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
Answers
Answer:
Step-by-step explanation:
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
16 divides n^2-n-2
n² - n - 2
= n² -2n + n - 2
= n(n-2) + 1(n-2)
= (n+1)(n-2)
16 = 2 * 2 * 2 * 2
out of (n+1)& (n-2) only one can be even
=> n + 1 = 16k or n-2 = 16k
=> n = 16k-1 or 16k+2
Values of n Between 100 & 300
111 , 127 , 143 , 159 , 175 , 191 , 207 , 223 , 239 , 255 , 271 , 287
113 129 145 161 177 193 209 , 225 , 241 ,257 ,273 , 289
25 divides (n^2 + 2n – 3)
n² +3n - n - 3
= (n+3)(n-1)
25 = 5 * 5
out of (n+3)& (n-1) only one can be 5 multiple
=> n + 3 = 25k or n -1 = 25k
=> n = 25k + 3 or n = 25k +1
Values of n Between 100 & 300
122 147 172 197 222 247 272 297
101 126 151 176 201 226 251 276
There is no common integer in both
Answer:
Step-by-step explanation:
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
16 divides n^2-n-2
n² - n - 2
= n² -2n + n - 2
= n(n-2) + 1(n-2)
= (n+1)(n-2)
16 = 2 * 2 * 2 * 2
out of (n+1)& (n-2) only one can be even
=> n + 1 = 16k or n-2 = 16k
=> n = 16k-1 or 16k+2
Values of n Between 100 & 300
111 , 127 , 143 , 159 , 175 , 191 , 207 , 223 , 239 , 255 , 271 , 287
113 129 145 161 177 193 209 , 225 , 241 ,257 ,273 , 289
25 divides (n^2 + 2n – 3)
n² +3n - n - 3
= (n+3)(n-1)
25 = 5 * 5
out of (n+3)& (n-1) only one can be 5 multiple
=> n + 3 = 25k or n -1 = 25k
=> n = 25k + 3 or n = 25k +1
Values of n Between 100 & 300
122 147 172 197 222 247 272 297
101 126 151 176 201 226 251 276
some one did this and forgot to write other numbers denoted by 16k+2 when k=14 n=226 and for 25k+1 k=9 n=226 226 is common answer is 1