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Zorluk: ZorNumber Bases and Conversions

If 24x×13x=345x24_x \times 13_x = 345_x, where xx represents a positive integer base, find the value of xx.

Cevap: 7

Cevap

The value of the base xx is 7.
Expanding 24x24_x, 13x13_x, and 345x345_x into base 10 yields (2x+4)(x+3)=3x2+4x+5(2x + 4)(x + 3) = 3x^2 + 4x + 5. Expanding the left side gives 2x2+10x+122x^2 + 10x + 12. Equating and simplifying gives x26x7=0x^2 - 6x - 7 = 0, which factors as (x7)(x+1)=0(x - 7)(x + 1) = 0. The positive integer solution greater than 5 is x=7x = 7.

Adım Adım Çözüm

1
Convert all base xx numbers to decimal (base 10) expressions.
24x=2x+424_x = 2x + 4, 13x=x+313_x = x + 3, and 345x=3x2+4x+5345_x = 3x^2 + 4x + 5.
Place-value expansion allows algebraic manipulation in standard base 10.
2
Multiply the expanded factors on the left-hand side.
(2x+4)(x+3)=2x2+10x+12(2x + 4)(x + 3) = 2x^2 + 10x + 12.
Applying the distributive property of multiplication.
3
Equate the expanded left-hand side to the right-hand side and rearrange into standard quadratic form.
3x2+4x+5(2x2+10x+12)=0    x26x7=03x^2 + 4x + 5 - (2x^2 + 10x + 12) = 0 \implies x^2 - 6x - 7 = 0.
Setting the quadratic expression equal to zero enables factoring.
4
Factor the quadratic equation and select the valid base.
(x7)(x+1)=0    x=7(x - 7)(x + 1) = 0 \implies x = 7 (rejecting x=1x = -1).
A number base must be a positive integer strictly greater than any individual digit in the given numbers (max digit is 5).

Anahtar Kavram

Solving polynomial equations derived from number base expansion.
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