



 
Hi Kelly, Suppose that $a$ is one of the two numbers then 7 is a factor of $a$ so there is an integer $k$ so that $a = 7 \times k.$ But you also know that 20 is a multiple of $a$ so there is an integer $c$ so that $a \times c = 20.$ Putting these two statements together you get that \[a \times c = 7 \times k \times c = 20.\] But this is impossible since 7 divides $7 \times k \times c$ but 7 does not divide 20. Thus there are no integers that have 7 as their HCF and 20 as their LCM. Penny  


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