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If x and y are positive integers such that the product of x [#permalink]
 19 Feb 2010, 12:02
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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 will wait for the few replies before posting the OA OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-positive-integers-such-that-the-product-of-x-142992.html
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A.
If the product of two integers x,y is a prime number p, then one of the integers is p and the other is 1, clearly.
Thus either x=1 and y=p or x=p and y=1.
By fermat's little theorem, for any constant integer a, a^(p-1) = 1 (mod p)
(1) 24 < y < 32
Implies that x=1 and y= p with p>2 (clearly p is either 29 or 31, but it only matters that p is odd).
Thus 7^x + 9^y = 7 + 9^y
But by Fermat's little theorem, 9^(y-1) = 1 (mod y).
Thus 9^y = 9 (mod y).
And 7 + 9^y = 7 + 9 = 16 (mod y)
So we have that the units digit of 7^x + 9^y is 6.
SUFFICIENT
(2) x=1
Implies that y=p.
Take y=2, then 7^x + 9^y = 7+81 = 88
Now, take y=3, then 7^x + 9^y = 736
Thus the units digit is either 8 if y=2 or 6 if y>2.
INSUFFICIENT
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I got it so far that y would be either 29 or 31 and x would be 1. However is there a way to know without calculation that any integer to power of 29 and 31 (for example) would result in the same units digit.
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claytont wrote: By fermat's little theorem, for any constant integer a, a^(p-1) = 1 (mod p)
(1) 24 < y < 32
Implies that x=1 and y= p with p>2 (clearly p is either 29 or 31, but it only matters that p is odd).
Thus 7^x + 9^y = 7 + 9^y
But by Fermat's little theorem, 9^(y-1) = 1 (mod y).
Thus 9^y = 9 (mod y).
And 7 + 9^y = 7 + 9 = 16 (mod y)
So we have that the units digit of 7^x + 9^y is 6.
First, you certainly do not need Fermat's Little Theorem on the GMAT. Second, it's being misapplied here. Yes, it is true that 9^(y-1) = 1 mod y, if y is prime, but let, for example, y be 29. If you apply Fermat's Little Theorem, you can be certain that 9^28 = 1 mod 29, but that tells you nothing about the units digit of 9^28; we want to know what 9^28 is equal to mod 10, not mod 29. It is pure coincidence that Fermat's Little Theorem appears to give the correct answer here, as you will find if you try to apply the same reasoning with a different base. Try 3^y instead with y=31, for example, to see why FLT does not tell you about units digits: the units digit of 3^31 is 7, not 3. We do not need any special theorems to answer the question. shuj00 wrote: 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
We can be certain that one of x or y is prime, and the other is 1. From Statement 1, y is prime and x is 1. Since y is a prime greater than 2, y is odd, and for any positive odd integer y, 9^y will end in 9 (since if y is odd, y = 2k+1 for some integer k, and thus 9^y = 9^(2k + 1) = (9^2k)(9^1) = (81^k)(9), and the units digit of 81^k is 1). So 7^x + 9^y = 7 + 9^y, which will end in 6. Statement 2 is almost sufficient. We know that x=1 and y is prime. If y is an odd prime, then 7^1 + 9^y will end in 6. However, y might be equal to 2, in which case 9^y= 9^2 will end in 1, not 9, and 7^x + 9^y will end in 8.
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Thanks it makes sense now.
I get the point that 81^y for any value of y will result in 1 as units digit. So why are we seeking for a negative exponent. How does that help
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shuj00 wrote: 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
(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
I will wait for the few replies before posting the OA IMO A.. Given x*y is prime.. means either x = 1 or y = 1 S1: 24<y<32... this gives x = 1... and y = 29 or 31 (2 primes in the range) Unit digit of 7^x + 9^y = 7^1 + 9^{29} or 7^1 + 9^{31}Cyclicity of 9 is 2.... therefore unit digit for 9^29 = 9^1 = 9 and 9^31 = 9^1 = 9 and unit digit for 7^1 = 7... Therefore unit digit of Unit digit of 7^x + 9^y = 6 SUFF S2: x = 1.... gives y is the prime number.... but this is INSUFF.....
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jeeteshsingh wrote: shuj00 wrote: 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
(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
I will wait for the few replies before posting the OA IMO A.. Given x*y is prime.. means either x = 1 or y = 1 S1: 24<y<32... this gives x = 1... and y = 29 or 31 (2 primes in the range) Unit digit of 7^x + 9^y = 7^1 + 9^{29} or 7^1 + 9^{31}Cyclicity of 9 is 2.... therefore
unit digit for
9^29 = 9^1 = 9 and
9^31 = 9^1 = 9and unit digit for 7^1 = 7... Therefore unit digit of Unit digit of 7^x + 9^y = 6 SUFF S2: x = 1.... gives y is the prime number.... but this is INSUFF..... Could you please explain what you mean by cyclicity of 9 being 2? I dont understand your logic leap here in part B, if you could maybe write it out a little bit more I would be interested to see the solution construction.
