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If a and b are distinct positive integers, what is the units digit of

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If a and b are distinct positive integers, what is the units digit of [#permalink]

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If a and b are distinct positive integers, what is the units digit of 2^a*8^b*4^(a+b)?

(1) b = 24 and a < 24
(2) The greatest common factor of a and b is 12
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 May 2015, 08:58, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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Re: If a and b are distinct positive integers, what is the units digit of [#permalink]

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The given expression can be simplified to 2^(a+3b+2a+2b) = 2^(3a+5b)

Using statement (1), the units digit will change depending on the value of a. Insufficient.
Using statement (2), both a and b have to be multiples of 12, which means that 3a+5b will always come out to be a multiple of 4. Using cyclicity, we can then say that the units digit of 2^(3a+5b) = units digit of 2^4 = 6. Sufficient.

(B) it is.
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Re: If a and b are distinct positive integers, what is the units digit of [#permalink]

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If a and b are distinct positive integers, what is the units digit of [#permalink]

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New post 28 May 2015, 00:24
I am still not able to get why it is B.. request someone help in getting the same understand...

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Re: If a and b are distinct positive integers, what is the units digit of [#permalink]

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sagarag wrote:
I am still not able to get why it is B.. request someone help in getting the same understand...


Hi sagarag,

The question here is testing about the knowledge of cyclicity of the number 2. We know that cyclicity of 2 is 4 i.e. the unit's digit of 2 would repeat itself after every 4 powers. For example:

\(2^1 -> 2\)
\(2^2 -> 4\)
\(2^3 -> 8\)
\(2^4 -> 16\)
\(2^5 -> 32\). Unit's digit of \(2^1\) and \(2^5\) is the same, i.e. they repeat after every 4 powers

Since the question prompt can be simplified to \(2^{3a +5b}\), we just need to find what values can \(3a + 5b\) take.

St-II tells us that GCD(a, b) =12. As 12 is the factor of both \(a\) and \(b\) we can write \(a\) and \(b\) as \(12x\) and \(12y\) respectively.

So \(3a + 5b = 12(3x +5y) = 4(9x + 15y)\). This expression is always divisible by 4. So we can write \(2^{3a + 5b} = 2^{4n}\) where \(n = 9x + 15y\)

We know that \(2^4\), \(2^8\), \(2^{12}\).....\(2^{4n}\) have the same units digit as 6.

Since st-II tells us that \(3a +5b\) is a multiple of 4, we can definitely find the unit's digit of the expression \(2^{3a +5b}\)

Hope this helps :)

Regards
Harsh
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Re: If a and b are distinct positive integers, what is the units digit of [#permalink]

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New post 28 May 2015, 08:00
Can you explain this portion of your explanation....

So 3a+5b=12(3x+5y)=4(9x+15y). This expression is always divisible by 4. So we can write 2^3a+5b=2^4n where n=9x+15y

Couldn't 12(3x+5y) also be = 6(6x+10y) so then the expression is always divisible by 6 so we can rewrite as 2^3a+5b=2^6n which wouldnt that give us a different unit digit than 2^4n?

Thanks for clarifying!

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Re: If a and b are distinct positive integers, what is the units digit of [#permalink]

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New post 28 May 2015, 21:28
EgmatQuantExpert wrote:
sagarag wrote:
I am still not able to get why it is B.. request someone help in getting the same understand...


Hi

The question here is testing about the knowledge of cyclicity of the number 2. We know that cyclicity of 2 is 4 i.e. the unit's digit of 2 would repeat itself after every 4 powers. For example:

\(2^1 -> 2\)
\(2^2 -> 4\)
\(2^3 -> 8\)
\(2^4 -> 16\)
\(2^5 -> 32\). Unit's digit of \(2^1\) and \(2^5\) is the same, i.e. they repeat after every 4 powers

Since the question prompt can be simplified to \(2^{3a +5b}\), we just need to find what values can \(3a + 5b\) take.

St-II tells us that GCD(a, b) =12. As 12 is the factor of both \(a\) and \(b\) we can write \(a\) and \(b\) as \(12x\) and \(12y\) respectively.

So \(3a + 5b = 12(3x +5y) = 4(9x + 15y)\). This expression is always divisible by 4. So we can write \(2^{3a + 5b} = 2^{4n}\) where \(n = 9x + 15y\)

We know that \(2^4\), \(2^8\), \(2^{12}\).....\(2^{4n}\) have the same units digit as 6.

Since st-II tells us that \(3a +5b\) is a multiple of 4, we can definitely find the unit's digit of the expression \(2^{3a +5b}\)

Hope this helps :)

Regards
Harsh



Can you please explain this part above :-

So \(3a + 5b = 12(3x +5y) = 4(9x + 15y)\). This expression is always divisible by 4. So we can write \(2^{3a + 5b} = 2^{4n}\) where \(n = 9x + 15y\)

We know that \(2^4\), \(2^8\), \(2^{12}\).....\(2^{4n}\) have the same units digit as 6.

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Re: If a and b are distinct positive integers, what is the units digit of [#permalink]

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healthjunkie wrote:
Can you explain this portion of your explanation....

So 3a+5b=12(3x+5y)=4(9x+15y). This expression is always divisible by 4. So we can write 2^3a+5b=2^4n where n=9x+15y

Couldn't 12(3x+5y) also be = 6(6x+10y) so then the expression is always divisible by 6 so we can rewrite as 2^3a+5b=2^6n which wouldnt that give us a different unit digit than 2^4n?

Thanks for clarifying!


Hi healthjunkie,

Let's assume an example of a number \(2^{12}\) for which we need to find its units digit. \(2^{12} = 2^6 * 2^6\). We know that cyclicity of 2 is 4, so \(2^6\) will have the same units digit as \(2^2\) which is 4. So, units digit of \(2^{12} = 4 * 4\) which will give us 6.

