if that gives you a way to generalize it.
In other words, the way to solve a problem is to take little steps and discover what you can do, not to beg others for hints when you don't yet see a solution. Just keep thinking.
And as you show further attempts, you may get further hints.