We could do it by first writing
[tex]7\cdot15=(4+3)\cdot(4+11)=4^2+14\cdot4+33[/tex]
so that
[tex]7\cdot15\equiv4^2+14\cdot4+33\equiv33\pmod4[/tex]
and since [tex]33=32+1=8\cdot4+1[/tex], we have
[tex]7\cdot15\equiv33\equiv1\pmod4[/tex]
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Alternatively, we can just compute the product and take it mod 4:
[tex]7\cdot15=105=104+1=26\cdot4+1[/tex]
and so we get the same result of 1.