For a number \(2^x\), units digit is 6 when \(x\) is a multiple of 4. It does not matter to us if \(x\) is also a multiple of 6, 12, 24 or any other number greater than 4. We can express all of them with 4 as the base. So

\(2^5 = 2^{4 +1} =\) Units digit same as \(2^1\)
\(2^6 = 2^{4+2} =\) Units digit same as \(2^2\)
\(2^7 = 2^{4 +3} =\) Units digit same as \(2^3\)
\(2^8 = 2^{4 + 4} =\) Units digit same as \(2^4\)
\(2^9 = 2^{2*4 +1} =\) Units digit same as \(2^1\)
and the same pattern continues

Given the cyclicity of a number, try to express the power of the number in terms of the cyclicity of the number to find its units digit. In other words assume power as the dividend, cyclicity as the divisor, the remainder will decide the units digit of the expression.

Hope it's clear :)

Regards
Harsh
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Re: If a and b are distinct positive integers, what is the units digit of [#permalink]

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New post 28 May 2015, 21:50
usadude05 wrote:
EgmatQuantExpert wrote:
sagarag wrote:
I am still not able to get why it is B.. request someone help in getting the same understand...


Hi

The question here is testing about the knowledge of cyclicity of the number 2. We know that cyclicity of 2 is 4 i.e. the unit's digit of 2 would repeat itself after every 4 powers. For example:

\(2^1 -> 2\)
\(2^2 -> 4\)
\(2^3 -> 8\)
\(2^4 -> 16\)
\(2^5 -> 32\). Unit's digit of \(2^1\) and \(2^5\) is the same, i.e. they repeat after every 4 powers

Since the question prompt can be simplified to \(2^{3a +5b}\), we just need to find what values can \(3a + 5b\) take.

St-II tells us that GCD(a, b) =12. As 12 is the factor of both \(a\) and \(b\) we can write \(a\) and \(b\) as \(12x\) and \(12y\) respectively.

So \(3a + 5b = 12(3x +5y) = 4(9x + 15y)\). This expression is always divisible by 4. So we can write \(2^{3a + 5b} = 2^{4n}\) where \(n = 9x + 15y\)

We know that \(2^4\), \(2^8\), \(2^{12}\).....\(2^{4n}\) have the same units digit as 6.

Since st-II tells us that \(3a +5b\) is a multiple of 4, we can definitely find the unit's digit of the expression \(2^{3a +5b}\)

Hope this helps :)

Regards
Harsh



Can you please explain this part above :-

So \(3a + 5b = 12(3x +5y) = 4(9x + 15y)\). This expression is always divisible by 4. So we can write \(2^{3a + 5b} = 2^{4n}\) where \(n = 9x + 15y\)

We know that \(2^4\), \(2^8\), \(2^{12}\).....\(2^{4n}\) have the same units digit as 6.


Hi usadude05,

Please go through the above post and let me know if you need clarification on any other point.

Hope this helps :)

Regards
Harsh
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Re: If a and b are distinct positive integers, what is the units digit of [#permalink]

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New post 27 Sep 2017, 07:44
Postal wrote:
If a and b are distinct positive integers, what is the units digit of 2^a*8^b*4^(a+b)?

(1) b = 24 and a < 24
(2) The greatest common factor of a and b is 12


This question certainly takes some drilling- multiples of 2 have units digit that follow the cycle 2 4 8 6 - we can also rewrite and simplify the stimulus before going into the statements- a must

2^(4b + 4a)

Statement 1

Several possibilities

insuff

Statement 2

Although we cannot identify a or b it doesn't matter because we know they are positive integers and if they are factors of 12 then the only possibilities are 1 and 12, 2 and 6, 4 and 3. If we go back and plug these numbers in we will have three different sums: 28, 32, 52; however, these numbers are all divisible by 4 - and if we understand how to find the units digit of a number then we can clearly see that with all these combinations the units digit of 2^(4b + 4a) is bound to be 6 Bunuel am I right?

B

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Re: If a and b are distinct positive integers, what is the units digit of [#permalink]

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New post 30 Sep 2017, 02:48
EgmatQuantExpert wrote:
sagarag wrote:
I am still not able to get why it is B.. request someone help in getting the same understand...


Hi sagarag,

The question here is testing about the knowledge of cyclicity of the number 2. We know that cyclicity of 2 is 4 i.e. the unit's digit of 2 would repeat itself after every 4 powers. For example:

\(2^1 -> 2\)
\(2^2 -> 4\)
\(2^3 -> 8\)
\(2^4 -> 16\)
\(2^5 -> 32\). Unit's digit of \(2^1\) and \(2^5\) is the same, i.e. they repeat after every 4 powers

Since the question prompt can be simplified to \(2^{3a +5b}\), we just need to find what values can \(3a + 5b\) take.

St-II tells us that GCD(a, b) =12. As 12 is the factor of both \(a\) and \(b\) we can write \(a\) and \(b\) as \(12x\) and \(12y\) respectively.

So \(3a + 5b = 12(3x +5y) = 4(9x + 15y)\). This expression is always divisible by 4. So we can write \(2^{3a + 5b} = 2^{4n}\) where \(n = 9x + 15y\)

We know that \(2^4\), \(2^8\), \(2^{12}\).....\(2^{4n}\) have the same units digit as 6.

Since st-II tells us that \(3a +5b\) is a multiple of 4, we can definitely find the unit's digit of the expression \(2^{3a +5b}\)

Hope this helps :)

Regards
Harsh


Can we take 3 common and apply the divisibility rule of 3?

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Re: If a and b are distinct positive integers, what is the units digit of   [#permalink] 30 Sep 2017, 02:48
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