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Find the number of integers N from 1 to 100 such that \(N^2 + 8N + 15 \) is a multiple of 7.

 
 Jun 23, 2022

Best Answer 

 #1
avatar+589 
+2

We can factor N^2+8N+15 into (N+3)(N+5). So if we want N^2+8N+15 to be divisible by 7, we need either n+3 or N+5 divisible by 7. The list of N such that N+3 is divisible by 7 is: 4,11,18,25,...,95, which is (95-4)/7+1=14 numbers. The list of N such that N=5 is divisible by 7 is 2, 9, 16, 23,...,100 which is (100-2)/7+1=15 numbers. 14+15=\(\boxed{29}\)

 
 Jun 23, 2022
 #1
avatar+589 
+2
Best Answer

We can factor N^2+8N+15 into (N+3)(N+5). So if we want N^2+8N+15 to be divisible by 7, we need either n+3 or N+5 divisible by 7. The list of N such that N+3 is divisible by 7 is: 4,11,18,25,...,95, which is (95-4)/7+1=14 numbers. The list of N such that N=5 is divisible by 7 is 2, 9, 16, 23,...,100 which is (100-2)/7+1=15 numbers. 14+15=\(\boxed{29}\)

 
SparklingWater2 Jun 23, 2022
 #2
avatar
+1

1 = 2
2 = 4
3 = 9
4 = 11
5 = 16
6 = 18
7 = 23
8 = 25
9 = 30
10 = 32
11 = 37
12 = 39
13 = 44
14 = 46
15 = 51
16 = 53
17 = 58
18 = 60
19 = 65
20 = 67
21 = 72
22 = 74
23 = 79
24 = 81
25 = 86
26 = 88
27 = 93
28 = 95
Total== 28 integers

 
 Jun 23, 2022
edited by Guest  Jun 23, 2022
 #3
avatar
0

You forgot to include 100

 
Guest Jun 23, 2022

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