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

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12 Oct 2013, 07:36

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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

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

1. b/2 is an odd integer. 2. the remainder of b/10 is b

Here is my explanation on this,

Problem statement, 10^a + b is this divisible by 3. In other words, the question is whether the sum of digits that 10^a + b add up to 3. Here whatever value 'a' takes 10^a will only be 1, especially since 'a' is always a positive integer. So this mean, once you add 1 with b's digits, you can solve the question .

Option A,

b/2 is odd, so this means b is 2*(y) where y could be any odd number - 1, 3, 5, 7, 9, etc.... so different values of b are 2, 6, 10, 14, 18... etc... now, find out the sum of digits of 10^a + b ... we know a is always 1, so when b is 1, the sum of digits is 3, when b is 6, value is 7, when b is 10, value is 11, when b is 14, value is 15... so here we notice 6 and 15 are divisible by 3 and rest are not. So this means A is insufficient. <INSUFFICIENT>

Option B,

Remainder of b/10 is b, this means the only possibility is b is less than 10, (check when b is 11, it gives remainder of 1, so the value should only be 9 and less).. so the different values of b are 9, 8, 7, 6, 5.. to 1.... now finding value of sum of digits of 10^a + b, where a is always 1.. so when b is 1, sum is 2, b is 2, sum is 3, when b is 3, sum is 4, when b is 4, sum is 5... etc... If we notice we have values divisible by 3 and those that are not... so this option is insufficient <INSUFFICIENT>

At this point option A, B and D are ruled out

Combine option A and B, this is simple, simply check common values of b for option A and B .. so possible values of b are 2 and 6 only... this makes things easier, we have just calculate sum of digits of 10^a+b for these 2 values.. we had already done this calculation for A and the two sums of digits are 3 and 7 respectively. 3 is divisible by 3 but 7 is not, hence combining also hasn't given us an answer. <INSUFFICIENT>

Hence option is E...

--- Kudos if this explanation helped ------
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If a and b are positive integers, is (10^a) + b divisible by 3?

1. b/2 is an odd integer. 2. the remainder of b/10 is b

Here is my explanation on this,

Problem statement, 10^a + b is this divisible by 3. In other words, the question is whether the sum of digits that 10^a + b add up to 3. Here whatever value 'a' takes 10^a will only be 1, especially since 'a' is always a positive integer. So this mean, once you add 1 with b's digits, you can solve the question .

Option A,

b/2 is odd, so this means b is 2*(y) where y could be any odd number - 1, 3, 5, 7, 9, etc.... so different values of b are 2, 6, 10, 14, 18... etc... now, find out the sum of digits of 10^a + b ... we know a is always 1, so when b is 1, the sum of digits is 3, when b is 6, value is 7, when b is 10, value is 11, when b is 14, value is 15... so here we notice 6 and 15 are divisible by 3 and rest are not. So this means A is insufficient. <INSUFFICIENT>

Option B,

Remainder of b/10 is b, this means the only possibility is b is less than 10, (check when b is 11, it gives remainder of 1, so the value should only be 9 and less).. so the different values of b are 9, 8, 7, 6, 5.. to 1.... now finding value of sum of digits of 10^a + b, where a is always 1.. so when b is 1, sum is 2, b is 2, sum is 3, when b is 3, sum is 4, when b is 4, sum is 5... etc... If we notice we have values divisible by 3 and those that are not... so this option is insufficient <INSUFFICIENT>

At this point option A, B and D are ruled out

Combine option A and B, this is simple, simply check common values of b for option A and B .. so possible values of b are 2 and 6 only... this makes things easier, we have just calculate sum of digits of 10^a+b for these 2 values.. we had already done this calculation for A and the two sums of digits are 3 and 7 respectively. 3 is divisible by 3 but 7 is not, hence combining also hasn't given us an answer. <INSUFFICIENT>

Hence option is E...

--- Kudos if this explanation helped ------

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Re: If a and b are positive integers
[#permalink]
28 Aug 2015, 12:44

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