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

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
_________________

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.

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

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!

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 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
_________________

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
_________________

We can rewrite this \(2^a8^b4^{a+b}\) to such form: \(2^a2^{3b}2^{2a+2b}\) -> \(2^{3a}2^{5b}\)
As we know unit digits of 2 in any exponent has a repeating pattern every 4 exponents:
2, 4, 8, 16, 32, 64, 128, 256....
So the question is what number is or is 3a+5b multiple of 2, 3, 4

1) We know what b multiple of 4 but know not enough about a
Insufficient

2) GCF(a, b) = 12 So we can infer that both this numbers are multiple of 4 and unit digits of this statement \(2^a8^b4^{a+b}\) will be equal to 6
Sufficient
Answer is B
_________________

Question : Unit digit of \(2^a8^b4^{a+b}\)?

Unit digit of any number depends on the Cyclicity i.e. the cycle of powers after which the Unit digits start repeating

Numbers with Unit digits 2, 3, 7, 8 have Cyclicity 4 i.e. their unit digit repeates after every 4th power
Numbers with Unit digits 0, 1, 5, 6 have Cyclicity 1 i.e. their unit digit repeates after every power
Numbers with Unit digits 4, 9 have Cyclicity 2 i.e. their unit digit repeats after every 2nd power

Here, \(2^a8^b4^{a+b}\) can be rewritten as \((2^a)(2^{3b})(2^2{(a+b)})\)

i.e. \(2^a8^b4^{a+b}\) = \(2^{a+3b+2a+2b}\) = \(2^{3a+5b}\)

Statement 1: b = 24 and a < 24

i.e. \(2^{3a+5b}\) = \(2^{3a+120}\) where a can be 1 or 2 or 3 etc. which will give us different uniti digits

Hence, NOT SUFFICIENT


Statement 2: The greatest common factor of a and b is 12

i.e. a and b both are multiple of 12

i.e. 3a+5b is a multiple of 4 as both a and b are multiple of 4

therefore, \(2^{3a+5b}\) = \(2^{4x}\) i.e. Unit Digit is always = unit digit of \(2^4\) i.e. 6

e.g. at a=b=12, \(2^{3a+5b}\) = \(2^{96}\) i.e. Unit digit = 6
e.g. at a=b=2,4 \(2^{3a+5b}\) = \(2^{}\) i.e. Unit digit = 6

Hence, SUFFICIENT

Answer: Option

MANHATTAN GMAT OFFICIAL SOLUTION:

This problem contains a common trap seen in many difficult DS questions. C is a tempting short-cut answer, as the combined statements would give us the values of both a and b, which we could simply plug into the expression and answer the question.

It is only by rephrasing that we can determine whether exact values for both a and b are truly required.

Manipulate the expression:
\(2^a8^b4^{a + b} = 2^a(2^3)^b(2^2)^{a + b}= 2^a(2^{3b})(2^{2a + 2b})= 2^{3a + 5b}\)

Remembering the units digit patterns for powers of 2 will help on this problem:
2^1 = 2
2^2 = 4
2^3 = 8
2^3 = 8
2^4 = 16
2^5 = 32
…etc.

The units digits for powers of 2 is a repeating pattern of [2, 4, 8, 6].

If we can determine the relationship of 3a + 5b to a multiple of 4 (i.e. where 2^(3a + 5b) is in the predictable 4-term repeating pattern of units digits), we will be able to answer the question. This question can be rephrased as “What is the remainder when 3a + 5b is divided by 4?”

(1) INSUFFICIENT: If b = 24, then 5b is a multiple of 4. However, we know only that a is an integer less than 24. Possible remainders when 3a is divided by 4 are 0, 1, 2, or 3.

(2) SUFFICIENT: If the greatest common factor of a and b is 12, then 12 must be a factor of both variables. That is, both a and b are multiples of 12 and thus also multiples of 4. As a result, 3a and 5b will be multiples of 4 as well, so the remainder will be 0 when 3a + 5b is divided by 4.

The correct answer is B.
_________________

Why is the inference that a & b = multiples of 4 sufficient to conclude b? Why could you not infer that they are also multiples of 3?

Thank you.

Hey Sisorayi01,

Perhaps, this might help address you concern.

Units digit of \(2^x\) follows this pattern - 2,4,8,6,2,4,8,6...

\(2^a*8^b*4^{a+b}\)
\(2^a*2^{3b}*2^{2(a+b)}\)
\(2^{3a+5b}\)

Statement 1: If b = 24, a = 1, then units digit = 2; If b = 24, a = 2, then units digit = 4. (Follow the pattern)

Insufficient.

Statement 2: GCF is 12, which means that A and B are multiples of 12 (you may assume A = 12 and B = 24 for convenience). This implies that xA + yB will be a multiple of 12. Therefore, 3A + 5B will be a multiple of 12. If 3A + 5B is a multiple of 12, then, according to the pattern, units digit will be 6.

Sufficient

Hence, B
_________________

If the last digit is 4 why aren’t we looking at cyclicity of 4 which is 2. If 4 is raised to an odd power then unit digit is 4 if to even =6. So can we just establish is a+b is even or odd? Sorry if a v silly q...I struggle with number properties

Posted from my mobile device

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________