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Divisibility and Power query

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Divisibility and Power query

by waltz2salsa » Tue Oct 19, 2010 5:33 am
i read in this forum itself that

'a' raise to the power 'm' raise to the power 'n' = a raise to the power ( m raise to the power n)
'a' raise to the power 'm' raise to the power 'n' NE (a raise to the power m) raise to the power n

if this be the case then is the solution mentioned in the 300+ question wrong.

Ques: What is the units digit of a^36?
a) a^2 has 9 as the units digit
b) a^3 has 3 as the units digit

Solution:

1) a^2 ends in 9
Well, a^36 = (a^2)^18
If we know that a^2 ends in 9, it's certainly possible to calculate the units digit of (...9)^18 - sufficient.

2) a^3 ends in 3
Well, a^36 = (a^3)^12
If we know that a^3 ends in 3, it's certainly possible to calculate the units digit of (...3)^12 - sufficient.
Each statement is sufficient alone: choose D.

Please clarify whether a raise to power 36 = (a raise to the power 2) raise to the power 18 or is 'a' raise to power 36 = (a raise to the power ( 2 raise to the power 18) ... I understand that the logic says the former to be correct and the latter downright improbable but please clarify the formula mentioned above with regard to this problem.

Regards,
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by Geva@EconomistGMAT » Tue Oct 19, 2010 5:57 am
The former is correct, of course.
a^36 is equal to (a^2)^18 - the parentheses make it clear that the a^2 itself is raised to the power of 18, and must be included.

if you see a^2^18 without parentheses, then you must deal with the exponent first: raise 2 to the power of 18, then raise a to the resulting power.

Remember order of operation PEMDAS: Parentheses, Exponents, Multiplication/Division, Addition/Subtraction. Exponents take precedence over everything except parentheses - means that you deal with the exponent first, unless the parentheses tell you otherwise. In the question above, the parentheses are implied by the fact that the statements talk about a^2's unit digit, not about a's unit digit: you need to take a^2 as a single whole 'entity' with a units digit of 9.
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by shovan85 » Tue Oct 19, 2010 5:59 am
waltz2salsa wrote:
'a' raise to the power 'm' raise to the power 'n' = a raise to the power ( m raise to the power n)
'a' raise to the power 'm' raise to the power 'n' NE (a raise to the power m) raise to the power n
(a^m)^n = a^(m*n)
a^m^n = a ^ (m^n)
a^m * a^n = a^(m+n)
waltz2salsa wrote: Ques: What is the units digit of a^36?
a) a^2 has 9 as the units digit
b) a^3 has 3 as the units digit
Another approach:

This question is basically testing the cyclicity of the unit digit.

Cyclicity means after a power the Unit digit will repeat the similar pattern. As here after every 4 powers 3 will repeat similar pattern in the Unit place:
3^1 = 3 --> Unit place 3
3^2 = 9 --> Unit place 9
3^3 = 27 --> Unit place 7
3^4 = 81 --> Unit place 1
3^5 = 343 --> Unit place 3
3^6 --> Unit place 9
...

So it repeats after 4 th power

Option 1: a^2 has 9 as the units digit so a could be wither 3 or 7 (as 3^2 = 9 or 7^2 = 49)
When u take a = 3 then cyclicity of 3 is 4 (i.e. 3,9,7,1 )
when a^36 unit digit is 1.
When u take a = 7 then cyclicity of 7 is 4 (i.e 7,9,3,1)
So a^36 unit digit is 1.

Thus 1 is sufficient.

Option 2: a^3 has 3 as the units digit so a could be only 7.
Thus a^36 unit digit is 1.

So 2 is sufficient.

IMO
D
waltz2salsa wrote: Please clarify whether a raise to power 36 = (a raise to the power 2) raise to the power 18 or is 'a' raise to power 36 = (a raise to the power ( 2 raise to the power 18) ... I understand that the logic says the former to be correct and the latter downright improbable but please clarify the formula mentioned above with regard to this problem.
Your first opinion is correct. a raise to power 36 = (a raise to the power 2) raise to the power 18
We can say a^36 = (a^2)^18
If the problem is Easy Respect it, if the problem is tough Attack it
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