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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 1.) y can be 29 or 31 - insufficeint 2.) no value of y using both possible because x=1 and y has 2 values. so it is not possible to find the value... hope I am correct 
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FlimFlam wrote: jeeteshsingh wrote: shuj00 wrote: 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
(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
I will wait for the few replies before posting the OA IMO A.. Given x*y is prime.. means either x = 1 or y = 1 S1: 24<y<32... this gives x = 1... and y = 29 or 31 (2 primes in the range) Unit digit of 7^x + 9^y = 7^1 + 9^{29} or 7^1 + 9^{31}Cyclicity of 9 is 2.... therefore
unit digit for
9^29 = 9^1 = 9 and
9^31 = 9^1 = 9and unit digit for 7^1 = 7... Therefore unit digit of Unit digit of 7^x + 9^y = 6 SUFF S2: x = 1.... gives y is the prime number.... but this is INSUFF..... Could you please explain what you mean by cyclicity of 9 being 2? I dont understand your logic leap here in part B, if you could maybe write it out a little bit more I would be interested to see the solution construction. Cyclicity of a number is - basically the power of a number when the last digit starts to repeat itself... For e.g.... what is the cyclicity of 2.... 2^1 = 2... 2^2 = 4... 2^3 = 8 2^4 = 16... 2^5= 32... (last digit is 2... which is same as last digit of 2^1) Therefore cyclicity of 2 is 4 since... on 2^5... the last digit begins to repeat. More reference: math-number-theory-88376.html?view-post=666609#p666609 Section "Last digit of a power" Now if u consider 9.... 9^1 = 9 9^2 = 81 9^3 = 729... last digit same as 9^1... therefore cyclicity of 9 is 2.... If u know the cyclicity of a number... you can find the last digit of that number raise to power any figure. For e.g: 9^{3457} last digit would be calculated as follows: 1. Get the cyclicity of 9 i.e. 2 2. Get the remainder by dividing 3457 by the cyclicity number i.e 2. The remainder would be 1. 3. Therefore last digit would be 9^remainder = 9^1 = 9 In case the remainder comes 0.. then last digit = 9^cyclicity number itself. Now in the given problem we have 7^1 + 9^29 or 7^1 + 9^31. In both the cases we would get the last digit as 7 + 9 (31 or 29 would give 1 as remainder when divided by cyclicity of 9) = 16.. i.e. 6 as the last digit. Hope it's clear now.
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This seems to be more of 700 level question.
Whatever is the difficulty level, sound understanding of concepts is a must.
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IMO A
From the given part of the problem we can derive that one of the integers should be 1 and the other prime in order the product of these two positive integers to be prime. We don't wich one is one and which one is prime.
From the first statement we can derive that x is 1, and y is prime number. Y is not 2, so it is certainly odd. All odd powers of 9 have the units digit of 9. 7^1 ends with 7. SUFFICIENT
From the second statment we can derive that 7 is 1, and that y may be any prime number including 2, and all the other odd primes. Here we have two possible results so it is INSUFFICIENT
This problem is a little bit tricky. First I thought the answer was D becuase I solved the problem in my head rather than on the scrap. But when I began to analyze it before posting I realized that I overlooked the possibility of 2 as a value of y in the second statement. Hope will be more careful on the actual test.
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i get A as well..
stem says X*Y=prime number..which means either x or y equal 1..
so if y=29, 31 then x is=1..this is sufficient as 9^odd gives unit digit of 9..
stmt 2) is insuff..cause of if x=1 then y=2 or 3..i.e 9^even=1 unit or 9^odd=9 unit..
this is insuf..
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Re: If x and y are positive integers such that the product of x [#permalink]
22 Jan 2013, 00:36
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Isn't this supposed to be D?
Statement 1: As you all have proven is SUFFICIENT
Statement 2: x=1
This means x*y = prime which make y a prime
9^1 = 9
9^2 = 1
9^3= 9
...
9^7= 9
9^11 = 9
9^13 = 9
9^17= 9
It seems that 9^y is always 9 and we know that 7^1 is 7
Then the units digit is always 6.
Answer: D
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Re: If x and y are positive integers such that the product of x [#permalink]
22 Jan 2013, 04:59
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Re: If x and y are positive integers such that the product of x
[#permalink]
22 Jan 2013, 04:59
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