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16 Sep 2026, 10:44
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45% (03:01) correct
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In how many ways can 480 be written as the product of three distinct positive even numbers in increasing order?
A. 8
B. 9
C. 10
D. 11
E. 12
A. 8
B. 9
C. 10
D. 11
E. 12
16 Sep 2026, 19:18
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480 can be written as 2^5 x 3 x 5. Now creating cases with this:
2 x 12 x 20
2 x 10 x 24
2 x 6 x 40
2 x 8 x 30
2 x 4 x 60
4 x 10 x 12
4 x 6 x 20
6 x 8 x 10
so, total 8 cases.
2 x 12 x 20
2 x 10 x 24
2 x 6 x 40
2 x 8 x 30
2 x 4 x 60
4 x 10 x 12
4 x 6 x 20
6 x 8 x 10
so, total 8 cases.
16 Sep 2026, 20:37
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\(480=2^5×3×5\)
Let the factorization be a×b×c, where a<b<c.
Since all three are even, take out one 2 for each of a, b and c, that is a=2p, b=2q and c=2r.
The remaining product after taking out three 2s is 2×2×3×5 as \(480=2^5×3×5=a×b×c=2p×2q×2r=8pqr=8×(2×2×3×5)\).
If p=1, then qr=2×2×3×5 and q<r.
The max value of q will be less than square root of 2×2×3×5 or 60.
Now \(7^2<60<8^2\).
So q can take any value greater than 1 and less than 8.
Factors of 60 less than 8 are 2, 3, 4, 5 and 6.
Thus, when a=2, b can take five values(2×2,2×3,2×4,2×5,2×6) and c will take values according to b.
5 ways
If p=2, then qr=2×3×5 and q<r.
The max value of q will be less than square root of 2×3×5 or 30.
Now \(5^2<30<6^2\).
So q can take any value greater than 2 (as p is 2) and less than 6.
Factors of 30 greater than 2 but less than 6 are 3 and 5.
Thus, when a=4, b can take two values(2×3,2×5) and c will take values according to b.
2 ways
If p=3, then qr=2×2×5 and q<r.
The max value of q will be less than square root of 2×2×5 or 2p.
Now \(4^2<20<5^2\).
So q can take any value greater than 3 and less than 5.
Factor of 20 greater than 3 and less than 5 is 4.
Thus, when a=6, b can take five values(2×4) and c will take values according to b.
1 way.
If p=4, then qr=3×5 and q<r.
The max value of q will be less than square root of 3×5 or 15.
Now \(3^2<15<4^2\).
No values exist as p becomes greater than q.
total : 5+2+1 or 8 ways.
Let the factorization be a×b×c, where a<b<c.
Since all three are even, take out one 2 for each of a, b and c, that is a=2p, b=2q and c=2r.
The remaining product after taking out three 2s is 2×2×3×5 as \(480=2^5×3×5=a×b×c=2p×2q×2r=8pqr=8×(2×2×3×5)\).
If p=1, then qr=2×2×3×5 and q<r.
The max value of q will be less than square root of 2×2×3×5 or 60.
Now \(7^2<60<8^2\).
So q can take any value greater than 1 and less than 8.
Factors of 60 less than 8 are 2, 3, 4, 5 and 6.
Thus, when a=2, b can take five values(2×2,2×3,2×4,2×5,2×6) and c will take values according to b.
5 ways
If p=2, then qr=2×3×5 and q<r.
The max value of q will be less than square root of 2×3×5 or 30.
Now \(5^2<30<6^2\).
So q can take any value greater than 2 (as p is 2) and less than 6.
Factors of 30 greater than 2 but less than 6 are 3 and 5.
Thus, when a=4, b can take two values(2×3,2×5) and c will take values according to b.
2 ways
If p=3, then qr=2×2×5 and q<r.
The max value of q will be less than square root of 2×2×5 or 2p.
Now \(4^2<20<5^2\).
So q can take any value greater than 3 and less than 5.
Factor of 20 greater than 3 and less than 5 is 4.
Thus, when a=6, b can take five values(2×4) and c will take values according to b.
1 way.
If p=4, then qr=3×5 and q<r.
The max value of q will be less than square root of 3×5 or 15.
Now \(3^2<15<4^2\).
No values exist as p becomes greater than q.
total : 5+2+1 or 8 ways.
kevincan
