Question

The number of integers n with 100 < n < 300 such that 16 divides (n^2-n-2) and 25 divides (n^2 2n 3) is

Answer 1

Answer:

1

Step-by-step explanation:

Given The number of integers n with 100 < n < 300 such that 16 divides (n^2-n-2) and 25 divides (n^2+ 2n - 3) is

Now n^2 – n – 2 can be split into (n + 2)(n – 1)  

Similarly second equation can be split into (n + 3)(n – 1)

According to question 100 < n < 300

So 16 should divide (n + 2)(n – 1)

Now the numbers 101 to 299 divisible by 16 will be

112 128 144 160 176 192 208 224 240 256 272 288

So either n + 2 is divisible or n – 1

So n + 2 values are

114 130 146 162 178 194 210 226 242 258 274 290

Similarly for n – 1 we get

111 127 143 159 175 191 207 223 239 255 271 287

Now similarly we need to calculate 25 for equation (n + 3)(n – 1)

We get 1 as answer.