Finding Consecutive Natural Numbers with a Given Product
Let the two consecutive natural numbers be represented by $n$ and $n+1$. The problem states that their product is 182.
Formulating the Equation
We can write the relationship as an equation:
$ n(n+1) = 182 $Solving for the Numbers
Expand the equation:
$ n^2 + n = 182 $Rearrange it into a standard quadratic equation:
$ n^2 + n - 182 = 0 $We can solve this quadratic equation by factoring. We need two numbers that multiply to -182 and add to 1. These numbers are 14 and -13.
Factor the quadratic equation:
$ (n + 14)(n - 13) = 0 $This gives two possible values for $n$:
- $n + 14 = 0 \implies n = -14$
- $n - 13 = 0 \implies n = 13$
Since we are looking for natural numbers (positive integers), we choose $n = 13$.
Identifying the Greater Number
The two consecutive natural numbers are $n = 13$ and $n+1 = 14$.
We can verify their product: $13 \times 14 = 182$.
The question asks for the greater of the two numbers.
The greater number is $14$.
This matches Option A.
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(Where N - Natural Numbers
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Q - Rational Numbers
Z - Integers)
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How many zeroes are there at the end of the following product?
1 x 5 x 10 x 15 x 20 x 25 x 30 x 35 x 40 x 45 x 50 x 55 x 60
Let XYZ be a three-digit number, where (x + y + Z) is not a multiple of 3. Then (XYZ + YZX + ZXY) is not divisible by
