I am a number greater than 40,000 and less than 60,000:
40,000 < n < 60,000
This means that:
n = 10,000n₁ + 1,000n₂ + 100n₃ + 11n₄
And also:
4 ≤ n₁ < 6
0 ≤ n₂ ≤ 9
0 ≤ n₃ ≤ 9
0 ≤ n₄ ≤ 9
My ten thousands digit is 1 less than 3 times the sum of my ones digit and tens digit:
n₁ = 3*2n₄ - 1
n₁ = 6n₄ - 1
This means that:
n = 10,000*(6n₄-1) + 1,000n₂ + 100n₃ + 11n₄
n = 60,000n₄ - 10,000 + 1,000n₂ + 100n₃ + 11n₄
n = 60,011n₄ - 10,000 + 1,000n₂ + 100n₃
<span>My thousands digit is half my hundreds digit, and the sum of those two digits is 9:
n</span>₂ = 1/2 * n₃
<span>
n</span>₂ + n₃ = 9
<span>
Therefore:
n</span>₂ = 9 - n₃
<span>
Therefore:
9 - n</span>₃ = 1/2 * n₃
<span>
9 = 1/2 * n</span>₃ + n₃
<span>
9 = 1.5 * n</span>₃
<span>
Therefore:
n</span>₃ = 6
<span>
If n</span>₃=6, n₂=3.
<span>
This means that:
</span>n = 60,011n₄ - 10,000 + 1,000*3 + 100*6
n = 60,011n₄ - 10,000 + 3,000 + 600
n = 60,011n₄ - 6,400
Therefore:
0<n₄<2, so n₄=1.
If n₄=1:
n = 60,011 - 6,400
n = 53,611
Answer:
53,611
What we know:
we know that there are <u>75</u> chicks in the hen house, we also know that in all there are 108 chicks
What we need to find:
we need to find out how many chicks were outside
Answer equation:
108 - 75 = 33
Answer: 5
Step-by-step explanation:
