Given: XZ bisects ZYXW and ZYZW Prove: AXYZ 2 AXWZ
1 answer:
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Answer:
There are 16 oddly powerful integers less than 2010
Step-by-step explanation:
∵ b is an odd integer
∵ b > 1
∴ The first value of b is 3
∵ a is an integer
- We can use a = 1, 2, 3, ..........
∵
∵ n < 2010
- Let a = 1, 2, ............... 12 because 12³ is greatest integer < 2010
∵ 1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64, 5³ = 125, 6³ = 216, 7³ = 343,
8³ = 512, 9³ = 729, 10³ = 1000, 11³ = 1331, 12³ = 1728
∴ There are 12 oddly powerful integers with b = 3
Now the second value of b is 5
but we took 1 before so we will start with 2
∵ ,
,
- is the greatest integer < 2010
∴ There are 3 oddly powerful integers with b = 5
Now the third value of b is 7
∵
- is the greatest integer < 2010
∴ There is 1 oddly powerful integers with b = 7
Now the fourth value of b is 9
∵
- is the greatest integer < 2010
- But we used 512 before
∴ There is no oddly powerful integers with b = 9
- 9 is the greatest value of b which makes
∵ 12 + 3 + 1 = 16
∴ There are 16 oddly powerful integers less than 2010