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Re: No. Properties [#permalink]
06 Nov 2008, 10:08

for a number to be divisible by 3, sum of the digits have to be divisible by 3. So I'd try looking to get at the sum of digits from \(10^a + b\) \(10^a\) gives us just 1+how many ever 0s. so the options for sum of the digits of b ought to be 2, 4, 8 etc..

From (1) we are told b/2 is odd.. Take for example b = 6, so b/2 = 3. But then sum of digits yields 1+3, which is not divisible by 3. Alternatively, take b=10, b/2 = 5. Then sum of digits yields 5+1 = 6, which is divisible by 3. Since we have conflicting results from (1), it is not sufficient by itself. Cross out A and D.

(2) says remainder of b/10 is b. That means, b < 10. We'll get conflicting results if we were to chose b=5 or b = 4. So (2) alone is not sufficient enough. Cross out B.

(1) and (2) together, says b/2 is odd and b < 10, it gives only choice for b as 6. that is conclusive enough. \(10^a + 6\) is not divisible by 3. So my pick for the answer is C _________________

excellence is the gradual result of always striving to do better

Re: If a and b are positive integers, is 10^a + b divisible by [#permalink]
12 Oct 2013, 06:36

4

This post received KUDOS

study wrote:

If \(a\) and \(b\) are positive integers, is \(10^a + b\) divisible by 3?

1. \(\frac{b}{2}\) is an odd integer 2. the remainder of \(\frac{b}{10}\) is \(b\)

what positive integer divided by 10 gives a remainder of itself?

Hi all, let me try to explain this one.

So we have 10^a + b /3 what is the remainder?

First of all note that you are told that a,b are positive integers. THEY ALWAYS TELL YOU THIS FOR SOME REASON, DON'T IGNORE IT

So we have that 10^a = (10,100 etc...)

Now, what is the divisibility rule for multiples of 3? That's right the digits of the number must add to a multiple of 3 We already have the 1 on the 10^a. Let's see what we can get out of 'b'

Statement 1: b/2 is odd integer. So b could be (2,6,10,14 etc...) If b is 2 then remainder is 0 If b is 6 then remainder is 1 If b is 10 then remainder is 2 Not good enough

Statement 2: b/10 remainder is b. Now wait a second. What this is telling us is that b<10. But that could be (1,2,3,4,5 etc...) This is stil not enough

Statement (1) and (2) - Now going back to statement 1. We still have 2 or 6 as possible answer giving different remainders

Hence (E) is the correct answer

Kudos if you like!

gmatclubot

Re: If a and b are positive integers, is 10^a + b divisible by
[#permalink]
12 Oct 2013, 06:36

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...