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stich3 [128]
3 years ago
13

180 product of a prime factor using indices

Mathematics
2 answers:
zvonat [6]3 years ago
5 0
Start with 180. 
<span>Is 180 divisible by 2? Yes, so write "2" as one of the prime factors, and then work with the quotient, 90. </span>

<span>Is 90 divisible by 2? Yes, so write "2" (again) as another prime factor, then work with the quotient, 45. </span>

<span>Is 45 divisible by 2? No, so try a bigger divisor. </span>
<span>Is 45 divisible by 3? Yes, so write "3" as a prime factor, then work with the quotient, 15 </span>

<span>Is 15 divisible by 3? [Note: no need to revert to "2", because we've already divided out all the 2's] Yes, so write "3" (again) as a prime factor, then work with the quotient, 5. </span>

<span>Is 5 divisible by 3? No, so try a bigger divisor. </span>
Is 5 divisible by 4? No, so try a bigger divisor (actually, we know it can't be divisible by 4 becase it's not divisible by 2)
<span>Is 5 divisible by 5? Yes, so write "5" as a prime factor, then work with the quotient, 1 </span>

<span>Once you end up with a quotient of "1" you're done. </span>

<span>In this case, you should have written down, "2 * 2 * 3 * 3 * 5"</span>
Masteriza [31]3 years ago
5 0
The answer is 2•2•3•3•5
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