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If x and y are positive integers such that the product of x [#permalink]

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19 Sep 2010, 05:45

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A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

51% (02:20) correct
49% (01:06) wrong based on 37 sessions

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If x and y are positive integers such that the product of x and y is prime, what is the units’ digit of 7^x + 9^y ?

(1) 24 < y < 32 (2) x = 1

----- I get that just from the question stem itself, you know either x or y must be equal to 1, thus B cannot be the answer. I can't figure out how you can tell the units digit of 9 to the Y just from the statement one.

If x and y are positive integers such that the product of x and y is prime, what is the units’ digit of 7^x + 9^y ?

(1) 24 < y < 32 (2) x = 1

----- I get that just from the question stem itself, you know either x or y must be equal to 1, thus B cannot be the answer. I can't figure out how you can tell the units digit of 9 to the Y just from the statement one.

Help?

Since we know xy is a prime, it implies as you rightly pointed that one of them is 1.

(1) This implies y is not 1, hence x must be 1. Also, we know y must be a prime since xy is a prime. So y can only be 29 or 31. Either case, y is odd. What you need to note is that the units digit of 9^{Odd number} is always 9. So combining with x=1, I know the units digit of the number in question must be 7+9 or 6. Hence, sufficient

(2) Not sufficient, as I know nothing about y

So answer is (a)

Further explanation units digits of 9^x x=1 9 x=2 1 x=3 9 x=4 1 ... The pattern is easy to spot and imagine why it happens. As the last digit is always multiplied by 9, leaving either 1 or 9 itself
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Hi! I see that noone answer your question yet. I won't discuss about (2) because you already gave the solution.

(1) We know that 24 < y < 32 hence x = 1 ! As a matter fact, if we want a product to be prime we need to multiply 1 by a prime number. As y is much greater than 1, x = 1. Besides, I just said that has to be a prime number. So y = 29 or 31! Now let's see how it goes for the unit digit of th power of 9 :

9^0 = 1 9^1 = 9 9^2 = 81 9^3 = 729 9^4 = ...1

so 9^k if k is odd gives the unit digit as 9.... As y = 29 or 31 (odd numbers), (2) is SUFFICIENT

interestingly i think but for the fact that 2 is a prime and an even prime at that, we are not able to answer based on (2). so we know x=1, so y has to be prime. all primes are odds (but for 2), if not for that, would we not be able to say 7 + 9^odd = ends in 6
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when xy is some prime number, either x = 1 or y =1. 1. 24<y<32 ------> x =1 , y = 29 or 31 when y =29----> 7^1 + 9 ^ 29 = 7 + any number ending in 9 ----> unit digit 6 when y = 31 -----> 7^1 + 9 ^ 31 = 7 + any number ending in 9 ----> unit digit 6 1) is sufficient 2) not sufficient Ans is A
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one more thing to add here----- cyclicity of numbers is always good in GMAT to remember - any number ending with 0,1,5,6 raised to any power will have same unit digit as the number itself, cyclicity is 1 any number ending with 2,3,7,8 will have cyclicity 0f 4 for example 3^(n+1) will have same unit as 3^(n+5) or 3^(n+9) or so on.... where n >= 0 3 -> 9 -> 27 -> 81 ---- 243 -> 729 -> xxx7 -> xxx1 --- any number ending with 4,9 will have cyclicity of 2
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If x and y are positive integers such that the product of x and y is prime, what is the units’ digit of 7^x + 9^y ?

(1) 24 < y < 32 (2) x = 1

----- I get that just from the question stem itself, you know either x or y must be equal to 1, thus B cannot be the answer. I can't figure out how you can tell the units digit of 9 to the Y just from the statement one.

Help?

My approach was: 1. Y can be either 29 or 31. Now y^29 or y^31 will always end in 9 and watever the power of X is the units digit is going to be the same. So A.

2. X=1 is not enough for a conclusion.
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