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Re: If a is a positive integer and 20 < a < 100, is a 3 − a [#permalink]
29 Aug 2013, 08:04

1

This post received KUDOS

Expert's post

If a is a positive integer and 20 < a < 100, is a^3 − a divisible by 15?

In order an integer to be divisible by 15 it should be divisible by both 3 and 5.

(1) The units digit of a is 9 --> \(a^3-a=(a-1)a(a+1)\) --> out of 3 consecutive integers ((a-1), a, and (a+1)), one must be divisible by 3 and as the units digit of a is 9, then the units digit of a+1 is 0, so a+1 is divisible by 5. Thus a^3-a is divisible by 15. Sufficient.

(2) The tens digit of a is 3. If a=30, then the answer is YES but if a=32 ((a-1)a(a+1)=31*32*33), then the answer is NO. Not sufficient.

Re: If a is a positive integer and 20 < a < 100, is a^3 − a [#permalink]
04 Mar 2014, 03:08

(1) units digit of a is 9. If you raise 9 to some power, it repeats this pattern: 9^1=9 9^2= 81 9^3 =...9. Hence, if we raise a number with the units digit 9 to the power of 3, the outcome will also have the units digit of 9. If we then substract the number with the units digit 9, the outcome will have the units digit 0. Hence div. by 15, SUFF.

(2) Tens digit a three: It's thus 31,32,33...or 39. 31^3-31 would be div. by 15, 32^3 - 32 not. IS.

Re: If a is a positive integer and 20 < a < 100, is a^3 − a [#permalink]
04 Mar 2014, 04:26

From S1:We can infer that a^3-a=(a-1)*a*(a+1) which are 3 consecutive integers.So they're always divisible by 3 in the range of 'a' given.We just need to prove that any one or more of these will have 5 as a factor too. S1 gives that 'a' ends with 9 so (a+1) will end with 0.always div by 5.Hence suff.

From S2:we get that 'a' will be in the range of 30s.We could consider 2 simple egs. (30,31,32)-Ans Yes. While (31,32,33)-And No.Not sufficient.

Re: If a is a positive integer and 20 < a < 100, is a^3 − a [#permalink]
19 Mar 2015, 11:14

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