Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**4littlepiggiesmom****Member**- Registered: 2006-01-09
- Posts: 42

What is the largest power k such as 3^k divides easily into 40? With out anyremainders?

a. 1

b. 4

c. 13

d. 17

e. none of these

I know a can't be right as 40 is to even of a number to be that only.

but do I use the k like tha other problem with 1/2^3, 1/2^4...and so on? Or is there another one???

Offline

**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

I think you need to replace k with those numbers from a, b, c and d.

3^1 = 3

3^4 = 81

3^13 = 1594323

3^17 = 129140163

Since none of these divide into 40 without a remainder, I'd say the answer is e.

Aloha Nui means Goodbye.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,102

rickyoswalidow is absolutely right!

3^k is a power of 3 and the factors of this number are 1, 3 and the higher powers of 3 up to 3^(k-1). For a number to be divisible by 40, the number should have three 2s and 5 as prime factors. Since 3^k does not have these prime factors,it is not divisble by 40 for any value of k!

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

Pages: **1**