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Zorluk: Çok zorFactors, Multiples, and Prime Factorization

Let N=2a×3b×7cN = 2^a \times 3^b \times 7^c be a positive integer, where aa, bb, and cc are positive integers. The number of positive factors of NN that are divisible by 66 is 3636, the number of positive factors of NN that are divisible by 1414 is 3232, and the number of positive factors of NN that are divisible by 2121 is 3030. What is the total number of positive factors of NN?

Cevap: 60

Cevap

The total number of positive factors of NN is 60.
By setting up the system of equations based on exponent constraints for each divisor (6,14,216, 14, 21), we find a=4,b=3,c=2a=4, b=3, c=2. Substituting these into the total factor formula (a+1)(b+1)(c+1)(a+1)(b+1)(c+1) yields 5×4×3=605 \times 4 \times 3 = 60.

Adım Adım Çözüm

1
Formulate algebraic expressions for the constrained factor counts
ab(c+1)=36ab(c+1) = 36, ac(b+1)=32ac(b+1) = 32, and bc(a+1)=30bc(a+1) = 30
A factor 2x×3y×7z2^x \times 3^y \times 7^z is divisible by 6=21×316 = 2^1 \times 3^1 if x≥1x \ge 1, y≥1y \ge 1, z≥0z \ge 0, yielding a⋅b⋅(c+1)a \cdot b \cdot (c+1) factors. Similarly, divisibility by 1414 requires x≥1,z≥1x \ge 1, z \ge 1, giving a⋅c⋅(b+1)a \cdot c \cdot (b+1), and divisibility by 2121 requires y≥1,z≥1y \ge 1, z \ge 1, giving b⋅c⋅(a+1)b \cdot c \cdot (a+1).
2
Expand equations and substitute P=abcP = abc
ab=36−Pab = 36 - P, ac=32−Pac = 32 - P, bc=30−Pbc = 30 - P
Expanding ab(c+1)=abc+ab=36ab(c+1) = abc + ab = 36 gives ab=36−Pab = 36 - P. Applying the same expansion to the other two equations gives expressions for acac and bcbc in terms of PP.
3
Multiply the three equations together and solve for PP
(ab)(ac)(bc)=(abc)2=P2⇒(36−P)(32−P)(30−P)=P2⇒P=24(ab)(ac)(bc) = (abc)^2 = P^2 \Rightarrow (36-P)(32-P)(30-P) = P^2 \Rightarrow P = 24
Testing positive integer values of PP that divide into products reveals that for P=24P = 24, (36−24)(32−24)(30−24)=12×8×6=576=242(36-24)(32-24)(30-24) = 12 \times 8 \times 6 = 576 = 24^2.
4
Calculate the values of abab, acac, bcbc, and solve for aa, bb, cc
ab=12ab = 12, ac=8ac = 8, bc=6bc = 6, giving a=4a = 4, b=3b = 3, c=2c = 2
a2=(ab)(ac)bc=12×86=16⇒a=4a^2 = \frac{(ab)(ac)}{bc} = \frac{12 \times 8}{6} = 16 \Rightarrow a = 4. Similarly, b2=12×68=9⇒b=3b^2 = \frac{12 \times 6}{8} = 9 \Rightarrow b = 3, and c2=8×612=4⇒c=2c^2 = \frac{8 \times 6}{12} = 4 \Rightarrow c = 2.
5
Compute the total number of positive factors of NN
(4+1)(3+1)(2+1)=5×4×3=60(4+1)(3+1)(2+1) = 5 \times 4 \times 3 = 60
The total number of factors of N=2a×3b×7cN = 2^a \times 3^b \times 7^c is (a+1)(b+1)(c+1)(a+1)(b+1)(c+1).

Anahtar Kavram

Constrained Factor Counting and Prime Factorization Systems
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Konular

Factors, Multiples, and Prime Factorization

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